INSTRUCTOR’S SOLUTIONS MANUAL A PATHWAY TO INTRODUCTORY STATISTICS SECOND EDITION. JAMES LAPP

Table of Contents Chapter 1: Performing Operations and Evaluating Expressions 1.1 Variables, Constants, Plotting Points, and Inequalities ................................................................ 1 1.2 Expressions................................................................................................................................... 4 1.3 Operations with Fractions and Proportions; Converting Units ..................................................... 7 1.4 Absolute Value and Adding Real Numbers................................................................................ 11 1.5 Change in a Quantity and Subtracting Real Numbers ................................................................ 14 1.6 Ratios, Percents, and Multiplying and Dividing Real Numbers ................................................. 16 1.7 Exponents, Square Roots, Order of Operations, and Scientific Notation ................................... 20 Review Exercises ....................................................................................................................... 24 Chapter Test ............................................................................................................................... 27 Chapter 2: Designing Observational Studies and Experiments 2.1 Simple Random Sampling .......................................................................................................... 31 2.2 Systematic, Stratified, and Cluster Sampling ............................................................................. 34 2.3 Observational Studies and Experiments ..................................................................................... 36 Review Exercises ....................................................................................................................... 40 Chapter Test ............................................................................................................................... 43 Chapter 3: Graphical and Tabular Displays of Data 3.1 Frequency Tables, Relative Frequency Tables, and Bar Graphs ................................................ 45 3.2 Pie Charts and Two-Way Tables ................................................................................................ 49 3.3 Dotplots, Stemplots, and Time-Series Plots ............................................................................... 52 3.4 Histograms.................................................................................................................................. 58 3.5 Misleading Graphical Displays of Data...................................................................................... 64 Review Exercises ....................................................................................................................... 66 Chapter Test ............................................................................................................................... 71 Chapter 4: Summarizing Data Numerically 4.1 Measures of Center ..................................................................................................................... 75 4.2 Measures of Spread .................................................................................................................... 80 4.3 Boxplots ..................................................................................................................................... 85 Review Exercises ....................................................................................................................... 90 Chapter Test ............................................................................................................................... 92 Chapter 5: Computing Probabilities 5.1 Meaning of Probability ............................................................................................................... 95 5.2 Complement and Addition Rules ............................................................................................... 97 5.3 Conditional Probability and the Multiplication Rule for Independent Events .......................... 100 5.4 Discrete Random Variables ...................................................................................................... 103 5.5 Finding Probabilities for a Normal Distribution ....................................................................... 106 5.6 Finding Values of Variables for Normal Distributions ............................................................ 110 Review Exercises ..................................................................................................................... 112 Chapter Test ............................................................................................................................. 114 Chapter 6: Describing Associations of Two Variables Graphically 6.1 Scatterplots ............................................................................................................................... 117 6.2 Determining the Four Characteristics of an Association .......................................................... 121 6.3 Modeling Linear Associations .................................................................................................. 125 Review Exercises ..................................................................................................................... 130 Chapter Test ............................................................................................................................. 133

Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 7.1 Graphing Equations of Lines and Linear Models ..................................................................... 135 7.2 Rate of Change and Slope of a Line ......................................................................................... 138 7.3 Using Slope to Graph Equations of Lines and Linear Models ................................................. 142 7.4 Functions .................................................................................................................................. 148 Review Exercises ..................................................................................................................... 152 Chapter Test ............................................................................................................................. 156 Chapter 8: Solving Linear Equations and Inequalities to Make Prediction 8.1 Simplifying Expressions ........................................................................................................... 159 8.2 Solving Linear Equations in One Variable ............................................................................... 161 8.3 Solving Linear Equations to Make Predictions ........................................................................ 163 8.4 Solving Formulas ..................................................................................................................... 170 8.5 Solving Linear Inequalities to Make Predictions ...................................................................... 175 Review Exercises ..................................................................................................................... 181 Chapter Test ............................................................................................................................. 186 Chapter 9: Finding Equations of Linear Models 9.1 Using Two Points to Find an Equation of a Line ..................................................................... 191 9.2 Using Two Points to Find an Equation of a Linear Model ....................................................... 193 9.3 Linear Regression Model ......................................................................................................... 198 Review Exercises ..................................................................................................................... 204 Chapter Test ............................................................................................................................. 207 Chapter 10: Using Exponential Models to Make Predictions 10.1 Integer Exponents ..................................................................................................................... 209 10.2 Rational Exponents................................................................................................................... 211 10.3 Graphing Exponential Models .................................................................................................. 213 10.4 Using Two Points to Find an Equation of an Exponential Model ............................................ 216 10.5 Exponential Regression Model ................................................................................................. 221 Review Exercises ..................................................................................................................... 226 Chapter Test ............................................................................................................................. 230

Chapter 1: Performing Operations and Evaluating Expressions 1

Chapter 1: Performing Operations and Evaluating Expressions Homework 1.1 2. A constant is a symbol that represents a specific number.

4. Data are quantities or categories that describe people, animals, or things.

6. In 2017, about 37% of children aged 6–12 participated in a team sport (organized or unorganized) on a regular basis.

8. The temperature is 10F . That is, the temperature is 10 degrees below 0 (in Fahrenheit).

10. The statement t  3 represents the year 2012 (3 years before 2015).

12. Answers may vary. Example: Let s be the annual salary (in thousands of dollars) of a person. Then s can represent the numbers 25 and 32, but s cannot represent the numbers 15 and 9 .

14. Answers may vary. Example: Let n be the number of students enrolled in a prestatistics class. Then n can represent the numbers 15 and 28, but n cannot represent the numbers 20 or 0.5.

16. Answers may vary. Example: Let T be the temperature (in degrees Fahrenheit) in an oven. Then T can represent the numbers 300 and 450, but T cannot represent the numbers 300 or 450 .

18. a. Answers may vary. Some possible answers are shown below.

b. In the described situation, the symbols W and L are variables. Their values can change. c. In the described situation, the symbol A is a constant. Its value is fixed at 36 square feet.

20. a. Answers may vary. Some possible answers are shown below.

b. In the described situation, the symbols W, L, and A are all variables. All their values can change. c. In the described situation, none of the symbols are constants. All their values can change.

30. The integers between 6 and 3, inclusive, are 6,  5,  4,  3,  2,  1, 0, 1, 2, and 3.

22.

24.

26.

32.

28. The counting numbers between 1 and 5 are 2, 3, and 4.

34. The positive integers between 4 and 4 are 1, 2, and 3.

36. Answers may vary. Example: 2, 5 and 40 .

38. Answers may vary. Example: 2.1, 2.3, and 2.8 .

40. The temperature at the top of a skyscraper can be positive or negative, depending on the location of the skyscraper and the time of year. Temperature is not usually reported using fractions. So, among the choices, the integers are the smallest group of number that contains possible data.

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2 ISM: A Pathway to Introductory Statistics

42. The commute time of an employee cannot be negative, but it can be measured in fractions. So, among the choices, the nonnegative real numbers are the smallest group of numbers that contains possible data.

44. McDonald’s sells hamburgers every day of every year and there is never just a portion of a hamburger sold. So, among the choices, the counting numbers is the smallest group of numbers that contains possible data.

50.

46.

48.

52. a. b. The number of hours of video uploaded to YouTube per minute increased between 2009 and 2014. The number of hours of video uploaded to YouTube per minute went up each year. c. The annual increases in the number of hours of video uploaded to YouTube per minute increased between 2009 and 2014. The annual increases are shown below.

Years 2009 to 2010 2010 to 2011 2011 to 2012 2012 to 2013 2013 to 2014

Increase 25  14  11 48  25  23 73  48  25 100  73  27 300  100  200

54. a. b. The number of microbreweries increased from 2013 to 2017. c. The increases in the number of microbreweries stayed approximately constant from 2013 to 2017. The annual increases are shown below.

Years

Increase

2013 to 2014 2014 to 2015

2.1  1.5  0.6 2.6  2.1  0.5

2015 to 2016

3.2  2.6  0.6

2016 to 2017

3.8  3.2  0.6

56. – 68.

70. The y-coordinate is 4 .

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Chapter 1: Performing Operations and Evaluating Expressions 3 72. Point A is 2 units to the left of the origin and 4 units down. Thus, its coordinates are (2, 4) .

Point B is 3 units to the left of the origin on the x-axis. Thus, its coordinates are (3, 0) .

Point C is 5 units to the left of the origin and 4 units up. Thus, its coordinates are (5, 4) .

Point D is 4 units to the right of the origin and 2 units up. Thus, its coordinates are (4, 2) .

Point E is 3 units below the origin on the y-axis. Thus, its coordinates are (0, 3) .

Point F is 3 units to the right of the origin and 2 units down. Thus, its coordinates are (3, 2) .

74. True. The number 2 lies to the right of 6 on a number line.

76. False. 5  5 , thus 5 is not strictly greater than 5 .

88. Inequality: x  3

78.

Interval notation:  ,3

80.

Graph:

82.

90. Inequality: x  1

84.

Interval notation:  1,  

86. Inequality: x  5

Graph:

Interval notation:  5,   Graph: 92.

Inequality

numbers less than or equal to 6 numbers greater than 1

x  6

, 6

x 1

1,  

numbers greater than or equal to 4 numbers less than 5

x  4

 4,  

x5

(,5)

94.

Graph

Interval Notation

In Words

98.

96.

100.

In Words numbers between –3 and 0 numbers between 1 and 4, as well as 1 numbers between –3 and 1, as well as 1 numbers between –4 and –1, inclusive

Inequality 3  x  0

Graph

Interval Notation (3, 0)

1 x  4

[1, 4)

3  x  1

(3,1]

4  x  1

[4, 1]

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4 ISM: A Pathway to Introductory Statistics 102. The student completes the homework assignment in 30 or more minutes.

104. Inequality: h  44

110. The average gas mileage of a car on highways is between 35 and 40 miles per gallon.

Interval notation:  44,   Graph:

112. Inequality: 41  T  56 Interval notation:  41,56

106. Inequality: T  2 Interval notation:  , 2

Graph:

Graph: 114. Inequality: 140  w  145

108. Inequality: V  4.2

Interval notation: (140,145)

Interval notation:  4.2,  

Graph:

Graph:

116. No. Answers may vary. Example: The numbers 2 and 5 are not “between 2 and 5.” The integers between 2 and 5 are simply 3 and 4.

118. The ordered pairs selected and plotted points may vary. The points will lie on the same horizontal line.

Answers may vary.

120. Answers may vary. The inequality represents “4 is less than or equal to 4,” and 4 is equal to 4.

122. The types of numbers discussed in this section are real numbers, rational number, irrational numbers, integers, and counting numbers (or natural numbers). Answers may vary.

Homework 1.2 2. We evaluate an expression by substituting a number for each variable in the expression and then calculating the result.

4. The quotient of a and b is a/b, where b is not zero.

6. Substitute 6 for x in 5  x : 5  6  11

12. Substitute 6 for x in 30  x : 30  (6)  5

8. Substitute 6 for x in x  4 : 6  4  2

14. Substitute 6 for x in x  x : 6  6  0

10. Substitute 6 for x in x(9) : 69  54

16. Substitute 6 for x in x  x : 6  6  1

18. Substitute 47 for r in r  29 : 47  29  76 . So, if 47% of Republicans favor gays to marry legally in 2017, then in that same year, about 76% of Democrats favor gays to marry legally.

20. Substitute 13.5 for U in U  6 : 13.5  6  7.5 . So, in 2016 if the average daily shipping volume for UPS was 13.5 million packages, in that same year, the average daily shipping volume for FedEx was about 7.5 million packages.

22. Substitute 17 for n in 599.99n : 599.99 17  10,199.83 . So, if 17 thousand Fender Standard Jazz Electric Bass Guitars with maple fingerboards are sold, the total revenue is about $10,200,000.

24. Substitute 328 for T in T  4 : 328  4  82 . So, if a student earns a total of 328 points on four tests, the student’s average test score is 82 points.

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Chapter 1: Performing Operations and Evaluating Expressions 5 26. a.

Speed Limit (miles per hour)

Driving Speed (miles per hour)

35

35  5

40

40  5

45

45  5

50

50  5

s

s5

The expression s  5 represents the driving speed if the speed limit is s miles per hour. b. Substitute 65 for s in s  5 : 65  5  70 . So, if the speed limit is 65 miles per hour, the person will be driving 70 miles per hour.

28. a. Number of Shares

Total Value (dollars)

1

74.74 1

2

74.74  2

3

74.74  3

4

74.74  4

n

74.74n

The expression 74.74n represents the total value of the shares. b. Substitute 7 for n in 74.74n : 74.74 7   523.18 . So, the total value of 7 shares is $523.18.

30. a. Number of Siblings

Share of Cost (dollars)

2

3000  2

3

3000  3

4

3000  4

5

3000  5

n

3000  n

The expression 3000  n represents each sibling’s share of the cost in dollars. b. Substitute 6 for n in 3000  n : 3000  6  500 . So, the share of each sibling’s cost is $500.

32. a. We can write an expression 10  v to represent the total cost of parking and money spent on a vase. b. Substitute 25 for v in the expression 10  v : 10  25  35 . So, if $10 is spent on parking then the total cost of parking and money spent on a vase is $35.

34. a. We can write an expression r  2 to represent the net price of a shaver whose retail price is r dollars. b. Substitute 6 for r in the expression r  2 : 6  2  4 . So, if the retail price of a shaver is $6, then the net price is $4.

36. a. We can write an expression 105c to represent the total cost of tuition when enrolling in c credits of classes. b. Substitute 15 for c in the expression 105c : 105 15  1575 . So, if a student enrolls in 15 credits of classes, then the total cost of tuition is $1575.

38. a. We can write an expression 420  n to represent the equal share each of n siblings will receive of the inheritance. b. Substitute 3 for n in the expression 420  n : 420  3  140 . So, each of 3 siblings will receive an equal share of $140,000 of a $420,000 inheritance.

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6 ISM: A Pathway to Introductory Statistics 40. 8  x ; substitute 8 for x in 8  x : 8  8  0 .

50. The quotient of 6 and the number

42. 6  x ; substitute 8 for x in 6  x : 6  8  14.

52. Two less than the number

44. x  15; substitute 8 for x in x  15: (8)  15  23.

54. The sum of 4 and the number 56. The product of the number and 5

46. x  7 ; substitute 8 for x in x  7 : 8  7  1 .

58. The sum of the number and 3

48. 5 x ; substitute 8 for x in 5 x : 5 8  40 .

60. The quotient of the number and 5

62. Substitute 6 for x and 3 for y in the expression y  x : 3  6  9 64. Substitute 6 for x and 3 for y in the expression xy : 63  18.

66. Substitute 6 for x and 3 for y in the expression x  y : 6  3  2.

68. x  y ; substitute 9 for x and 3 for y in the expression x  y : 9  3  12.

70. x  y ; substitute 9 for x and 3 for y in the expression x  y : 9  3  3.

72. Substitute 90.0 for c and 104.8 for r in the expression c  r : 90.0  104.8  194.8. So, in 2015 the average annual per-person consumption of chicken and red meat was 194.8 pounds.

74. Substitute 11.26 for w and 19.98 for a in the expression a  w : 19.98  11.26  8.72. So, in 2015 the college enrollments of all students who were not women was 8.72 million.

76. Substitute 2.5 for N and 1.8 for A in the expression NA : 2.5 1.8  4.5. So, in 2016 the average number of AP exams taken was 4.5 million.

78. Substitute 205,200 for s and 3.6 for n in the expression s  n : 205, 200  3.6  57, 000. So, in 2014 the average money earned by a teacher was about $57,000.

80. a. Substitute 4 for x in the expression x  2 : 4  2  6. Substitute 5 for x in the expression x  2 :

5  2  7 . Substitute 6 for x in the expression x  2 : 6  2  8 . b. Substitute 4 for x in the expression 2 x: 2  4  8 . Substitute 5 for x in the expression 2 x : 2 5  10 .

Substitute 6 for x in the expression 2 x: 2 6  12 . c. Observe the values after substitution are different for the two expressions.

82. a. n 1 2 3 4

x

x2

2x

4

42  6

2(4)  8

5

5 2  7

2(5)  10

6

62  8

2(6)  12

3n 3 1  3 3 2  6 33  9 3  4  12

The price of bread is $3, $6, $9, and $12 for 1, 2, 3, and 4 loaves, respectively. b. The cost per loaf of bread is $3. The cost per loaf is a constant while the number of loaves is a variable. In the expression 3n, the constant is 3 and the variable is n. c. Answers may vary. Example: For each additional loaf bought, the total price increases by $3.

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Chapter 1: Performing Operations and Evaluating Expressions 7 84. a. t 1 2 3 4

2t 2 1  2 22  4 23  6 24  8

The elevator rises are 2 yards, 4 yards, 6 yards, and 8 yards for every 1, 2, 3, and 4 seconds, respectively. b. The elevator is rising at a speed of 2 yards per second. The distance risen is a constant amount of 2 yards while the number of seconds is a variable. In the expression 2t , the constant is 2 and the variable is t. c. Answers may vary. Example: For each second that passes, the distance the elevator rises is another 2 yards.

86. Answers may vary.

88. Answers may vary.

Homework 1.3

2. The reciprocal of

a b is . b a

4. If an object is made up of two or more parts, then the sum of their proportions equals 1.

6. The numerator of

2 is 2.

5

8. 18  2  9  2  3  3  2  3  3

32.

4 8 4 3 43 223 3 3        7 3 7 8 7  8 7  2  2  2 7  2 14

34.

4 4 1 4 1 22 2 2 2      9 9 2 9  2 33 2 33 9

36.

2 8 2  8 10 2  5 2      15 15 15 15 3  5 3

38.

13 9 13  9 4 22 2 2       18 18 18 18 2  3  3 3  3 9

10. 24  4  6   2  2   2  3  2  2  2  3 12. 27  3  9  3  3  3  3  3  3 14. 105  5  21  5  3  7   3  5  7 16.

10 2  5 2 5 5     14 2  7 2 7 7

40. The LCD is 9:

18.

27 333 333 1 1     54 3  3  3  2 3  3  3 2 2

42. The LCD is 24:

20.

9 33 33 1 1 1      81 3  3  3  3 3  3 3  3 3  3 9

22.

15 35 3 5 5 5      18 3  3  2 3 3  2 3  2 6

44. The LCD is 7: 2 

24.

5 4 5  4 5  2  2 20     7 9 7  9 7  3  3 63

46. The LCD is 4:

26.

2 5 25 25 5 5      3 6 3 6 3 2 3 33 9

48. The LCD is 42:

28.

5 5 2 5 2 5 5 2      12 12 1 2  2  3 2  3 6

30.

7 2 7 3 7 3     12 3 12 2 12  2 7 3 7 7    2  2 3 2 2  2  2 8

1 5 1 3 5 3 5 8        3 9 3 3 9 9 9 9 3 1 3 3 1 4 9 4        8 6 8 3 6 4 24 24 13  24

3 1 3 1 2 3 2 1        4 2 4 2 2 4 4 4 5 4 5 7 4 6 35 24        6 7 6 7 7 6 42 42 11  42

50. The LCD is 7: 1 

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3 2 7 3 14 3 17       7 1 7 7 7 7 7

9 1 7 9 7 9 2       7 1 7 7 7 7 7 2  7

8 ISM: A Pathway to Introductory Statistics 25 25 15 25 3 553 5 56. 9       15 9 3 9 15 3  3  3  5 9 3

3 3 7 3 5 15 52. 4      7 4 5 4 7 28 5 5 5 20 5 21 53 7 7     54. 3   20 3 21 3 20 3  2  2  5 4 21

58. Substitute 3 for x and 12 for z in the expression

z 12 3  2  2 2  2 4 :    4 x 3 3 1 1 1

60. Substitute 4 for w, 3 for x, 5 for y, and 12 for z in the expression 62. Substitute 3 for x, 5 for y, and 12 for z in the expression The LCD is 12: 64.

67 381   0.90 71 399

66.

149 31   1.16 215 52

y w 5 4 5 2 2 5 5  :     z x 12 3 2  2  3  3 3  3 9

y y 5 5  :  x z 3 12

5 5 5 4 5 20 5 25        3 12 3 4 12 12 12 12

68.

614 391   1.85 701 400

70. Answers may vary. Example:

72. In 2018, since 10 of the top 40 songs sold on iTunes were pop songs, we can write a proportion of the songs 10 1  . that were pop songs as 40 4 74. The whole survey group consists of the proportions of the three political parties, so the sum of the proportions equals 1.

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Chapter 1: Performing Operations and Evaluating Expressions 9 76. The category of American adults who picked football as their favorite sport to watch OR who picked basketball as their favorite sport to watch is the category of adult Americans who picked football together with 4 1 the adults who picked basketball. So, we add the fractions  .

11 9 4 1 4 9 1 11      11 9 11 9 9 11 36 11   99 99 47  99 78. Proportion of employees who spend at least $101 on commuting to work: 80. Proportion of the disk that is orange: 1 

1 1 7 5 12     5 7 35 35 35

2 7 2 5    7 7 7 7

82. Proportion of Hispanic adults that do not use at least one social media site: 1  84. Proportion of the disc that is red and blue: Proportion of the disc that is yellow: 1 

8 11 8 3    .

11 11 11 11

1 1 3 2 5     2 3 6 6 6

5 6 5 1    6 6 6 6

86. Proportion of Hispanic and Caucasian undergraduates:

2 1 4 7 11 .     7 2 14 14 14

Proportion of undergraduates of ethnicities other than Hispanic and Caucasian: 1 

11 14 11 3 .    14 14 14 14

88. Let m be the proportion of income for mortgage and f be the proportion of income for food. The proportion 1 1 remaining is given by the expression 1  m  f . Substitute for m and for f in the expression.

3 6 1 1 1 m  f  1  3 6 1 6 1 2 1      1 6 3 2 6 6 2 1    6 6 6 6  2 1  6 3  6 1  2 1 So, of the income remains.

2

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10 ISM: A Pathway to Introductory Statistics 90. a. i. 2370 out of 3180 degrees were bachelor’s degrees. Proportion of bachelor’s degrees:

2370  0.745 .

3180

2370 3180 2370 810     0.255 3180 3180 3180 3180 iii. 496 + 84 = 580 degrees were master’s and doctoral degrees. Proportion of master’s and doctoral 580  0.182 . degrees: 3180 b. The six exact proportions consist of all the degrees the university awards, so the sum of the exact proportions equals 1. This may not be the case for the sum of the approximations. Rounding may cause the sum to differ slightly from 1.

ii. 1 

92.

23 centimeters 1 inch   9.06 inches 1 2.54 centimeters

94.

113 kilometers 1 mile   70.19 miles per hour 1 hour 1.61 kilometers

96.

42.5 milligrams 1 gram 16 ounces 0.68 grams    = 0.68 grams per pound 1 ounce 1000 milligrams 1 pound 1 pound

98.

25 meters 3600 seconds 1 kilometer 1 mile     55.90 miles per hour 1 second 1 hour 1000 meters 1.61 kilometers

100.

2250 milligrams 1 gram 16 ounces   = 3.6 grams per pound 10 ounces 1000 milligrams 1 pound

1 26, 000 cup milligrams 26 grams 1000 milligrams 8 8     3250 milligrams per ounce 102.

1 cup 1 gram 1 ounce 1 ounce

104. Answers may vary. Example: In this case, Student 2 actually did better. When you compare the proportion of 82 41 43 with the proportion of question right for Student 2, , we see that question right for Student 1,  100 50 50 43 41 .

Student 2 did better since  50 50 106. a. i.

2 3 23 6    1 3 2 3 2 6

ii.

4 7 4  7 28    1 7 4 7  4 28

1 6 1 6 6    1 6 1 6 1 6 b. Answers may vary. Example: The product of a fraction and its reciprocal equals 1.

iii.

108. Answers may vary. Example: The student should have only multiplied the numerator by 2. Rewrite 2 as

2 and 1

3 2 3 23 6 then multiply across. 2      5 1 5 1 5 5

110. Answers may vary. Example: The denominator of a fraction is the name of the things it represents. The numerator of a fraction is the number of those things it represents. When we add two fractions with the same denominator, we keep the same denominator, or name, and add the two numerators, or number of things.

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Chapter 1: Performing Operations and Evaluating Expressions 11 Homework 1.4 2. The absolute value of a number is the distance the number is from 0 on the number line.

4. False. The sum of –4 and 2 is negative: 4  2  2 . The sum of 5 and –1 is positive: 5  (1)  4 .

8.     2    2  2

6.  9  9

10. 6  6 because 6 is a distance of 6 units from 0 on a number line.

12. 1  1 because 1 is a distance of 1 unit from 0 on a number line.

14.  5   5  5

16.  9   9  9

18. The numbers have different signs, so subtract the smaller absolute value from the larger.

5  3  5  3  2 Since 5 is greater than 3 , the sum is positive.

5   3  2

20. The numbers have the same sign, so add the absolute values. 3  2  3  2  5 The numbers are negative, so the sum is negative. 3   2  5 22. The numbers have different signs, so subtract the smaller absolute value from the larger. 9  6  9  6  3 Since 9 is greater than 6 , the sum is negative.

6   9  3

24. The numbers have different signs, so subtract the smaller absolute value from the larger.

4  3  4  3  1 Since 4 is greater than 3 , the sum is positive. 3  4  1 26. The numbers have the same sign, so add the absolute values. 9  5  9  5  14 The numbers are negative, so the sum is negative. 9   5  14 28. The numbers have different signs, so subtract the smaller absolute value from the larger.

8  2  8  2  6 Since 8 is greater than 2 , the sum is positive.

8   2  6

30. 8   8  0 because the numbers are opposites and the sum of opposites is 0.

32. 7  7  0 because the numbers are opposites and the sum of opposites is 0.

34. The numbers have different signs, so subtract the smaller absolute value from the larger.

17  14  17  14  3 Since 17 is greater than 14 , the sum is positive.

17   14  3 Copyright © 2021 Pearson Education, Inc.

12 ISM: A Pathway to Introductory Statistics 36. The numbers have different signs, so subtract the smaller absolute value from the larger. 89  57  89  57  32 Since 89 is greater than 57 , the sum is negative. 89  57  32 38. The numbers have the same sign, so add the absolute values. 347  594  347  594  941 The numbers are negative, so the sum is negative. 347   594  941 40. 127,512   127,512  0 because the numbers are opposites and the sum of opposites is 0.

42. The numbers have the same sign, so add the absolute values. 3.7  9.9  3.7  9.9  13.6 The numbers are negative, so the sum is negative. 3.7   9.9  13.6 44. The numbers have different signs, so subtract the smaller absolute value from the larger.

7  0.3  7  0.3  6.7 Since 7 is greater than 0.3 , the sum is positive. 0.3  7  6.7 46. The numbers have different signs, so subtract the smaller absolute value from the larger.

37.05  19.26  37.05  19.26  17.79 Since 37.05 is greater than 19.26 , the sum is positive.

37.05   19.26  17.79

48. The numbers have different signs, so subtract the smaller absolute value from the larger.

2 1 2 1 1     5 5 5 5 5 2 1 is greater than  , the sum is positive.

Since 5 5 2  1 1     5  5 5 50. The numbers have different signs, so subtract the smaller absolute value from the larger.

5 1 5 1 4 2       6 6 6 6 6 3 5 1 , the sum is negative.

Since  is greater than 6 6 5 1 2    6 6 3 52. The numbers have the same sign, so add the absolute values.

2 5 2 5 2 2 5 4 5 9 3            3 6 3 6 3 2 6 6 6 6 2 The numbers are negative, so the sum is negative.

2  5 3         3 6 2

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Chapter 1: Performing Operations and Evaluating Expressions 13 54. The numbers have different signs, so subtract the smaller absolute value from the larger.

3 2 3 2 3 3 2 4 9 8 1            4 3 4 3 4 3 3 4 12 12 12 3 2 , the sum is negative.

Since  is greater than 4 3 2  3 1      3  4 12 56. 7498.34  6435.28  1063.06

60. 

37  25      0.08 642  983 

58. 38, 487.26   83, 205.87   121, 693.13

62. The balance is 112.50  170 dollars. The numbers have different signs, so subtract the smaller absolute value from the larger.

170  112.50  170  112.50  57.50 Since 170 is greater than 112.50 , the sum is positive: 112.50  170  57.50 So, the balance is $57.50.

64. We can find the final balance by finding the balance after each transaction.

Transaction Paycheck

Balance= 135.00  549.00  414.00

FedEx Kinko's

414.00  10.74  403.26

ATM

403.26  21.50  381.76

Barnes and Noble 381.76  17.19  364.57 So, the final balance is $364.57.

66. The new balance is 2739  530 . The numbers have different signs, so subtract the smaller absolute value from the larger. 2739  530  2739  530  2209 Since 2739 is greater than 530 , the sum is negative. 2739  530  2209 So, the new balance is 2209 dollars.

68. The balance after sending the check is 873  500  373 .

The balance after buying the racquet is 373   249  622 .

The balance after buying the outfit is 622   87   709 .

So, the final balance is 709 dollars.

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14 ISM: A Pathway to Introductory Statistics 70. The current temperature is 12  8 . The numbers have different signs, so subtract the smaller absolute value from the larger. 12  8  12  8  4 Since 12 is greater than 8 , the sum is negative. 12  8  4 So, the current temperature is 4 F .

72. If a is positive and b is negative (but with a larger absolute value), the sum a  b will be negative. b units

a units

b

a+b



a

74. If a  b is positive, then both numbers are positive, or the numbers have opposite signs but the number with the larger absolute value is positive.

76. a. Substitute 2 for a and 5 for b: a  b   2   5  3

d. Substitute 4 for a and 9 for b: a  b  4   9  13 b  a  9   4  13 The results are the same. e. Answers may vary.

b. Substitute 2 for a and 5 for b: b  a  5   2   3

c. The results are the same. f. Yes; when adding two quantities, the order of the addition does not matter.

78. Answers may vary. Example: The value of a stock investment can be measured in gains and losses. It is possible to assign a as a variable to represent the value of a stock that suffers a loss (when the value of the stock falls below the price of purchase) and to assign b as a variable that represents the value of a stock that experiences a gain (when the value of the stock rises above the price of purchase). Suppose you have two stocks, a and b, in a portfolio and you want to determine the value of the portfolio at the conclusion of a particular day. If on that day a  $300.00 and b  $500.00 you can find the value of the portfolio by combining a and b: 300  500  200 . So, the value of the portfolio on that day is $200.

Homework 1.5 2. To subtract a number, add its opposite.

4. True. A decreasing quantity has negative change.

6. 3  7  3   7   4

26. 1.7  7.4  1.7   7.4  9.1

8. 3  9  3   9  12

28. 3.1   3.1  3.1  3.1  6.2

10. 5   1  5  1  6

30. 159.24   7.8  159.24  7.8  151.44

12. 7   3  7  3  4

1 4 1  4 5 32.            1   5 5 5 5 5

14. 4  7  4   7   11 16. 4   7   4  7  3

4  7 4 7 3 1 34.           9  9 9 9 9 3

18. 7  7  7   7   14

36.

20. 100  257  100   257   357 22. 1939   352  1939  352  1587 24. 5.8  3.7  5.8   3.7   2.1

5  1 5 1 5 1 2 5 2 7            12  6  12 6 12 6 2 12 12 12

2 2 2  2 2 5  2 3 38.                 3 5 3  5 3 5  5 3 

10  6  16      15  15  15

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Chapter 1: Performing Operations and Evaluating Expressions 15 40. 3  9  6

54. 83, 451.6  (408.549)  83,860.15

42. 4  (3)  4  3  1 5 1 4 2 44.       6 6 6 3

46. 6.4  3.5  2.9 56. 

48. 5  (8)  13

49 85   1.75 56 97

50. 5  9  5  (9)  4 52. 6178.39  52.387  6230.78

58. 12  18  6; So, the current temperature is 6 F .

60. 13   2  13  2  11; The change in temperature is 11 F .

62. a. 9   6  9  6  15; The change in temperature is 15 F . b. To estimate the change in temperature over the past hour, we divide the change over three hours by 3.

15  5; The estimated change in temperature over the past hour is 5 F .

3 c. Answers may vary. Example: The change in temperature is affected by the time of day in addition to the weather conditions. Thus, temperature change need not be uniform.

64. 29, 035   1312  29, 035  1312  30,347; The change in elevation is 30,347 feet.

66. a. Year 2011 2012 2013 2014 2015 2016 2017

Population

Change in Population

98



83

83  98  15

95

95  83  12

104

104  95  9

99

99  104  5

108

108  99  9

97

97  108  11

b. The population increased the most from 2012 to 2013. The change in population was 12. c. The population decreased the most from 2011 to 2012. The change in population was 15 . d. No; the change in population is the difference between births and deaths. An increase of 12 wolves means there were 12 more births than deaths.

68. a. Add the changes in the number of Patriot Groups from 2010 to 2016: 824  450  86  (264)  (222)  124  (375)  623 So, there were 623 Patriot Groups in 2016. b. An increasing number of groups is indicated by positive changes. Thus, the number of Patriot Groups was increasing from 2010 to 2011, from 2011 to 2012, and from 2014 to 2015. c. A decreasing number of groups is indicated by negative changes. Thus, the number of Patriot Groups was decreasing from 2012 to 2013, from 2013 to 2014, and from 2015 to 2016.

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16 ISM: A Pathway to Introductory Statistics 70. Evaluate a  c for a  5 and c  7:  5   7   12 72. Evaluate c  a for a  5 and c  7:  7    5  7  5  2 74. Evaluate b  a for a  5 and b  2: 2  5  2  5  7 76. x  4 ; Evaluate the expression for x  5: 5  4  5   4  9 78. x  5 ; Evaluate the expression for x  5: 5  5  5   5  10 80. x   6 ; Evaluate the expression for x  5: 5   6  5  6  1 82. The student changed the order of subtraction without changing the sign of the result.

2  6  2   6  4 84. a. i.

2  8  2   8  6

ii. 3  9  3   9  6 iii. 1  5  1   5  4 b. Answers may vary. Example: Since the quantity decreased, the final number is smaller than the beginning number. When finding the change in quantity, we subtract the beginning number from the final number.

Since the final number is smaller, the result will be negative.

86. Answers may vary. Example: It is impossible to find the sign. If x is greater than y , then x  y is negative.

If x is less than y , then x  y is positive.

88. Answers may vary. Example: To subtract a negative number from another number, take the opposite of the negative number and add it to the number. For instance, to subtract 6 from 4, we write 4  (6)  4  6  10 .

Homework 1.6 2. One hundred percent of a quantity is all of the quantity.

4. True. The product or quotient of two numbers that have different signs is negative.

6.

8.

9 3  15 5 3.7 million viewers 1.32  ; There were about 1.32 times as many viewers of Good Morning America as of 2.8 million viewers 1 CBS This Morning.

10. a.

b.

2 c mushrooms 0.5 c mushrooms  ; For each cup of cooked noodles, a half cup of sliced 4 c cooked noodles 1 c cooked noodles mushrooms is required.

4 c cooked noodles 2 c cooked noodles  ; For each cup of sliced mushrooms, 2 cups of cooked noodles 2 c mushrooms 1 c mushrooms are required.

9 1.80  5 1

Scherzer:

18 2.57  7 1

20 2.86  7 1

Strasburg:

10 1.43  7 1

12. a. Kershaw: Kluber: Sale:

12 3.00  4 1 Copyright © 2021 Pearson Education, Inc.

Chapter 1: Performing Operations and Evaluating Expressions 17 12. (continued) b. The pitcher with the largest unit ratio of wins to losses is Sale. The pitcher with the smallest ratio of wins to losses is Strasburg. c. No, the person is not correct. Answers may vary. Example: Even though Sale’s wins are less than Scherzer’s wins, Sale had fewer losses than Scherzer’s which means that Sale’s ratio will be higher than that of Scherzer’s.

155 0.96  161 1

Scherzer:

300 1.36  220 1

222 1.03  215 1

Strasburg:

156 1.20  130 1

d. Kershaw: Kluber:

237 1.50  158 1 e. The pitcher with the second largest ratio of strikeouts to innings is Scherzer. The pitcher with the second largest unit ratio of wins to losses is Kluber. The unit ratios differ since the values upon which the ratios are based are not linked. That is, a player’s wins and losses stand independent of a player’s strikeouts versus innings played—they are not directly proportional.

Sale:

14. a. b.

19,849,399 2.20  ; The population of New York is about 2.20 times larger than that of New Jersey.

9, 005, 644 1 571,951 3.67  ; The land area of Alaska is about 3.67 times larger than that of California.

155,959 1

c. Alaska:

739, 795 1.29  571,951 1

New Jersey:

California:

39,536, 653 253.51  155,959 1

Michigan:

9,962,311 175.38  56,804 1

New York:

9, 005, 644 1214.19  7417 1

19,849,399 420.41  47, 214 1

d. The state with the greatest population density is New Jersey. The state with the least dense population is Alaska. e. The person is not correct. Answers may vary. Example: Although Michigan has a larger population than New Jersey, it also has a larger land area which serves to lower its population density.

16. 91%  91.0%  0.91

20. 4%  4.0%  0.04

18. 0.01  1%

22. 0.089  8.9%

24. The proportion of books purchased in stores in 2017 was 0.62.

26. The proportion of teenagers who consider Snapchat their favorite social network is 0.47.

28. 37% of 304 executives said they would quit their job and be a stay-at-home parent if they could afford it.

30. Of the 46.9 million Americans who traveled at least 50 miles from home during Independence Day holiday weekend in 2018, 8.1% traveled by air.

287  0.14 . Approximately 14% of 2048 surveyed adults do not have a will because they 2048 do not like thinking about death.

32. The proportion is

34. 0.67 4500  3015 ; so, 67% of 4500 cars is 3015 cars.

36. 0.03(125.35)  3.7605  3.76 ; so, the sales tax is $3.76.

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18 ISM: A Pathway to Introductory Statistics 38. 0.111(24,503)  2719.833  2720 ; so, there were 2720 undergraduate business majors.

40.

42.

2.39  2.12  0.127; So, the percent change in the average price of regular gasoline from 2016 to 2017 is 2.12 about 12.7%. This means the average price of regular gasoline increased by 12.7%.

26.5  32.9  0.195; So, the percent change in viewership for the Academy Awards is 19.5%. This means 32.9 the viewership for the Academy Awards decreased by about 19.5%.

44. a. McDonald’s: 160.84  156.69  4.15; McDonald’s stock increased by $4.15. b. La-Z-Boy: 32.75  30.60  2.15; La-Z-Boy’s stock increased by $2.15. c. McDonald’s:

4.15  0.0265; The percent change in McDonald’s stock was about 2.6%.

156.69

2.15  0.0703; The percent change in La-Z-Boy’s stock was about 7.0%.

30.60 e. Answers may vary. Example: Even though McDonald’s stock price increased by a greater amount than La-Z-Boy’s, La-Z-Boy’s stock is actually a better investment because its percentage increase is more than McDonald’s stock.

d. La-Z-Boy:

46. 4(5)  20 48. Since the numbers have the same sign, the product is positive: 8  9  72.

50. Since the numbers have different signs, the quotient is negative: 24   3  8.

52. Since the numbers have the same sign, the quotient is positive: 1   1  1.

54. Since the numbers have the same sign, the product is positive: 124  29  3596.

56. Since the numbers have different signs, the quotient is negative: 1008   21  48.

58. Since the numbers have the same sign, the product is positive: 0.3  0.3  0.09.

60. Since the numbers have different signs, the quotient is negative: 0.12  0.3  0.4.

62. Since the numbers have different signs, the quotient is negative:

9  9   3  3. 3

64. Since the numbers have the same sign, the quotient is positive:

72  72   8  9. 8

66. Since the numbers have different signs, the product is negative:

1 7 7     .

3 5 15

1  7   5  35  .

68. Since the numbers have the same sign, the product is positive:         25   21  525 15 5 15 5 8 40 8  .

70. Since the numbers have different signs, the quotient is negative:        7 8 7 15 105 21 3  9  3 20 60 5 72. Since the numbers have the same sign, the quotient is positive:          .

8  20  8 9 72 6

74. 9   4  13

76. 49  7   7

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Chapter 1: Performing Operations and Evaluating Expressions 19 78. 2  7  2   7   9

90. 489.2  8.39  4104.39

80. 5 9  45 3  1 3 1 82.         8  10  8 10 3 5 1 4     8 5 10 4 15 4   40 40 15  4  40 11  40

84. 

92. 64.958   3.716  17.48

22  33  22  18          9  18  9  33  2 11  2  9  9  3 11 22  3 4  3

86.

15 35 3   35 75 7

88.

35 7  5 5   21 7  3 3

94. 

169  64      0.87 175  71 

96. 

75 13   12.59 22 48

6810 dollars 681 2.31   2950 dollars 295 1 b. For each $1 he pays towards his Sears account, he should pay about $2.31 towards his Visa account.

98. a.

100. 0.35 1590  556.50 1590  556.5  1033.50 The new balance would be $1033.50.

102. 3 89.50  268.50 0  268.50  268.50 The new balance is $268.50 .

104. Answers may vary. Example: The percentage of women in the U.S. Senate is

24  100  24% . The 100

3  100  0.33  100 which is about 33%. Even though 9 there is a greater number of women in the U.S. Senate vs. the U.S. Supreme Court, the fact that 3 seats are taken up by women in the U.S. Supreme Court out of a total of 9 seats means there is greater representation of women there vs the number of women in the U.S. Senate. It would take as many as 33 women Senators to match the relative representation in the U.S. Senate as there is in the U.S. Supreme Court.

percentage of women in the U.S. Supreme Court is

106. a. Negative; the quotient of two numbers with opposite signs is negative. b. Negative; the quotient of two numbers with opposite signs is negative. c. No; the variables a and b can take on positive or negative values, so the sign of the result is not clear without knowing the signs of a and b.

108. Answers may vary. Example: 3  6  6   6   6  18

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20 ISM: A Pathway to Introductory Statistics 110. a. Answers may vary. Example: 1, 2, 3. b. Answers may vary. Example: 1 , 2 , 3 . c. For 2x to equal x, x must be 0.

112. Answers may vary. Example: When comparing the performance of two stocks in the past year, it is more helpful to compare the percent changes in value. Percentage change lets you measure the growth rate of stocks over a period of time, whereas a change in the price of a stock only lets you see the difference in its value. If you know how well one stock is growing compared to others, you can identify whether it is a better investment compared to others.

Homework 1.7

2. If a is a nonnegative number, then

a is the nonnegative number we square to get a.

4. To write 3.56  104 in standard decimal notation, we move the decimal point k places to the left.

6. 34  3  3  3  3  9  3  3  27  3  81

18. 42 

8. 53  5  5  5  25  5  125

1 4

1

10. 7 2   7  7   49

20. 61 

12. 7    7  7   49

22. 103 

2

1

6

3

3  3   3   3  27 14.           5  5   5   5  125

1 16



1 6

1 3

10



1 1000

16  4 , because 42  16 .

24.

26.  25  5

16. (5)0  1 28.



2

4 is not a real number, because the radicand –4 is negative.

30.  16 is not a real number, because the radicand –16 is negative.

32. The number 62 is not a perfect square, so 34. The number 81 is a perfect square, so

62 is irrational.

81 is rational.

62  7.87

81  9 , because 92  81 .

36. 8   2  6  8   4  32

54. 6  2  3  5  7  6 5  5  7  30  35  5

38. 2  83  8  10 5  50

56. 3   6  2  4  8  3  6  2  4

40.

3  5  4  2  6 20  4 5 5

42.

1 9 8 8 4    2   4  2  4 6 3

 3   6  8  3   2  1

58.

44. 2 41  8  2 49  2(7)  14

5 2 2 5 2 5 5 10 15 5          6 3 5 6 3 2 6 6 6 2

60. 8  32  8  9  1

46. 1  9   4  1   36  35

62. 8  2  8  8  64 3

48. 16   4  2  4  2  8

64. 2  4  3  4  7  2 16  3  4  7 2

50. 3  7  1  4  1  3 52. 2  4 9  6  2  4 3  2  12  10

Copyright © 2021 Pearson Education, Inc.

 32   12  7  20  7  13

Chapter 1: Performing Operations and Evaluating Expressions 21 66. 9  7    3  24   2   3  24 2

2

74.

 4  3  16  12  16

84.7  82.9  89.3  80.1 337.0   84.25 4 4

0.35(1  0.35)  50

76.

 12   16

70.

(4) 2  (1) 2  12  42 4 1 (4)(4)  (1)(1)  1(1)  4(4)  4 1 16  1  1  16 34   4 1 3

78.

951  944 7  24 24 75 75  7

75

75 24 5 3  7 24 35 3   2.53 24

(2) 2  (1) 2  (3) 2 3 1 (2)(2)  (1)(1)  (3)(3)  3 1 4  1  9 14   7 2 2 

(5.8  6.2)2  (9.4  6.2)2  (3.4  6.2)2 3 1

80.

6 2 82 6 2 82   10  2  9 2 9 2 6(6) 8(8)  10  2  9 2  10  2

24

 7

(1  3)2  (2  3) 2  (6  3) 2 3 1

72. (30  40)  2

0.2275 50

 0.00455  0.07

 28

68.

0.35(0.65)  50

36 64  9 2

 10  2 4  32



(0.4) 2  (3.2) 2  (2.8) 2 2



(0.4)(0.4)  (3.2)(3.2)  ( 2.8)( 2.8) 2



0.16  10.24  7.84 2



18.24  9.12  3.02 2

 10  2 36  10  2(6)  10  12  2

82. Evaluate x  zs for x  15 , z  2 , and s  5 : 15  (2)(5)  15  (10)  5 84. Evaluate

xx 5  11 6   3 for x  5 , x  11 , and s  2 : 2 2 s

86. Evaluate

y2  y1 3  (5) 2 2 for x1  3 , x2  8 , y1  5 , and y2  3 :   8  (3) 5 x2  x1 5

88. Evaluate x  t

s n

 8  8 for x  25 , t  2 , s  8 , and n  4 : 25  2   25  2    25  2  4  25  8  17 2  4 

1 1 8 1 8 90. Evaluate ab x for a  8 , b  2 , and x  3 : 8(2) 3  8  3   8       1 8 1 8 8 2 

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22 ISM: A Pathway to Introductory Statistics x   98.26  95.77 8.41 s  92.

20 n 2.49  8.41 20 8.41  2.49  20

94.

pˆ  p p(1  p ) n

0.68  0.65 0.65(1  0.65) 2000

 0.03 

0.65(1  0.65) 2000

0.2275 2000 1  0.03   2.80 0.0107  0.03 

20 8.41 2 5  2.49  8.41  1.32  2.49 

98. 3 



96.

The answer is about 2.88.

8 8  3  2  3  2  1 ; evaluate the expression for x  4 : 3  x 4

100. x  x  5 ; evaluate the expression for x  4 : 4   4 5  4  20  16 Number of unprovoked shark attacks 102. a. Years since 1990 0 2.1  0  19 1 2.1 1  19 2 2.1  2  19 3 2.1  3  19 4 2.1  4  19 t 2.1  t  19 From the last row of the table, we see that the expression 2.1t  19 represents the number of unprovoked shark attacks in the United States t years after 1990.

b. Substitute 26 for t in 2.1t  19 : 2.1 26  19  54.6  19  73.6 . So, in 2016 (26 years after 1990) the number of unprovoked shark attacks in the United States was about 73.6.

104. a. Years since 2014 0 1 2 3 4 t

Percent 84  1.5  0 84  1.5 1 84  1.5  2 84  1.5  3 84  1.5  4 84  1.5t

From the last row of the table, we see that the expression 84  1.5t represents the percent of U.S. households that use pay TV, t years after 2014. b. Substitute 4 for t in 84  1.5t : 84  1.5(4)  84  6  78 . So, in 2018 (4 years after 2014) the percent of U.S. households that use pay TV will be 78%.

106. The increase in Amazon.com’s net sales from 2016 to 2017 was 0.308(136.0) billion dollars. To find Amazon.com’s net sales (in billions of dollars) in 2017, we add 0.308(136.0) to 136.0: 136.0  0.308 136.0  136.0  41.888  177.888. Amazon.com’s net sales were about $177.9 billion in 2017.

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Chapter 1: Performing Operations and Evaluating Expressions 23 108. The decrease in the U.S. revenue from newspapers from 2016 to 2017 was 0.098(18.3) billion dollars. To find the revenue from newspapers (in billions of dollars) in 2017, we subtract 0.098(18.3) from 18.3: 18.3  0.098(18.3)  18.3  1.7934  16.5066 . The revenue from newspapers was about $16.5 billion in 2017.

110. The leftmost blue bar has width 2 and height 0.07, so its area is 2(0.07)  0.14.

Each of the other two blue bars have width 2 and height 0.12, so the area of one of these blue bars is 2(0.12)  0.24.

The total area of all three blue bars is 0.14  0.24  0.24  0.62.

Because the area of the entire object is 1, the area of the orange bar is 1  0.62  0.38.

112. The leftmost blue bar has width 4 and height 0.03, so its area is 4(0.03)  0.12.

The rightmost blue bar has width 4 and height 0.04, so its area is 4(0.04)  0.16.

The total area of the outside blue bars is 0.12  0.16  0.28.

Because the area of the entire object is 1, the remaining area of the other two bars is 1  0.28  0.72.

Since the remaining two bars are equal in area, to get the area of the orange bar, we calculate:

0.72  0.36.

2

114. For 8.31  106 , the decimal point must move six places to the right. Thus, the standard decimal is 8,310,000.

116. For 6.488  105 , the decimal point must move five places to the left. Thus, the standard decimal is 0.00006488.

118. For 8.7  102 , the decimal point must move two places to the left. Thus, the standard decimal is 0.087.

120. For 280,000, the decimal point needs to be moved five places to the left so that the new number is between 1 and 10. Thus, the scientific notation is 2.8  105.

122. For 0.000023, the decimal point needs to be moved five places to the right so that the new number is between 1 and 10. Thus, the scientific notation is 2.3  105.

124. For 0.0004, the decimal point needs to be moved four places to the right so that the absolute value of the new number is between 1 and 10. Thus, the scientific notation is 4  104.

2  0.00002 100, 000 3 3E-4   0.0003 10, 000 1.14E7  1.14 10, 000, 000  11, 400, 000 1.71E8  1.71 100, 000, 000  171, 000, 000

126. 2E-5 

130. 1012  1  1012 

1 1, 000, 000, 000, 000

 0.000000000001 watt/m 2

132. 25, 000,000,000,000  2.5  10,000,000,000,000  2.5  1013 miles

128. 2.389  105  2.389 100, 000  238,900 miles 134. 0.0000000317 

3.17  3.17  108 year 100,000,000

136. In the first line, the student only squared 3 instead of 3 . The correct expression is

32  4 3  5  9  12  5  3  5  2.

138. The student did not perform multiplication and division in the correct order (from left to right).

16  2  4  8  4  32 140. Answers may vary.

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24 ISM: A Pathway to Introductory Statistics Chapter 1 Review Exercises 1. The total box office gross from U.S. and Canada movie theaters was $11.4 billion in 2016.

2. Answers may vary. Example: Let p be the percentage of students who are full-time students. Then p can represent the numbers 60 and 70, but p cannot represent the numbers 12 and 107.

3.

4. The negative integers between 5 and 5 are 4, 3, 2, and 1.

5. The numbers listed (in millions) are 2, 4, 1 , and 3.

6. The amount of time a student plays a video game cannot be negative, but it can be measured in fractions. So, among the choices, the nonnegative real numbers are the smallest group of numbers that contains possible data.

8. Inequality: c  24

7.

Interval notation:  24,   Graph: 9. A teenager spends between 2 and 5 hours, inclusive, playing a video game.

10. Substitute 20.7 for F and 9.6 for X in the expression F  X : 20.7  9.6  11.1 . So, in 2017 Fujifilm’s annual revenue was $11.1 billion more than Xerox’s annual revenue.

11. Proportion of consumers who prefer to pay for purchases and other transactions with debit cards OR credit 2 1 8 5 13   . cards is the sum of the proportions:   5 4 20 20 20 12. Proportion of the disc that is red and blue: Proportion of the disc that is green: 1  13.

2 1 8 3 11     3 4 12 12 12

11 12 11 1    12 12 12 12

121 gallons 4 quarts 4 cups 1 year     5.3 cups per day 1 year 1 gallon 1 quart 365 days

14. 7   10  3

15. 3   5  3  5  8

16. Since the numbers have different signs, the product will be negative: 5  9  45 17. Since the numbers have different signs, the quotient will be negative: 8   2  4 18. 24  10  2  24  8  3

22. 4  2  6  4   12  16

19.  2  65  8   4 3  12

23. 8   2  5   4  5  20

20.

5  2  7  6 20  5 4 4

21.

28 2  8 6 6 3     3   1 3  1 4 4 2

24. 2  4  7   8  2  2  3  6   6  6

  6   6  12 Copyright © 2021 Pearson Education, Inc.

Chapter 1: Performing Operations and Evaluating Expressions 25 25. 14   7   3 1  5  14   7   3  4  2  3  4 

36. 3  2  4  2  1 2

 3  2  2  4  2  1

 2   12

 3  4  4  2  1

 2  12

 12  8  1

 14

 20  1

26. 5  3  2 1  7   5  3  2  6  5  3  12

 19

37. 7 2  3 2  5   3 2

 5   9

 7 2  3  3   3 2

 5  9

 7  7   3  3 3   3

4

 49  27   3

27. 4.2   6.7   4.2  6.7  10.9

 49   9

 8   16   8   25  28.                15   25   15   16 

 49  9  58

8 25   15 16 200  240 5  6

29.

38.

(2) 2  (1) 2  (3) 2 3 1 (2)(2)  (1)(1)  (3)(3)  3 1 4  1  9 14   7 2 2 

5  2 5 2 5 2 7        9  9 9 9 9 9

30. 82   8  8  64 31. 43 

1 3

4



(2  4) 2  (3  4) 2  (7  4) 2 3 1

39. 5.7  2.34   9.4  8.68

1 1  4  4  4 64

32.  49  7 33. 6 3  6 3  3  6 9  54 2

34. 2  4  2   2 2 2  4  2 3

 8  4  2



 8   8  8  8 0

35.

17   3

2

17   3 3

 5   4  4 5  42 17  9 8 8    5  16 11 11

0.15(1  0.15)  200

40.

41.

0.15(0.85) 200

0.1275  0.0006375  0.03 200

28 4  7 7   40 4 10 10

15 15 9 15 10 3  5 2  5 5  5 25    42. 8      9 8 10 8 9 2  4 3  3 3  4 12 10

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26 ISM: A Pathway to Introductory Statistics 43. 4789  800  102.99  3.50  4789  800  106.49  3989  106.49  4095.49 The student now owes the credit card company $4095.49.

44. a. 8  4  12; The change in temperature is 12 F . b. Divide the change for the past three hours by 3 to estimate the change over 1 hour. 12  4; The estimated change for the past hour is 4 F .

3 c. Answers may vary. Example: Temperature need not change uniformly.

45. a. 722  745  23; The change in total fundraising for the Democratic nominee from 2008 to 2012 was $23 million. b. 368  367  1; The change in total fundraising for the Republican nominee from 2004 to 2008 was $1 million. c. 745  328  417; The greatest change in total fundraising for the Democratic nominee occurred between 2004 and 2008. The change was $417 million. d. 450  368  82; The greatest change in total fundraising for the Republican nominee occurred between 2008 and 2012. The change was $82 million.

46.

47.

65 billion messages per day 1.86  ; The number of messages sent per day in 2018 is 1.86 times larger than 35 billion messages per day 1 the number of messages sent per day in 2015.

863  0.5600  56.0%; 56% of adults surveyed think climate change is the biggest threat to the United 1541 States.

48. 0.137(45,500)  6233.5; In a survey of 45,500 high school students, about 6234 used marijuana in the past month.

49.

34  74  0.5405; So, the percent change in the number of cases of worldwide polio decreased by about 74 54.1%.

50. 0.20(5493)  1098.6; The person pays off $1098.60 of the balance. 5493  0.20(5493)  5493  1098.6  4394.4 So, after paying off 20% of the balance, the person has a balance of –4394.40 dollars.

51. Evaluate b  c 2 for a  2 , b  5 , c  4 , and d  10 : (5)  (4) 2  5  16  11 52. Evaluate

ab 2  (5) 7 1   for a  2 , b  5 , c  4 , and d  10 : 4  10 14 2 cd

53. Evaluate x  t 

 12   12  for x  30, t  2, s  12, and n  9: 30  2    30  2  3   30  2  4  30  8  22  9 n

s

54. 7  x ; substitute 3 for x in 7  x : 7   3  7  3  4 55. 1 

24 24 24  1 8  9 ; substitute 3 for x in 1  : 1 3 x x

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Chapter 1: Performing Operations and Evaluating Expressions 27

56. a. Years since 2010

Foreclosure Inventory (millions of houses)

0

1.18  0.13  0

1

1.18  0.13 1

2

1.18  0.13  2

3

1.18  0.13  3

4

1.18  0.13  4

t

1.18  0.13t

From the last row of the table, we see that the expression 1.18  0.13t represents the foreclosure inventory in millions of houses, t years after 2010. b. Substitute 6 for t in 1.18  0.13t : 1.18  0.13(6)  0.4 . So, the foreclosure inventory in the year 2016 (6 years after 2010) was 0.4 million houses.

57. The decrease in the number (in thousands) of Pfizer employees from 2016 to 2017 was 0.065(96.5). To find the number of employees (in thousands) in 2017, we subtract 0.065(96.5) from 96.5: 96.5  0.065(96.5)  96.5  6.2725  90.2275 . The number of employees was about 90.2 thousand in 2017.

58. Each of the two blue outside bars have width 2 and height 0.08, so the area of one of these blue bars is 2(0.08)  0.16.

The total area of the two outside blue bars is 2(0.16)  0.32.

Because the area of the entire object is 1, the remaining area of the other two bars is 1  0.32  0.68.

Since the remaining two bars are equal in area, to get the area of the orange bar, calculate 0.68  2  0.34.

59. For 3.85  104 , the decimal point must move four places to the left. Thus, the standard decimal is 0.000385.

60. For 54,000,000, the decimal point needs to be moved seven places to the left so that the new number is between 1 and 10. Thus, the scientific notation is 5.4  107 .

Chapter 1 Test 1. a. Answers may vary. Example:

b. In the described situation, the symbols W and L are variables. Their values can change. c. In the described situation, the symbol A is a constant. Its value is fixed at 36 square feet.

2. The integers between 4 and 2, inclusive, are 4, 3, 2, 1, 0, 1, and 2.

5. Inequality: w  30 3.

4.

Interval notation: ,30 Graph:

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28 ISM: A Pathway to Introductory Statistics 6. Substitute 16,873 for r and 378 for n in r  n : 16,873  378  44.64 . This means in 2015, the average monthly cell phone bill was $44.64.

7. Proportion of adults that gave the response of “Often justified” and “Sometimes justified”: Proportion of adults that gave other responses: 1  8.

1 3 2 3 5     4 8 8 8 8

5 8 5 3    8 8 8 8

0.62 pound 16 ounces 9.92 ounces   = 9.92 ounces 1 1 pound 1

9. 3  9   3  3   3  6

13. 5  2  10   4  5   8   4  5  2  3

10. 4  23  7    2 4  8

14. 20  5   2  9 3

11.

 20  5   7  3

4  7 3 1   1  5 6 2

 4  21  25

12. 3 32  4  3 36  3 6  18

15. Since the two numbers have different signs, the product will be negative: 0.4  0.2  0.08 16. 

27 18 27 75    10 75 10 18 333355  2 5 2 33 335  22 45  4

3 5 3 4 5 5 17.        10 8 10 4 8 5 12 25   40 40 12  25  40 13  40

18. 25 

1 5

2



1 1  2  2  2  2  2 32

19. 7  23  32  7   2  2  2  3  3  789  15  9 6

20. 1  3  7   10   5 2

 1   4  10   5 2

 1   4 4  10   5  1  16   2  15   2  17

21.

22.

(5) 2  (2) 2  22  52 4 1 (5)(5)  (2)(2)  2(2)  5(5)  4 1 25  4  4  25 58   4 1 3 84 2  2 3 7 21   16 2222 4

14 25  14  21 23.

21 25 20 20 14 20   25 21 27 2 25   55 3 7 222  53 8  15 Copyright © 2021 Pearson Education, Inc.

Chapter 1: Performing Operations and Evaluating Expressions 29 24.

25. a. 11.1  10.3  0.8; The change in the tax audit rate from 2009 to 2011 was 0.8 audit per 1000 tax returns. b. 9.6  11.1  1.5; The change in the tax audit rate from 2011 to 2013 was 1.5 audits per 1000 tax returns. c. From 2003 to 2005, the change was 3.2 audits per 1000 returns.

26.

27.

32.44 3.55  ; The average ticket price in 2018 was about 3.55 times the average price in 1991.

9.14 1 7175  6121  0.172; The percent change in the number of hate crime incidents from 2016 to 2017 is about 6121 17.2%.

28. Substitute 6 for a, 2 for b, and 5 for c in the expression ac 

65 

6 6  30  2 2

a : b

 30  3  30   3  33

29. Substitute 6 for a, 2 for b, and 5 for c in the expression a  b3  c 2 :

6  23  52  6  222  5  5  6   8  25  14  25  11

30. Evaluate

x 4  10 6   3 for x  4 ,   10 , and s  2: s 2 2

31. 2 x  3 x ; evaluate the expression for x  5: 2  5  3  5  10   15  10  15 5

32.

10 10  6 ; evaluate the expression for x  5:  6  2  6  4 x 5

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30 ISM: A Pathway to Introductory Statistics 33. a. Years since 2012 First-Class Mail Volume (billions of pieces) 0 1.9 0  68.8 1 1.9 1  68.8 2 1.9 2  68.8 3 1.9 3  68.8 4 1.9 4  68.8 1.9t  68.8 t From the last row of the table, we see that the expression 1.9t  68.8 represents the U.S. Postal Service first-class mail volume (in billions of pieces) in the year that is t years since 2012. b. Substitute 5 for t in 1.9t  68.8 : 1.9(5)  68.8  9.5  68.8  59.3. So, the first-class mail volume was 59.3 billion pieces in 2017.

34. The decrease in the number of Subway stores (in thousands) from 2016 to 2017 was 0.03(26.7). To find the number of Subway stores (in thousands), we subtract 0.03(26.7) from 26.7: 26.7  0.03 26.7   26.7  0.801  25.899 The number of Subway stores in 2017 was about 25.9 thousand.

35. For 0.0000678, the decimal point needs to be moved five places to the right so that the new number is between 1 and 10. Thus, the scientific notation is 6.78  105.

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Chapter 2: Designing Observational Studies and Experiments

31

Chapter 2: Designing Observational Studies and Experiments Homework 2.1 2. A sample is the part of a population from which data are collected.

4. The symbol p̂ “p hat” stands for a sample proportion.

6. A sampling method that consistently underemphasizes or overemphasizes some characteristic(s) of the population is said to be biased.

8. a. Ward A. Brockman, Robert Butts, Jerry Heidler, Richard Sealey, and Nicholas Tate. b. County of conviction, race, current age (in years), and time served (in years). c. County of conviction: Muscogee, Baldwin, Toombs, Clayton, and Paulding. Race: Caucasian, African American, Caucasian, African American, and Caucasian. Current age, all in years: 47, 41, 41, 54, and 38.

Time served, all in years: 28, 20, 19, 16, and 13. d. Ward A. Brockman: 47  28  19 years. Robert Butts: 41  20  21 years. Jerry Heidler: 41  19  22 years. Richard Sealey: 54  16  38 years. Nicholas Tate: 38  13  25 years. e. The youngest when convicted was Ward A. Brockman, at age 19 years.

10. a. Mars Polar Lander, Opportunity, Mars Reconnaissance Orbiter, Yinghuo-1, and Mars Orbiter Mission. b. Mission, outcome, launch mass (in pounds), and cost (in millions of dollars). c. Mission: lander, rover, orbiter, orbiter, and orbiter. Outcome: failure, success, success, failure, and success. Launch mass, all in pounds: 640, 408, 4810, 29,100, and 2948. Cost, all in millions of dollars: 110, 400, 720, 163, and 74. d. Mars Polar Lander, 110 / 640  0.172 ; Opportunity, 400 / 408  0.980 ; Mars Reconnaissance Orbiter, 720 / 4810  0.150 ; Yinghuo-1, 163 / 29,100  0.006 ; Mars Orbiter Mission, 74 / 2948  0.025 ; all in millions of dollars per pound. e. Yinghuo-1, 0.006 million dollars per pound.

12. a. The variable is whether American adults disapprove of the Affordable Care Act. b. The sample is the 1037 American adults who were surveyed. c. The population is all American adults.

14. a. The variable is whether American adults in the United States think that the United States does too much in solving the world’s problems. b. The sample is the 2008 American adults who were surveyed. c. The population is all American adults.

16. a. The variable is whether American adults think religion is very important in their lives. b. The sample is the 1025 American adults who were surveyed. c. The population is all American adults. d. Words: The proportion of the 1025 respondents who said religion is very important in their lives.

Symbol: p̂ Number: 0.51 e. Words: The proportion of all American adults who say religion is very important in their lives.

Symbol: p Number: unknown 18. a. The variable is whether parents will limit their children’s choices of college based on cost. b. The sample is the 1000 parents who were surveyed. c. The population is all parents with college-bound teenagers ages 16 to 18 years.

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32 ISM: A Pathway to Introductory Statistics

18. (continued) d. Words: The proportion of the 1000 parents with college-bound teenagers ages 16-18 who will limit their children’s choices of college based on cost.

Symbol: p̂ Number: 0.48 e. Words: The proportion of all parents with college-bound teenagers ages 16-18 who will limit their children’s choices of college based on cost.

Symbol: p Number: unknown 20. a. The researchers were trying to answer whether simvastatin heals ulcers. b. The sample is the 66 ulcer patients who were tested. c. The population is all patients with ulcers. d. The researchers concluded that the drug heals ulcers. The study is part of inferential statistics because it uses sample data to draw a conclusion about a population.

22. a. The researchers were trying to answer whether autistic adults are less able to process social rewards than monetary rewards. b. The sample is the 20 adults who were in the study. c. The population is all adults. d. The conclusion is that adults with autism are less able to process social rewards than adults without autism. It is part of inferential statistics because it uses data from a sample to make a statement about the population.

24. a. The researchers were trying to answer whether women who are more sexually confident are also more likely to achieve sexual satisfaction. b. The sample is the 45 women who took the online survey. c. The population is all women. d. The conclusion is that women who are more sexually confident are also more likely to achieve sexual satisfaction. It is part of inferential statistics because it draws a conclusion about a population, based on data taken from a sample.

26. a. statistics question b. variable e. statistics question

c. variable d. variable

28. a. Answers may vary. Example: 1. Which smartphone brand is most common?

2. Do people use certain brands of smartphone for taking pictures more so than others?

3. Is smartphone usage (in minutes) greater for men or for women on any given day? b. Answers may vary. Example: 1. Number of phones belonging to each smartphone brand 2. Number of pictures taken on each smartphone 3. Report of daily smartphone usage for women (in minutes); report of daily smartphone usage data for men (in minutes) 30. Using a TI-84: Mariah, Rani, May, and Brenton. Using StatCrunch: Rani, Brenton, Kali, and Shea.

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Chapter 2: Designing Observational Studies and Experiments 33 32. a. Using a TI-84: Samuel, Paola, Joshua, Win, and Phoebe. Using StatCrunch: Win, Taja, Nathan, Samuel, and Jeffrey.

3 2 b. Using a TI-84: 3  5  . Using StatCrunch: 2  5  .

5 5 Using this result to describe the sample is part of descriptive statistics because it does not draw conclusions about a larger group. c. Using a TI-84: Samuel, Jeffrey, Phoebe, Win, and Arnold. Using StatCrunch: Karen, Win, Monique, Arnold, and Jeffrey.

3 2 d. Using a TI-84: 3  5  . Using StatCrunch: 2  5  .

5 5 e. Using a TI-84: Yes. Using StatCrunch: Yes. However, not all randomly selected samples of size 5 will give the same results. Because of the randomness of choosing the sample, different samples could be collected.

34. a. Using a TI-84: Dimitrios, Aksana, Jessica, Luis, Fan, Chris, and Gauri. Using StatCrunch: Gauri, Chris, Aksana, Fadi, Devin, Jose, and Julia.

4 5 b. Using a TI-84: 4  7  . Using StatCrunch: 5  7  .

7 7 9 14 d. (Either technology) No. The difference between the answers in b. and c. is due to sampling error. e. For many random samples of size 7, the proportion of students who think it is more important to improve student success would not be the same on each sample and would not all be the same as the proportion for all 14 students. This is due to sampling error.

c. 9  14 

36. a. 14, 650  28, 614  0.512; So p  0.512. This result is a parameter since the full population is 28,614. b. 527  1000  0.527; So pˆ  0.527. This result is a statistic since the result is based on the sample 1000. c. No, the result from part (b) does not equal the result from part (a). This is due to sampling error. d. It would be inferential statistics because it draws a conclusion about a population based on data from a sample.

38. Do you access Facebook every day or not access Facebook every day?

40. Do you have a regular exercise program or not have one?

42. The method favors students who take evening classes, so it has sampling bias.

44. The wording of the question is not clear (since it asks whether they post daily and also if they like Facebook), so it has response bias.

46. Because 9 out of 12 subjects did not respond, the method has nonresponse bias. It also has response bias because an adult who neglects their children is unlikely to say “yes.”

48. Because 93% of those who were contacted did not give a response, the method has nonresponse bias.

50. The method has response bias because the scale of numbers for the response is not consistent.

52. This method has sampling bias because it favors cars that pass by during the morning rush hour.

54. The method has response bias because the question addresses more than one issue. It also has sampling bias, because it excludes people who do not watch this TV show.

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34 ISM: A Pathway to Introductory Statistics 56. a. The survey is likely to have nonresponse bias because participation is voluntary. It probably also has sampling bias, since it favors diners who want to complain about their experience. b. The survey likely has less nonresponse bias because of the incentive. It likely has less sampling bias because of the incentive. It may have more response bias, since the future discount likely improves the customer’s satisfaction with the restaurant.

58. Answers may vary. Example: The five steps of statistics are often called a cycle because once the steps are followed for one question, the results often lead to more questions, which prompts the five steps to be completed again and again.

60. Answers may vary.

62. Answers may vary.

64. Answers may vary. Example: A statistic is a numerical summary of a sample—it describes a sample. A parameter is a numerical summary of a population. After analyzing a sample, the value you would know is the statistic.

66. Sampling error refers to the random nature of the sample; nonsampling error refers to the design of the sampling process.

Homework 2.2 2. We should always round down when calculating k for systematic sampling.

4. False. Convenience sampling should never be used because such samples usually do not represent the population well.

6. Cluster sampling is the method because the 40 blocks are randomly selected, but every adult resident of each block is surveyed.

8. Systematic sampling is the method because every 100th car fuel tank after the first selected tank is tested.

10. Convenience sampling is the method because the employee only surveys the Americans whom she can contact easily.

12. Stratified sampling is the method because registered voters are randomly sampled within each of three strata: Republicans, Democrats, and Independents.

14. Simple random sampling is the method because sample members are selected at random from the whole population.

16. The method is systematic sampling because the pollster surveys every 10th person after the first to be selected.

18. The method is simple random sampling because members are randomly selected from all the paying guests in the past month.

20. a. 420  50  8.4; round down to 8. b. Using a TI-84: 4. Using StatCrunch: 5. c. Using a TI-84: 4, 4  8  12, 12  8  20, 20  8  28, 28  8  36 Using StatCrunch: 5, 5  8  13, 13  8  21, 21  8  29, 29  8  37 22. a. 88,110  300  293.7 ; round down to 293. b. Using a TI-84: 120

Using StatCrunch: 155 c. Using a TI-84: 120, 120  293  413, 413  293  706, 706  293  999, 999  293  1292. d. Using StatCrunch: 155, 155  293  448, 448  293  741, 741  293  1034, 1034  293  1327

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Chapter 2: Designing Observational Studies and Experiments 35 24. From the police department, survey 0.6170  42.7 or 43 employees. From the fire department, survey 0.35 70  24.5 or 25 employees. From the justice department, survey 0.04 70  2.8 or 3 employees.

26. The total number of students in the four schools is 1788  1690  1053  157  4688 . The proportions are: Franklin High School, 1788  4688  0.381 ; Centennial High School, 1690  4688  0.360 ; Fred J. Page High School, 1053  4688  0.225 ; Middle College High School, 157  4688  0.033 . The numbers of students in the sample from each high school, respectively, are: 0.38150  19.05 or 19; 0.360 50  18 ; 0.225 50  11.25 or 11; 0.033 50  1.65 or 2.

28. The total number of applicants to the five graduate business majors is 109  149  108  63  74  503 . The proportions are: Accounting, 109  503  0.217 ; Finance, 149  503  0.296 ; Information Risk and Operations Management, 108  503  0.215 ; Management, 63  503  0.125 ; Marketing, 74  503  0.147.

The numbers of applicants in the sample from each major, respectively, are: 0.217 100  21.7 or 22; 0.296 100  29.6 or 30; 0.215 100  21.5 or 22; 0.125 100  12.5 or 13; 0.147 100  14.7 or 15.

30. The proportions of each of the strata: Female undergraduate, 10,588  29,135  0.363; female graduate, 4475  29,135  0.154; female professional, 1421  29,135  0.049; male undergraduate, 7762  29,135  0.266; male graduate, 3736  29,135  0.128; male professional, 1153  29,135  0.040.

The numbers of students in the sample from each of the strata, respectively: 0.363 1200  436; 0.154 1200  185; 0.049 1200  59; 0.266 1200  319; 0.128 1200  154; 0.040 1200  48. Because

of rounding, the sample would actually have 1201 students.

32. Using a TI-84:

So, out of the frame of 17 Republicans, the Arizona Senate members selected would be: Ugenti-Rita, Gowan, Leach, Carter, and Boyer. Out of the frame of 13 Democrats, the Senate members selected would be: Contreras, Gonzales, Bowie, and Steele.

Using StatCrunch: Republicans: Gowan, Allen, Pratt, Boyer, and Pace; Democrats: Bowie, Alston, Steele, and Quezada.

34. The number of clusters is 75  25  3. Using a TI-84: Red Sox, Royals, Athletics. Using StatCrunch: Royals, Tigers, Indians.

36. Stratified sampling is being used, where the strata are farmers and city or suburban residents because each of the two strata is sampled separately.

38. Cluster sampling would require the least money and effort because surveying each resident on a selected block involves less travel time than a simple random sample. The city would decide on a sample size, identify a frame of all the blocks in Los Angeles, then divide the desired sample size by the smallest number of residents per block. The required number of blocks would be randomly selected, and then every resident on the selected blocks would be surveyed.

40. Simple random sampling is the best method, since Barnes & Noble® has a frame and the surveying can be done using e-mail. The company would choose a desired sample size, then randomly select that many online customers from the frame.

42. a. If each city block is treated as a cluster, the city would decide on a sample size, identify a frame of all the city blocks in Kansas City, and then divide the desired sample size by the smallest number of residents per block. That many blocks would be randomly selected, and every resident on the selected blocks would be interviewed in person.

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36 ISM: A Pathway to Introductory Statistics 42. (continued) b. To conduct stratified sampling, the data collectors would first choose a total sample size, then identify what proportions of registered voters are Democrats, Republicans, Independents, and so on, and compute the sample size for each of the strata by multiplying the total sample size by the respective proportions.

The required number for each of the strata would then be randomly selected. c. Cluster sampling would be easier than stratified because the data collectors would only need to visit the selected blocks in person. d. Stratified sampling would probably give better results if the sample size is small because it is more likely to get a sample that represents the whole city.

44. a. The police used systematic sampling when the traffic was heavier because they stopped every fourth car. b. Sampling every third and fourth car is not systematic sampling because it violates the pattern of selecting every kth person, animal, or thing. c. In lighter traffic, the police could have pulled over every other car. They would still be stopping two cars out of every four, but they would be using systematic sampling.

46. The type of sampling method used is voluntary response sampling. Answers may vary. Example: Sampling bias has occurred since it is impossible to know who is completing the survey since the survey is both voluntary and online. In other words, there is no way to ensure that adults from the United States are answering the survey. A better sampling method to use might be cluster sampling. Cluster sampling would allow a better representation of the population, which is adult Americans. Through this method, the blogger could select for individuals 21 and over (assuming 21 is the legal drinking age) and also the country of origin (to ensure that the poll is completed by Americans in this case).

48. Answers may vary.

50. Answers may vary.

52. Answers may vary. Example: A merchandise store has a counting machine set up near its exit, with buttons to press to indicate whether an adult has had a good experience or a poor experience during their visit. This is an example of voluntary sampling and illustrates sampling bias since there is no way to know who is pressing the buttons and how many times. It could be that a child presses the “poor experience” button several times, making it appear that more adults have had poor experiences than is actually the case. A better sampling method to use would be cluster sampling or systematic sampling so that it is possible to select for adults.

54. Answers may vary.

Homework 2.3 2. In a double-blind study, neither the individuals nor the researcher in touch with the individuals know who is in the treatment group(s) and who is in the control group.

4. A lurking variable is a variable that causes both the explanatory and response variables to change during the study.

6. a. The treatment groups are the second and third groups because they receive training in addition to that which the first group receives. b. The study is an experiment because each participant is assigned to one of the treatment or control groups. c. Random assignment means that the researchers use random sampling to decide which participants are in which groups. For example, the researchers could create a frame of all 50 older adults, randomly choose 17 of them to be in the second group, randomly choose another 17 for the third group, and assign the other 16 to the first group. d. The sample is the 50 older adults in the study. The population is all older adults.

8. a. The explanatory variable is the type of training that participants did. The response variable is walking speed when an older person is performing a mental task at the same time. b. The researchers concluded that the training methods for improving walking speed that include both physical and mental tasks are more effective than those used in the first group when an older adult is performing a mental task at the same time. Causality can be concluded because the participants were randomly assigned to the treatment and control groups.

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Chapter 2: Designing Observational Studies and Experiments 37 8. (continued) c. The first group’s confidence may have increased because the group’s training was easier than that of the second and third groups.

10. a. It makes sense that the study is observational because the researchers cannot randomly assign anyone to have a major bone fracture. b. It would be unethical to randomly assign an older adult to “treatment” when that treatment requires a major bone fracture. c. The sample is the people whose records were studied. The population is all adults over 60. d. The explanatory variable is whether or not the person had a major bone fracture. The response variable is the death rate. e. The conclusion is that the death rate for older adults who have had a major fracture is higher than the death rate for older adults who have never had a major fracture. Only an association can be concluded because there was no random assignment to treatment or control groups.

12. a. The study is observational because there is no random assignment. b. Since a placebo has no proven medical effect, it would be unethical for the doctors to administer it instead of prescribing an effective remedy for an acute cough. c. The sample is the 241 children in the study. The population is all children with an acute cough. d. The researchers concluded that children who took levodroprophizine recovered better from coughs than children who took other cough syrups. Only an association can be concluded because there was no random assignment to treatment and control groups. e. Researchers could be influenced, consciously or unconsciously, by earning a salary from the company that manufactures levodroprophizine. The two researchers’ disclosing that they work for the company encourages other researchers who do not work for the company to repeat the experiment and see if they get similar results.

14. a. The treatment group is the one that received a gift card plus monetary rewards based on their class work.

The control group is the one that received only a gift card. b. The study is an experiment because the researchers randomly assigned students to the treatment and control groups. c. Random assignment means that the researchers chose some students at random to receive the treatment and others to be the control group. d. The sample is the 1019 students in the study. The population is all low-income community college students who are parents.

16. a. It would be impossible to use a placebo for a monetary reward. A participant would quickly discover whether they have real or fake money. b. In order to be double-blind, the participants would have to not know whether they will receive monetary rewards, which would remove the incentive to earn more credits. c. The explanatory variable is whether or not students would receive an additional monetary reward based on the credits they earn. The response variable is the number of credits the students earned. d. The researchers concluded that monetary rewards increase the number of credits earned by low-income community college students who are parents. Causality can be concluded because students were randomly assigned to the treatment and control groups. e. Mistakenly giving monetary rewards to some of the control group introduces a possible lurking variable; believing that they will be rewarded no matter how many classes they pass could decrease students’ motivation to do well.

18. a. This is an observational study because there is no differentiation into treatment and control groups and no random assignment. b. The sample is the 210 motorists whose behavior was observed. The population is all motorists in Chicago.

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38 ISM: A Pathway to Introductory Statistics 18. (continued) c. The explanatory variable is whether the crosswalk was marked or unmarked. The response variable is whether or not the motorist stopped. d. The researchers concluded that motorists are more likely to follow the Must Stop law at marked than at unmarked crosswalks. Only association can be concluded because there was no random assignment into treatment and control groups. e. We cannot assume the conclusion is also true in Prairie City because the habits of motorists could be very different in a small farm town than in a large city. Motorists are also more likely to be acquainted with the pedestrians in a small town, which could influence their behavior.

20. Using a TI-84:

So we select the ultrabooks with positions 3, 5, 6, and 2 and in the frame for the treatment group: HP Spectre 13, Acer Aspire S 13, Acer S3-391-9606, and Surface Laptop 2.

Using StatCrunch: HP Spectre 13, Surface Laptop 2, Asus ZenBook 3, ThinkPad X260.

This is an example of random assignment because the ultrabooks were randomly assigned to the treatment and control groups.

22. Using a TI-84:

So we select the 4-year colleges with positions 8, 11, 12, 10, 7, and 4 in the frame for the treatment group: Palm Beach State College, Virginia Military Institute, Kean University, SUNY Oneonta, Angelo State University, Langston University.

Using StatCrunch: Eastern Illinois University, Lander University, Boise State University, Langston University, Kean University, Oakland University.

This is an example of random assignment because the colleges were randomly assigned to the treatment and control groups.

24. Answers may vary. Example: If the students and the researcher knew which students had consumed caffeine this would flaw the experiment since the students may be biased toward thinking caffeine really does improve their performance. In that case, the outcome might be due to a perception on the part of the students about caffeine rather than an actual physiological change. It is also the case that students may perform better on an exam, in the treatment group, owing to having studied for the test, which would result in a better outcome. So, preparedness could be a lurking variable. To improve the study, the researcher should ensure that all students equally prepare for the test and that all students are roughly equal in terms of their overall performance in the course. The students should then be randomly assigned to the treatment or control group (with the students unaware to which group they are assigned). The students should be administered caffeine in a form that does not enable them to know if they are in the treatment group or the control group. To reduce the bias that the researcher him/herself might impart on the study, the study should be designed as double-blind. Otherwise, it is possible that the researcher might influence the outcome of the results if he/she is observing the students while they take the test. Knowing which students consumed caffeine and which did not, the researcher might focus on watching certain students to see if they exhibit any physical changes during test-taking, which could impact the students’ performance.

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Chapter 2: Designing Observational Studies and Experiments 39 26. a. No, it would not be good practice for the treatment group to consist of the 25 lightest participants and the control group to consist of the 25 heaviest participants. It is possible that the lightest participants might be affected by alcohol in a shorter amount of time than the heaviest participants. So, weight could be a lurking variable. b. No, it would not be good practice for the treatment group to consist of the 25 youngest participants and the control group to consist of the 25 oldest participants. It is possible that the age affects reflex reactions, and therefore driving ability. So, age could be a lurking variable. c. Answers may vary. Example: One way to form the treatment and control group is to ensure that the treatment group and the control group contain participants of similar age and weight. Random assignment could then be used to select 25 participants for the treatment group, with the other 25 going to the control group. This method is reasonable because it addresses the influence of confounding and lurking variables.

28. a. Because this is an observational study, the student cannot conclude causality. There is likely to be response bias; people may overstate the amount of exercise they get. Also, motivation to stay healthy could be a lurking variable affecting whether or not people smoke (or quit smoking) and whether or not they exercise. b. The student could find people who currently smoke and randomly assign the smokers to one of two groups, treatment or control. The treatment group would exercise on a regular basis, while the control group would not. It would be impossible for the study to be double-blind since the participants know what treatment they have; however, it could be a blind study if the researcher(s) in contact with the participants do not know which is in each group.

30. a. Although there is an explanatory variable, there is no control group (or random assignment). This is an observational study, and causality cannot be concluded. There is also a confounding variable since the reward is not just monetary. Students could also be motivated to write a better project for the public honor of appearing in the newspaper. b. The students could be randomly assigned to one of two groups, treatment or control. The control group would have the same assignment but no prize for the best project. The treatment group will be told that the best project will win a $25 prize. It could be a blind study if the projects are graded by another teacher who does not know which group the students were in.

32. a. Although there is an explanatory variable, there is no control group (or random assignment). This is an observational study and causality cannot be concluded. There could be sampling bias and response bias because the survey is online; it could favor people with greater incomes who could afford newer cars, and voluntary response could favor people whose mileage improved after using the additive. The financial state of the respondents could also be a lurking variable, since it could influence both the ability of car owners to respond online and their ability to keep the car in good running condition. b. The magazine could select a random sample of drivers and randomly assign them to either a treatment group or a control group. The treatment group would be given the additive and asked to use it, while the control group would not. The gas mileage of each car would be recorded at the beginning of the study, and again a month afterwards.

34. Researchers could recruit a sample of volunteers who suffer from insomnia and randomly assign them to either a treatment or a control group. Volunteers in the treatment group would receive the drug, and volunteers in the control group would receive a sugar pill. The study could be double-blind, with one researcher labeling pill vials but not giving the code to the researchers in contact with the patients. The extent of insomnia would be measured at the beginning of the study and again a month later.

36. A sample of students who have the same class with the same professor could be randomly assigned to a treatment or control group. Each student could be given a recording of the professor’s lectures, but only the treatment group would be instructed to listen to the lectures while they sleep. The professor (or someone else who grades the tests) could be blind to which students are in which group.

38. With random assignment, the frame is a sample, and sampling divides the individuals into treatment group(s) and a control group. Stratified sampling does not involve treatment and control groups; it defines groups with similar characteristics (strata) that already exist in the population and creates frames for each of the strata.

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40 ISM: A Pathway to Introductory Statistics 40. The key difference in the designs of an experiment and an observational study is the presence of both treatment group(s) and a control group to which individuals are randomly assigned. Random assignment to one of the groups makes it possible to isolate the effects of the treatment from other factors.

42. Answers may vary.

44. Answers may vary. Example: A study that seeks to determine whether patients recovering from knee replacement surgery recover faster when they undergo aquatic-rehabilitation than patients who undergo more traditional means of rehabilitation. Since the control group is recovering from surgery, it is impossible to administer a placebo when the recovery of patients in the control group necessitates some form of rehabilitation.

46. The student is correct. The objects do not know whether they are in a treatment or control group.

48. A researcher wants to study whether American adults who take a Zinc supplement starting at the onset of a cold and for three days afterward experience a faster recovery of their colds. To conduct a well-designed experiment, the researcher selects 1000 participants: 500 for the control group and 500 participants for the treatment group who are similar in age, have similar medical backgrounds, be taking no other medications or supplements that might interfere with the results during the period of the study, have similar sleeping and eating habits, stress levels, family sizes, and activity levels. The study would be double-blind. The explanatory variable is adults who take a Zinc supplement at the start of a cold and for three day afterward and the response variable is shorter duration of the cold. Five hundred participants would be randomly assigned to the treatment group, but not aware whether the product they consume during the study is the Zinc or the placebo. By default, the other 500 participants would be in the control group. The results of the study would seek to determine whether the population—all American adults—could recover faster if they consumed a Zinc supplement at the onset of a cold and for three days afterward.

Chapter 2 Review Exercises 1. a. The individuals are the countries: Bahrain, Iraq, Iran, Kuwait, and Saudi Arabia. b. The variables are government, 2017 population (in millions), 2017 military expenditure (in millions of dollars), and oil production (in millions of barrels). c. For the variable government: monarchy, republic, republic, monarchy, and monarchy. For the variable 2017 population, all in millions: 1.5, 38.3, 81.2, 4.1, and 32.9. For 2017 military expenditure, all in millions of dollars: 1397, 7416, 14,548, 6831, and 69,413. For 2017 oil production, all in millions of barrels: 18, 1624, 1456, 1066, 3818. d. Bahrain, 1397  18  78 Iraq, 7416  1624  5 , Iran, 14,548  1456  10 , Kuwait, 6831  1066  6 , Saudi Arabia, 69, 413  3818  18 , all in dollars per barrel.

e. Bahrain has the greatest ratio of 2017 military expenditure to 2017 oil production, 78 dollars per barrel.

2. a. The variable is whether American adults experience a lot of happiness and enjoyment. b. The sample is the 500 American adults who were telephoned. c. The population is all American adults. d. Words: The proportion of the 500 American adults who say they experience a lot of happiness and enjoyment.

Symbol: p̂ Number: 0.46 e. words: The proportion of all American adults who say they experience a lot of happiness and enjoyment.

Symbol: p Number: This value is unknown.

3. a. Using a TI-84: Antoine, Jacob, Ruben, Sandra, Dante, Jose. Using StatCrunch: Mario, Sandra, Antoine, Jacob, Alyssa, John.

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Chapter 2: Designing Observational Studies and Experiments 41 3. (continued) 1 1 b. Using a TI-84: 3  6  . Using StatCrunch: 3  6  . Using this result to describe the sample is part of 2 2 descriptive statistics because it does not generalize the results of the sample to describe the population.

5  0.417.

12 d. Using a TI-84 or Using StatCrunch: No, the sample proportion who prefer comedies does not equal the population proportion who prefer comedies. The difference is due to sampling error. e. No, the proportion in each sample would not equal the proportion of all 12 students who prefer comedies.

The difference is due to sampling error. f. If two researchers perform the same study with different simple random samples of the same size, their inferences will not necessarily be the same because the sample data are not the same.

c. The proportion who prefer comedies is

4. Choose the number of questions you usually ask during one hour of your prestatistics class: 0, 1, 2, 3, or more than 3.

5. The method has sampling bias; the sampling favors people who visit the militia group site.

6. The method has sampling bias, response bias, and nonresponse bias. The sampling favors people who are often in the financial district; some people may exaggerate their salary; 55 of those who were approached declined to answer.

7. Answers may vary.

8. The method is cluster sampling because the researcher selects 50 blocks at random and then surveys each adult resident of those blocks.

9. The method is simple random sampling because Human Resources creates a frame of all U.S. employees and selects at random from that frame.

10. The method is convenience sampling because the pollster only surveys people who are easy to find, without attention to any random selection.

11. The method is stratified sampling because the researchers identify two strata (people with landlines and people with cell phones), and randomly select numbers in each of the strata.

12. The method is systematic sampling because the manager surveys every eighth person leaving the store after the first person is randomly selected.

13. a. 117,300  800  146.625 round down to 146. b. Using a TI-84: 64

c. Using a TI-84: 64, 64  146  210 , 210  146  356 , 356  146  502 , 502  146  648 Using StatCrunch: 52, 52  146  198 , 198  146  344 , 344  146  490 , 490  146  636 14. The total number of employments (in thousands) is 59  36  26  20  141 . The proportions are: commercial airplanes, 59  141  0.418 ; defense, space, and security, 36  141  0.255 ; corporate, 26  141  0.184 ; global services, 20  141  0.142 The numbers of employees in the sample from each group, respectively, are: 0.418 80  33 ; 0.255 80  20 ; 0.184 80  15 ; 0.142 80  11

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42 ISM: A Pathway to Introductory Statistics 15. Using a TI-84:

This means we select Democrats: Gopal, Diegnan, Weinberg, Cruz-Perez, and Rice; and Republicans: Corrado, O’Scanlon, and Pennacchio Using StatCrunch: Democrats: Smith, Pou, Codey, Cruz-Perez, and Lagana; Republicans: Kean, Thompson, and Corrado.

16. If the clusters are city blocks, cluster sampling would require the least time and effort. The city would create a frame of all the blocks in the city, select some at random, and then survey each resident of the selected blocks.

17. a. The treatment groups are the three groups who receive the drug; the control group is the group who receives a placebo. b. The study is an experiment because the researchers randomly assigned participants to one of the three treatment groups or to the control group. c. Random assignment means that the researchers randomly assigned patients to one of the four groups. To accomplish the random assignment, create a frame of the 560 MDD adults. For each treatment group, randomly select 140 different MDD adults to be in the group. The remaining 140 MDD adults are the control group. d. The sample is the 560 MDD adults in the study. The population is all adults with MDD. 18. a. The placebo could be a sugar pill. b. Neither the study participants nor the researchers in contact with them know which treatment (or placebo) the participants receive. One researcher could have labeled the pill vials with numbers to identify the pills, but not told the code to another researcher who was in contact with the participants. c. The explanatory variable is the dosage of the drug the person receives. The response variable is the person’s HRSD score. d. The conclusion of the study is that Lu AA21004 successfully lowers MDD adults’ HRSD scores. The researchers can conclude causality because adults were randomly assigned to the treatment and control groups. e. The researchers concluded that the drug tends to lower MDD adults’ HRSD scores, but that might not mean that the drug tends to reduce depression in MDD adults.

19. a. The study is observational because mothers were not randomly assigned to the group with eating disorders or the group without eating disorders. b. It would be impossible to use random assignment in this study because mothers could not start having an eating disorder, or stop having an eating disorder, due to a researcher telling them to. c. The sample is the mothers who were observed. The population is all mothers with first-born infants. d. The explanatory variable is whether or not the mother has an eating disorder. The response variable is the level of negative emotions expressed toward the infants during mealtimes. e. The conclusion of the study is that mothers with eating disorders express more negative emotions toward their first-born infants during mealtimes than mothers without eating disorders. It only describes an association; an observational study cannot conclude causality.

20. Using a TI-84: Elon University, Campbell University, Wellesley College, Columbia College, University of Mount Union. Using StatCrunch: Nichols College, Columbia College, Mills College, Rider University, Villanova University. Yes, it is an example of random assignment because the colleges were randomly assigned to the two groups.

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Chapter 2: Designing Observational Studies and Experiments 43 21. a. The coordinator did not use random assignment, so she cannot conclude causality. Also, the attendance should include a time requirement because a student who attended the math center for only five minutes once in the entire semester should not be considered a student who used the center. Motivation could be a lurking variable: students who attend the math center might be more motivated, and study harder, than other students. b. The coordinator could randomly assign some students to a treatment group and others to a control group.

The students in the treatment group would, for example, attend the math center for one hour per weekday during the entire semester, while the students in the control group would not attend the math center. After the semester is over, the coordinator could compare the proportion of the treatment group who passed their math classes that semester with the proportion of the control group who passed their math classes that semester.

22. The company could randomly assign some bald people to a treatment group and some to a control group. The treatment group would take the drug and the control group would take a sugar pill. The study could be doubleblind. The company would then measure the extent of the individuals’ hair growth after 8 months.

Chapter 2 Test 1. a. The individuals are Delaware, Hawaii, Mississippi, Texas, and Wisconsin. b. The variables are region, number of workers (in thousands), and number of workers in unions (in thousands). c. Region: East, West, South, South, and Midwest. Number of workers: 425, 605, 1106, 11,626, and 2778, all in thousands. Number of workers in unions: 45, 129, 59, 543, and 230, all in thousands.

Texas: 543  11, 626  0.0467  4.7% d. Delaware: 45  425  0.1059  10.6%

Hawaii: 129  605  0.2132  21.3%

Wisconsin: 230  2778  0.0828  8.3%

Mississippi: 59  1106  0.0533  5.3% e. Hawaii has the largest percentage of workers in unions, 21.3%.

2. a. The variable is whether the mainstream media divides people along racial, gender, and political lines. b. The sample is the 1024 American adults who were surveyed. c. The population is all American adults.

3. The study has response bias and nonresponse bias. The complex wording of the question may lead customers to give an answer that is not consistent with their opinion (response bias), and the nonresponse rate of 92% indicates nonresponse bias.

4. Using a TI-84: Jamie, Jared, Isabel, Lisa. Using StatCrunch: Jamie, Brianna, Dan, Michael.

5. The method is cluster sampling. The farmer randomly selects 8 subsections, then measures the total yield from each of those subsections.

6. a. k  500  80  6.25; round down to 6.

b. Using a TI-84: 1. Using StatCrunch: 2.

c. Using a TI-84: 1, 1  6  7, 7  6  13, 13  6  19, 19  6  25 Using StatCrunch: 2, 2  6  8, 8  6  14, 14  6  20, 20  6  26 7. The total number of students is given: 14,308. The proportions are: female undergraduates, 8262  14,308  0.577 ; female graduate students, 731  14,308  0.051 ; male undergraduates, 5078  14,308  0.355 ; male graduate students, 237  14,308  0.017 The numbers of students in the sample from each of the strata, respectively, are: 0.577 500  289 ; 0.051500  26 ; 0.355 500  178 ; 0.017 500  9

8. a. The treatment groups are the 4 groups taking different drug dosages; the control group is the group receiving a placebo. b. The study is an experiment because patients were randomly assigned to the treatment and control groups.

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44 ISM: A Pathway to Introductory Statistics 8. (continued) c. Random assignment means that the researchers randomly assigned the patients to the groups. To accomplish this, create a frame of the 361 patients. Then for each of the 4 treatment groups, randomly select 72 patients to be in the group. The remaining 73 adults should be in the control group. d. The sample is the 361 patients in the study. The population is all Japanese adults with type 2 diabetes.

9. a. The placebo could be a sugar pill. b. Neither the patients nor the researcher(s) in contact with the patients knew which patients were in each group; one researcher could have labeled the pill vials with numbers to identify the pills, but not tell the code to another researcher who was in contact with the individuals. c. The explanatory variable is the dosage of the drug. The response variable is the glycated hemoglobin level. d. The conclusion of the study is that the drug successfully lowers glycated hemoglobin levels in Japanese patients with type 2 diabetes. Because treatments and control were randomly assigned, the researchers can claim causality. e. Researchers could be influenced, consciously or unconsciously, by earning a salary from the company that manufactures ipragliflozin. Reporting that they work for the company encourages other researchers who do not work for the company to repeat the experiment and see if they get similar results.

10. a. The researcher did not use random assignment. The players who run every day may also practice basketball longer and harder than players who do not run every day; motivation may be a lurking variable. b. The researcher could randomly assign players to a treatment group and a control group. Players in the treatment group would run for one hour every day, and players in the control group would not run. The researcher could have an assistant oversee the training, so the researcher would be blind to which players are in which group. After the players have run daily for one month, the researcher would compare the scoring of the two groups in the next month, while the players in the treatment group continued to run daily.

11. The owner could randomly assign the sales force to a treatment group and a control group. The treatment group would attend a workshop about emotions for a weekend. The control group would not attend the workshop.

The owner could be blind to which employees attended the workshop. One month later, the owner would compare the monthly sales by the treatment and control groups.

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Chapter 3: Constructing Graphical and Tabular Displays of Data

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Chapter 3: Constructing Graphical and Tabular Displays of Data Homework 3.1 2. A numerical variable consists of measurable quantities that describe individuals.

4. The relative frequency of a category is the proportion of all the observations that fall in that category.

6. Numerical; a person’s height can be measured.

8. Categorical; the variable is a label applied to groups of persons.

10. Categorical; the area code is a label applied to groups of phone numbers.

12. Numerical; the cat’s weight (in pounds) can be measured.

14. We can make a frequency table.

Commute to School Tally Frequency Car

IIII

4

Bus

I

1

Bicycle Total

I

1 6

16. We can make a frequency and relative frequency table.

The frequency of the observation car is 4.

Majors

Tally Frequency Relative Frequency

business

II

2

art

I

1

psychology

I

1

Total

4

2  0.500 4 1  0.250 4 1  0.250 4 4 1 4

The relative frequency of the observation business is 0.500.

18. a. The variable is how often Facebook users visit the site. It is categorical because users are grouped into one of three categories Several times a day, Once a day, or Less often. b. Since the height of the “Less often” bar is approximately 354, 354 users visited the site less than once a day. c. The number who visited the site at most once a day is the number who visited once a day plus the number who visited less than once a day: approximately 313  354  667 Facebook users. d. The number of users who did not visit the site several times a day is the number of users who visited less than once a day subtracted from 1361: approximately 1361  354  1007 Facebook users. e. 1361  2002  0.680  68.0% of online users in the study used Facebook.

20. a. Since the bar for “Not worried at all” has height approximately at 267, about 267 of those surveyed are not worried at all about them or someone in their family becoming a victim of a mass shooting. b. The number who are somewhat worried OR not too worried is approximately 298  350  648 of the adults surveyed. c. The number who are NOT very worried is the total number surveyed minus the number who are very worried: approximately 1018  103  915 of those surveyed. d. The relative frequency is the number who is very worried divided by the total number surveyed: 103  1018  0.10.

22. The Fridays are dates 6, 13, 20, and 27.

24. The dates in the fourth week are 22, 23, 24, 25, 26, 27, and 28.

26. The dates that are Fridays OR the fourth week are 6, 13, 20, 22, 23, 24, 25, 26, 27, and 28.

28. There is only one date that is both a Friday AND in the fourth week, 27.

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46 ISM: A Pathway to Introductory Statistics 30. a. The proportion who read only print books is 0.38. b. The proportion who read only print books OR do not read books is the sum of the proportions: 0.38  0.26  0.64. c. The proportion who do NOT read only print books is 1 minus the proportion who read only print books: 1  0.38  0.62. d. The number who read only digital books is the proportion times the number of persons surveyed: 0.07  1520  106.4. e. There are no obvious sources of bias since most adults have either a landline or a cell phone (or both).

32. a. The proportion of those surveyed who heard offensive comments frequently was 0.06. b. The proportion of those surveyed who did not hear offensive comments frequently is the proportion who heard offensive comments frequently subtracted from 1: 1  0.06  0.94. c. The proportion of those surveyed who heard offensive comments occasionally OR rarely is the proportion who heard offensive comments occasionally along with those who heard offensive comments rarely: 0.25  0.47  0.72.

34. a. The sum of the relative frequencies is 0.06  0.25  0.47  0.22  1. This makes sense because the sum of the parts equals the whole. b. The frequencies are found by multiplying each relative frequency by the total number surveyed.

Category Frequently Occasionally Rarely Never Total

Frequency 0.06  3072  184 0.25  3072  768 0.47  3072  1444 0.22  3072  676 3072

c. Even if the survey was carried out well, the proportion for the population might be a bit different than the proportion for the sample due to sampling error. d. The percentage of the 54,806 randomly selected students who responded to the Web survey AND responded to the telephone interview is 0.49  (6928  54,806)  0.062 or about 6%. There is likely sampling bias since the survey was online. There also is likely response bias since the responses may not have reflected what the respondents really think.

36. a. The percentage of fathers in the 2012 survey who were unable to find work is the relative frequency times 100%, or 23%. b. Even if the survey was carried out well, the proportion for the population might be a bit different than the proportion for the sample due to sampling error. c. Most fathers in the 1989 study stayed at home because they were ill or disabled, 56% of the total. d. In 2012, a larger percentage of stay-at-home fathers were unable to find work, or chose to care for their home and their family. e. It is important that the vertical bar show relative frequencies and not frequencies because there may have been more fathers surveyed in 2012 than in 1989 (or vice versa).

38. a.

b.

c. The graphs are examples of descriptive statistics because the graphs describe the data that were collected.

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38. (continued) d. The proportion of the plays of the songs Delicate OR Going Bad is the sum of the proportions: 0.133  0.167  0.3. e. Of the top six songs played, the proportion of the plays of the songs that was not ZEZE is found by subtracting the proportion of plays of the song ZEZE from 1: 1  0.167  0.833.

40. a. The variable is most important part of the Super Bowl which is categorical. b.

c.

Category Game

Tally IIII IIII

Frequency 10

Friends

IIII

4

Commercials

IIII

5

Half-time show

II

2

Food Total

I

1 22

Category

Tally

Game

IIII IIII

Friends

IIII

Commercials

IIII

Half-time show

II

Food

I

Total

Relative Frequency 10  0.455 22 4  0.182 22 5  0.227 22 2  0.091 22 1  0.045 22 22 1 22

d. The proportion of respondents who said the most important part of the Super Bowl is NOT the game is found by subtracting the proportion of respondents who said the most important part of the Super Bowl is the game from 1: 1  0.455  0.545. e. The proportion of respondents who said the most important part of the Super Bowl is the game OR the half-time show is the sum of the proportions: 0.455  0.091  0.546.

42. a. Countries of Top 15 women in NYC Marathon

Tally

Frequency Relative Frequency

Ethiopia

I

1

Kenya

II

2

U.S.

IIII IIII

10

South Africa

I

1

Australia

I

1

Total

15

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1  0.067 15 2  0.133 15 10  0.667 15 1  0.067 15 1  0.067 15 15 1 15

48 ISM: A Pathway to Introductory Statistics 42. (continued) b.

c.

d. The percentage of the top 15 women who are NOT from the United States is the percentage of women who are from the United States subtracted from 1: 1  0.667  0.333 or about 33.3%. e. The percentage of the top 15 women who are from Kenya OR the United States is the sum of: 0.133  0.667  0.80 or about 80%.

44. a.

b.

c. The proportion who prefer electronic music is shown in the relative frequency bar graph, 0.100. d. The proportion who do NOT prefer electronic music is 1 minus the proportion who do prefer electronic music: 1  0.100  0.900. e. The proportion who prefer hip-hop OR rap is the sum of the proportions: 0.100  0.400  0.500.

46. a.

b.

c. The proportion who trust their spouse the most is shown in the relative frequency bar graph: 0.125. d. The proportion who do NOT trust their spouse the most is 1 minus the proportion who do trust their spouse the most: 1  0.125  0.875. e. For the surveyed adults, an adult is about equally likely to trust parents or a financial advisor. It is easier to tell this from the frequency bar graph, because it is easier to compare the heights of the bars than to count the observations in the table and then compare the frequencies.

48. Answers may vary.

52. Answers may vary.

50. Answers may vary.

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54. The proportion-of-the rest property reveals the proportion of a whole that remains after one of the proportions 3 that make up the whole is subtracted from the whole. For example, if a student eats of a whole pizza, then 8 using the proportion-of-the-rest property we find the proportion of the pizza that the student did not eat: 3 8 3 5 1    .

8 8 8 8 56. Answers may vary.

60. a. Answers may vary. b. Answers may vary. c. Answers may vary.

58. Answers may vary. d. Answers may vary. e. Answers may vary. f. Answers may vary.

Homework 3.2 2. The sum of all the percentages indicated by a pie chart is equal to 100%.

4. True. Simply adding a row total and a column total will double-count the value at the intersection of the row and the column.

6. a. The variable is whether respondents think high graduate starting salaries indicate that a university provides quality teaching. It is categorical because labels are applied to different categories. b. The proportion of students who strongly disagreed was 0.06. c. To find the proportion of students who did NOT strongly disagree, subtract the proportion of those who strongly disagree from 1: 1  0.06  0.94. d. The proportion of students who strongly agreed OR somewhat agreed is the sum of the proportions: 0.31  0.27  0.58. e. The student’s estimate of 31% might have a lot of error. The idea that starting salaries of high graduates is an indicator of quality teaching is based on worldwide data. The percentage could look differently if the survey was administered to only university students in the U.S. 8. a. The proportion of respondents who said Five Guys OR In-N-Out have the best burgers is the sum of the proportions: 0.15  0.09  0.24. b. Since each restaurant grouped under “Other” must have a percentage less than 7%, the number of restaurants must be at least 40  7  5.7, rounded up to 6. c.

d. Although a pie chart uses percentages and a relative frequency bar graph uses relative frequencies, the two diagrams contain the same information because percentages can be converted to relative frequencies and vice versa.

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50 ISM: A Pathway to Introductory Statistics 10. a. China emits the most carbon dioxide of the six regions, 29.2% of the worldwide total. b. China has the largest population of the six regions, 18.6% of the worldwide total. c. The United States emits 0.14 proportion of worldwide carbon dioxide emissions and has 4.3% of the worldwide population. One could argue that the United States emits more than its fair share because its proportion of carbon dioxide emissions is greater than its proportion of worldwide population. d. Using the argument of part (c), China, Germany, Russia, and Japan all emit more than their fair share of carbon dioxide. e. The list of regions in part (d) may not be complete because for the regions grouped under “Other,” we do not know whether their proportion of worldwide carbon dioxide emissions is greater than their proportion of worldwide population.

12. a.

b. The proportion of those surveyed who never unplug from technology can be read from the pie chart and converted from a percentage: 33.0  100  0.330. c. The proportion who do NOT unplug from technology daily is 1 minus the proportion who unplug daily: 1  0.160  0.84. d. The proportion who unplug from technology a few times a year OR a few times a month is the sum of the percentages converted to a proportion: (10.0  12.0)  100  0.220. e. Even if the survey was carried out well, the proportion for the population might be a bit different than the proportion for the sample due to sampling error.

14. a. The proportion of the adults who have poor health who are lower-income is 268  602  0.445. b. The proportion of the lower-income adults who have poor health is 268  663  0.404. c. The proportion of the middle-income adults who have poor health is 283  1287  0.220. d. The student is incorrect. On the basis of the results from parts (b) and (c), we conclude that lower-income adults are more likely to have poor health than middle-income adults. e. Response bias is likely to have occurred because not all the respondents were willing to give true answers to the survey questions.

16. a. The proportion of surveyed Caucasians who favored preference in hiring African-Americans is 134  929  0.144. b. The proportion of surveyed African-Americans who favored preference in hiring African-Americans is 77  180  0.428. This proportion is larger than the result of part (a) because African-Americans would stand to benefit more than Caucasians would. c. The student is incorrect. On the basis of the results from parts (a) and (b), we conclude that AfricanAmericans are more likely to favor preference than Caucasians. d. Even if the survey was carried out well, the proportion for all American adults might be a bit different than the proportion for the sample due to sampling error. e. Answers may vary.

18. a. The proportion of very-happy people who have excellent health is 271  551  0.492. b. The proportion of pretty-happy people who have excellent health is 247  551  0.448. c. The proportion of not-too-happy people who have excellent health is 33  551  0.060.

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18. (continued) d. The result in part (a) is greater than the result in part (b), which is greater than the result in part (c). The comparison shows that very-happy-people are more likely to have excellent health than pretty-happy people and that pretty-happy people are more likely to have excellent health than not-too-happy people. e. While it would appear from the survey and the results in parts (a), (b), and (c), that an increase in level of happiness has a possible impact on health, we cannot draw the conclusion that an increase in level of happiness causes an improvement in health. The reason is, the people who are very happy may engage in more health beneficial activities on a more frequent basis than people who are pretty happy or not too happy. Therefore, degree of happiness alone cannot be regarded as the cause for an improvement in health.

20. a. The variables described in the table are whether climate change will have a significant negative effect and political party affiliation. Both variables are categorical since labels are applied to different categories. b. The proportion of surveyed Democrats who think that climate change is going to have a significant negative effect in their lifetime is 258  344  0.75. c. The proportion of those surveyed who are Democrats AND think that climate change is going to have a significant negative effect in their lifetime is the proportions of those surveyed who are Democrats and think that climate change is going to have a significant negative effect in their lifetime: 258  1056  0.244. d. The questions in parts (b) and (c) involve two different groups. The group in part (b) involves just the group of Democrats (therefore, we use 344 to find the proportion) whereas the group in part (c) involves all those who were surveyed (therefore we use 1056 to find the proportion) who are Democrats who think climate change is going to have a significant negative effect. e. The proportion of those surveyed who are Democrats OR think that climate change is going to have a significant negative effect in their lifetime is the sum of all Democrats along with all those who think climate change is going to have a significant negative effect in their lifetime : 344  57  221  622 divided by the total number of respondents: 622  1056  0.589.

22. a. The proportion of participants with high fish consumption who had prostate cancer is 42  549  0.077. b. The proportion of participants with low fish consumption who had prostate cancer is 201  2621  0.077.

This result is the same as that in part (a). This means that the incidence of prostate cancer was about the same among those who consumed a high amount of fish as those who consumed a low amount of fish. c. The proportion of participants who did NOT have a high level of fish consumption is the proportion who had a high level of fish consumption subtracted from 1 (proportion with high level of fish consumption: 549  6272  0.088 ): 1  0.088  0.912. d. The proportion of participants who had prostate cancer OR moderate fish consumption is the sum of all those surveyed who had prostate cancer along with all those with moderate fish consumption: 14  201  209  42  2769  3235 divided by the total number of participants: 3235  6272  0.516. e. The proportion of participants who had prostate cancer AND moderate fish consumption is the proportion of all participants that had prostate cancer and ate fish at a moderate level: 209  6272  0.033.

24.

Working

Not Working Total

Female Male

4 3

2 1

6 4

Total

7

3

10

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52 ISM: A Pathway to Introductory Statistics 26. a. The proportion of the male students who were working is 11  17  0.647. b. The proportion of the female students who were working is 18  25  0.72. Female students were more likely than male students to be working. c. The proportion of students who were female AND working is 18  42  0.429. d. The student is incorrect. The result for part (b) is the proportion of female students who are working out of all the female students. The result for part (c) is the proportion of female students who were working, out of all the students, male or female. e. The proportion of the students who were female OR were working is 25  29  18  42  0.857.

28. A frequency and relative frequency table, a frequency or relative frequency bar graph, or a pie chart would be appropriate.

30. A two-way table or a multiple bar graph would be appropriate.

32. A frequency and relative frequency table, a frequency or relative frequency bar graph, or a pie chart would be appropriate.

34. The percentages do not add up to 100% because the variable, percentages of world population who have unfavorable views, is numerical, not categorical. For the same reason, it would not be appropriate to use a pie chart to describe the data.

36. In order to construct a pie chart for all the responses, it is necessary to know what the total number of responses was. To determine the percentage that each segment of the pie chart represents, it is necessary to know how many responses were collected in all.

38. You can use the information in a two-way table to construct multiple bar graphs by displaying the two categorical variables as two unique colors of bars. Bars can then be drawn to reflect the frequency of categories that are associated with each variable.

40. a. Answers may vary. b. Answers may vary. c. Answers may vary.

d. Answers may vary. e. Answers may vary. f. Answers may vary.

Homework 3.3 2. A continuous variable is a variable that can take on any value between two possible values.

4. A reasonable measure of the center of a distribution is the 50th percentile.

6. The variable, number of dogs, is discrete because there are gaps between successive possible values.

8. The speed (in miles per hour) of an airplane is continuous because it can take on any value between two possible values.

10. The weight (in ounces) of a 14-inch cheese pizza is continuous because it can take on any value between two possible values.

12. The number of string on a musical instrument is a discrete variable because there are gaps between successive possible values.

14. Of the 6 observations,

2  0.33 or about 33% were less than or equal to 2. So, the observation of 2 TVs is in 6

the 33rd percentile.

16. First, we find 75% of 8 observations: 6. Then we count the observations from left to right until we reach the 6th dot, which is 2. So, 2 hours is in the 75th percentile.

18. a. There are 8 homes with at least 1500 square feet, out of a total 30 homes, so the proportion is 8  30  0.267. b. There are 5 homes with square footages between 1500 and 2000 square feet, so the proportion is 6  30  0.2.

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18. (continued) c. The outliers are the two rightmost dots, with values approximately 2380 and 2490 square feet. d. It makes sense that the asking prices of the homes in part (c) are outliers because homes with large square footages tend to have high asking prices. e. Because the horizontal scale must be uniform, and the dots must be the same size, StatCrunch rounded the value to 1200.

20. a. The 50th percentile is the 12th smallest observation, 48, since half of the observations are less than or equal to 48 and the other half are greater than or equal to 48. b. The error is 48  52  4 years. c. It is not possible to accurately estimate a person’s age based on their appearance. Even if the students were skilled in estimating ages, each estimate would still be different. (Or, they could be trying to flatter their instructor by lowering their estimate.) d.

e. On the basis of the stemplot, 65 years appears to be the outlier because of the blank line above it. On the basis of the dotplot, 30 years appears to be the outlier because it is further from the center of the distribution than 65 years is. If there is one outlier, it would be 30 years because it is further from the center of the distribution than 65 years. f. Answers may vary.

22. a. The average spending in 2009 was approximately 668 dollars. b. Since the dot for 2013 corresponds to about 640 on the vertical axis, the average annual spending was approximately $640 in 2013. c. The change from 2012 to 2016 was approximately 663  579  84 dollars. d. The average annual spending has generally decreased from 2007 to 2016. e. Average spending on course materials could generally decrease if students purchase e-books instead of hardcover books, or if they rent books rather than purchase them.

24. a.

b. The frequency of the observation $5 is 4. This means that 4 of the 13 restaurants charge $5 for a burrito. c. The number of observations between $6 and $8 inclusive is 4. d. The proportion of the observations that are at most $6 is

6  0.462.

13

e. Of the 13 observations, 6 are less than or equal to $6. So the percent less than or equal to $6 is So $6 is at the 46th percentile.

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6  0.462.

13

54 ISM: A Pathway to Introductory Statistics 26. a.

b. The proportion of observations between 20 and 22 years, inclusive, is 11  26  0.423. c. The proportion that are at least 24 years is 4  26  0.154. d. The proportion that are no less than 21 years is 3  26  0.115. e. If there is an outlier, it would be 28 years. A student aged 28 years is much older than a typical student.

28. a. The variable is the number of rounds in the fight. It is discrete because there are gaps between successive possible values. b.

c. The value 12 is the 50th percentile since 15 (half) of the observations are less than or equal to 12 and the other 15 are greater than or equal to 12. d. Because of the round limits, it is possible that some fights ended after 12 rounds that would otherwise have gone longer. The effect of the limit is likely to be that there are fewer dots at the extreme right side. e. The least possible total number of rounds is the sum of the four smallest observations: 3  3  5  5  16 rounds. The largest possible total is the sum of the four largest observations: 15  15  15  15  60 rounds.

30. a.

b. The best-selling album sold 10.0 million copies. c. Multiply 30 minutes by 10.0 million and convert to years: 10, 000, 000  30 min. 

1 year 525, 600 min.

 570.78 years

d. The proportion of the 10 albums that have sales of at least 6.0 million albums is 5  10  0.50. e. First, we find 40% of 10 observations: 4. Then we count the observations from left to right using the stemplot until we reach the 4th observation, which is5.5. So, 5.5 million albums is in the 40th percentile.

32. a. The variable is highway gas miles (in miles per gallon). It is continuous because the miles per gallon can take on any value between any two real numbers. b.

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32. (continued) c. The frequency of the observation 27 is 5; 5 of the Dodge car models had a highway mileage of 27 miles per gallon. d. The best mileage is 41 miles per gallon. e. The 50th percentile is 27 miles per gallon since 14 of the observations are less than or equal to 27 and the other 14 are greater than or equal to 27.

34. a.

b. The observation 148 minutes has frequency 2. Two of the women had a finishing time of 148 minutes. c. The observations 151 minutes has the greatest frequency, 6. d. There are 8 observations between 160 and 169 minutes, inclusive. e. There are 19 observations less than 160 minutes.

36. a.

b. The average daily time American Adults use mobile devices has increased. c. Year Average Daily Time 2013 82 2014 117 2015 144 2016 170 2017 192

Change in Average Daily Time

 117  82  35 144  117  27 170  144  26 192  170  22

The greatest change from one year to the next in the average daily time American adults use mobile devices is 35 minutes. d. The change from 2013 to 2017 in the average daily time American adults use mobile devices is 192  82  110 minutes. If the average daily time American adults use mobile devices changes by the same amount from 2017 to 2021, by 2021, the average daily time American adults will use mobile devices in 2021 will be 192  110  302 minutes. No, we cannot assume this will be true. A new factor or event could occur that would cause the time American adults to use mobile devices to decrease.

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56 ISM: A Pathway to Introductory Statistics 38. a.

b. The percentage of American adults who smoke has decreased. c. The change in percentage of American adults who smoke from 2000 to 2010 is 19.4  23.1  3.7 percent. d. The change in percentage of American adults who smoke from 2010 to 2017 is 13.9  19.4  5.5 percent.

If the percentage changes by the same amount from 2017 to 2024, by 2024, it will be 13.9  5.5  8.4 percent in 2024. We cannot assume this will be true. A new factor or event could occur that would cause the percentage to increase.

40. a. Looking at the two dotplots, the 50th percentile of the shooting distribution is greater than that of the 50th percentile of the gymnastics distribution. This means that the typical age of an athlete participating in a shooting event tends to be greater than the typical age of an athlete participating in a gymnastics event. b. The comparison in part (a) is not surprising since most athletes participating in gymnastics in the Olympics tend to be fairly young since the sport is physically demanding. The age of an athlete participating in a shooting even may be more varied than that of an athlete participating in gymnastics since the shooting event may not put the same physical demands on a participant as gymnastics.

42. a. The 50th percentile for the “Most Possible” distribution is 5 deaths since 37 of the observations are less than or equal to 5 and the other 37 are greater than or equal to 5. Using the same reasoning, the 50th percentile for the “For Certain” is 4 deaths. b. The difference between the centers of the distributions is 5  4  1 death. We would expect it to be positive since the “most possible” deaths are logically greater than or equal to the “for certain” deaths. c. The greatest number of militant deaths possible in one strike is the observation 26; the greatest number for certain is the largest observation, 16 deaths. We cannot assume that the difference is this great for a particular drone strike because we do not know if the observations 26 (most possible) and 16 (for certain) were associated with the same drone strike.

44. The 50th percentile is a reasonable measure of the center of distribution for a numerical variable since half of the observations are less than or equal to the observation at the 50th percentile and half are greater than that observation. So the 50th percentile is the observation at the center of all the observations. For example, if 9 test scores are given: 65, 68, 70, 70, 71, 80, 82, 90, 95, the observation 71 is at the 50th percentile since half of the observations are less than or equal to 71.

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46. a. The time-series plot is incorrect because the scaling on the vertical axis is not uniform.

b. No, the data points do not lie on a straight line. When a uniform scaling is used on the vertical axis, we see that the data points follow a curved shape. c. Yes, using non-uniform scaling distorts the shape of the distribution and therefore gives a false depiction of the changes over time.

48. A stemplot would not be the best type of diagram to describe the ages of the students because the students’ ages are all likely to be very similar, with some potential outliers. The stemplot would have a large number of repeat digits.

50. Using StatCrunch, the cutoff values are: A, 91; B, 75; C, 64; D, 46. The frequency distribution of the grades is Grade A B C D F

Frequency 5 11 7 6 3

52. a.

b. The private, not-for-profit colleges and universities tend to have larger tuitions and fees than the public colleges and universities. Most of the private, not-for-profit colleges and universities have higher tuitions than the most expensive public college. c. The smallest tuition and fees for a 4-year public college or university is Haskell Indian Nations University at $80. The largest tuition and fees for a 4-year public college or university is University of PittsburghMain Campus at $16,590. The difference is $16,510.

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58 ISM: A Pathway to Introductory Statistics 52. (continued) d. The smallest tuition and fees for a 4-year private, not-for-profit college or university is Turtle Mountain Community College at $2250. The largest is Landmark College at $49,793. The difference is $47,543. e. The difference between the smallest and largest tuition and fees is nearly three times as large for private, not-for-profit colleges and universities as it is for public colleges and universities.

Homework 3.4 2. For a density histogram, the area of a bar is equal to the relative frequency of the bar’s class.

4. If the right tail of a unimodal distribution is longer than the left tail, the distribution is skewed right.

6. The dotplot associated with given histogram is (a). The shape and spread of the distribution in the dotplot is most constant spread of the distribution in (a).

8. The dotplot associated with given histogram is (d). The shape and spread of the distribution in the dotplot is most consistent with the shape and spread of the distribution in (d).

10. The histogram associated with the given variable is (b). The age ranges and relative frequencies of those ages are consistent with that of students at the start of a school year at a certain high school.

12. The histogram associated with the given variable is (a). The age ranges and relative frequencies of those ages are consistent with that of homeowners in a certain city.

14. a. The variable is the time spent working per week (in hours). It is continuous because it can take on any values between two boundary values. b. The number who work between 9 and 25 hours per week is the sum of the frequencies of the classes 10–14, 15–19, and 20–24: 1  1  2  4. c. The number who work at least 15 hours per week is the sum of the frequencies of the classes 15–19, 20–24, and 25–29: 1  2  2  5. d. The proportion who work at most 5 hours per week is the frequency of the class 0–4 divided by the total number of observations: 4  10  0.4. e. The following are all in hours: 0, 1, 2, 3, 12, 16, 20, 24, 27, and 27; Answers may vary.

16. a. The histogram is skewed right. The 50th percentile belongs to the class 0–39 deaths. Therefore, the right tail is longer than the left tail. b. The number of fatal crashes in which no less than 120 people died is found by adding the frequencies of the classes 120–159, 160–199, and 240–279: 2  1  1  4. There were 4 fatal crashes in which no less than 120 people died. c. The class that represents the greatest number of deaths is 240–279. An estimate of the number of people who died is (240  280)  2  260. d. Since the class 0–39 contains more than half of the lowest observations (it contains 9 out of 18), the 50th percentile belongs to the class 0–39 deaths. Yes, observations in this class are typical values because over half of the observations are in the class 0–39. e. We cannot assume that the frequency of deaths resulting from crashes in any one year is representative of any other year. Because so many different factors impact flights, the conditions that result in a crash and fatalities one year are not likely to be repeated in another year.

18. a. The proportion whose ages were between 15 and 19 years, inclusive, is the relative frequency, 0.25. b. The proportion whose ages were under 29 is 0.03  0.25  0.20  0.16  0.64. c. Add the proportions of the classes with lower class limit greater than or equal to 0.20  0.16  0.11  0.08  0.06  0.04  0.04  0.02  0.01  0.72. d. We cannot assume the proportion of people stopped by police who were between the ages of 20 and 24 years, inclusive, would be the same amount on other days. In particular, the proportion on weekdays might be quite different than on weekends.

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Chapter 3: Constructing Graphical and Tabular Displays of Data

59

20. a. Both of the distributions are skewed right. The salaries of players that are greater than the 50th percentile are much more spread out than the salaries of those that are less than the 50th percentile. b. Read the relative frequencies from each graph: the proportion of the Patriots who earn less than $1 million is about 0.51, and the proportion of the Rams who earn less than $1 million is about 0.75. c. The proportion of the Rams who earn at least $1 million is about 1  0.51  0.49. The proportion of the Rams who earn at least $1 million is about 1  0.75  0.25. d. The top-paid Rams player appears to have the larger salary. The largest Patriots salary is in the class 8 to 8,999,999 million dollars whereas the largest Rams salary is in the class 14 to 14,999,999 million dollars. e. The 70th percentile of the Patriots salaries is larger. The 70th percentile for the Patriots salaries is between 1 and $1,999,999 million whereas the 70th percentile for the Rams salaries is between 0 and $999,999.

22. a. From left to right, the areas of the bars are 0.05, 0.15, 0.3, 0.3, 0.15, and 0.05. b. The sum of the areas is 0.05  0.15  0.3  0.3  0.15  0.05  1. c. For any density histogram, the area of a bar is equal to the relative frequency of the bar’s class. And the sum of the relative frequencies of all the classes is equal to 1. d. Answers may vary.

24. a. The 50th percentile is in the class 2–3 days. Observations in that class are typical values since the class contains almost a quarter (23%) of the observations. b. The response time was at least 2 days for the following proportion of requests: 0.23  0.15  0.11  0.08  0.06  0.04  0.03  0.70. c. The proportion of requests where the response time was less than 4 days is 0.05  0.25  0.23  0.15  0.68. d. The proportion of requests where the response time was less than 3 days is 0.05  0.25  0.23  0.53. e. We cannot assume the distribution in 2019 will be the same as it was for the period of time represented by the distribution.

26. a. The variable is tuitions of 2-year, public colleges. It is discrete because there are gaps between successive possible values. b. The proportion between $3000 and $3999 is 0.23. c. The proportion with tuitions less than $6,000 is 0.13  0.12  0.23  0.31  0.15  0.94. d. The 50th percentile is in the class $4000–$4999. Yes, observations in this class are typical values; there are more observations in the class $4000–$4999 than any other class. e. The proportion with tuitions at least $6000 is the proportion with tuitions less than $6000 (found in part (c)) subtracted from 1: 1  0.94  0.06.

28. a. From the histogram, we see that 25% (or 0.13  0.12  0.25 ) of the observations are at most $2999 so we estimate $2996 is in the 25th percentile. b. From the histogram, we see that 0.13  0.12  0.23  0.31  0.79 or about 79% of the observations are $4999, so we estimate $4980 is in the 79th percentile.

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60 ISM: A Pathway to Introductory Statistics 28. (continued) c. The 48th percentile 0.13  0.12  0.23  0.48 is in the class $3,000–$3999. We estimate the tuition at Lake Erie College to be $3999. d. The 94th percentile 0.13  0.12  0.23  0.31  0.15  0.94 is in the class $5000–$5999. We estimate the tuition at Southeast Technical Institute to be $5999.

30. a. Class (Number of Reports) Frequency 0  0.49 2 0.5  0.99 5 1.0  1.49 3 1.5  1.99 3 2.0  2.49 0 2.5  3.99 1 Total 14

Class

Relative

(Number of Reports)

Frequency

b.

0  0.49 0.5  0.99 1.0  1.49 1.5  1.99 2.0  2.49 2.5  2.99 Total

2  0.143 14 5  0.357 14 3  0.214 14 3  0.214 14 0 0 14 1  0.071 14 14 1 14

c. The number of observations that are no less than 1 report of cheating per 100 domestic students is the sum of the frequencies for the classes 1.0–1.49, 1.5–1.99, 2.0–2.49, and 2.5–2.99: 3  3  0  1  7. d. The proportion of observations that are at least 1.5 reports of cheating per 100 domestic students is the 3 0 1 4 sum of the proportions for the classes 1.5–1.99, 2.0–2.49, and 2.5–2.99:     0.286.

14 14 14 14 e. Since University of California Davis had the greatest number of reports of cheating per 100 students, we assume that the observation 2.6 represents University of California Davis since 2.6 is the greatest number among the data given. Since there were 2.6 reports of cheating per 100 students and during the 2018 fall domestic enrollment there were 31,333 students, we estimate the number of reports of cheating at University of California Davis during the 2018–2019 academic year to be 31,133  100  2.6  809.46 or about 809 reports. b.

32. a.

c. The distribution is skewed right.

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Chapter 3: Constructing Graphical and Tabular Displays of Data

61

32. (continued) d. The proportion of shark attacks that are no more than 24 shark attacks per year is the sum of the relative frequencies for the classes 10–14.99, 15–19.99, and 20–24.99: 0.353  0  0.235  0.588. e. The proportion of the observations that are at least 30 shark attacks per year is the sum of the relative frequencies for the classes: 30–34.99 and 35–39.99: 0.118  0.059  0.177.

34. a. From the histogram of Chile, we see that 59% (0.59) of the observations are in the class 0–2.99. So the 50th percentile is in the class 0–2.99. From the histogram of Japan, we see that about 80% (0.80) of the observations are in the class 0–9.99. So the 50th percentile is in the class 0–9.99. Since in each histogram the 50th percentile is in the first class and the relative frequency for the first bar in that class exceeds 50%, we are unable to pinpoint what the observation is for the 50th percentile. So, we cannot define which is larger. b. When we compare the two histograms, the spreads of the distributions are very similar. c. Both distributions are skewed right since the right tails are longer than the left tails. We interpret this to mean that most deaths caused by individual volcanic eruptions in Japan and Chile since 1900 tended to be low; however, there were some volcanic eruptions that occurred that resulted in significant fatalities, especially in Japan. d. Identify the class containing the highest number of fatalities. That is 140–149 deaths. To estimate the number of people who died, we find the average of the lowest and highest numbers of fatalities in the class: (140  150)  2  145. We estimate the number of people who died at Mount Tokachi in 1926 to be about 145. e. On the basis of the relative frequency histograms alone, we cannot conclude that more people died from volcanic eruptions in Japan than in Chile since 1900. Without knowing how many volcanic eruptions actually occurred in each country, we cannot determine the actual numbers of fatalities.

36. a.

Obesity Rate (Percent)

Frequency

20–21.9

1

22–23.9

1

24–25.9

4

26–27.9

5

28–29.9

3

30–31.9

2

32–33.9

2

34–35.9

2

Relative Frequency 1  0.05 20 1  0.05 20 4  0.20 20 52  0.25 20 3  0.15 20 2  0.10 20 2  0.10 20 2  0.10 20 20 1 20

Total

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62 ISM: A Pathway to Introductory Statistics 36. (continued) b.

c.

d. The distribution is unimodal, skewed right. e. The student is incorrect. In this case, the differences between the groups are not large enough for the combined distribution to be bimodal.

38. a. The best approximation is the average of 100 and 200, or 150. b.

c.

d. The proportion with between 400 and 799 Facebook friends is approximately 0.17  0.17  0.34. e. The proportion who have at most 799 Facebook friends is approximately 0.17  0.27  0.17  0.17  0.78.

40. a.

b.

c. The distribution is skewed right. d. The proportion with prices between $300 and $399, inclusive, is 0.1. e. The class $100–$199.99 has the largest relative frequency, 0.30. 30% of the selected smartphones are priced in this range, more than in any of the other ranges.

42. The sum of the heights of the bars in a relative frequency histogram equals 1 because the whole is the sum of the parts.

44. If a frequency histogram has equal class widths of 1, then the frequency of each class is the number of observations of the same value. This is the same information that is given in a dotplot.

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Chapter 3: Constructing Graphical and Tabular Displays of Data 63 46. A histogram with large class widths will have fewer bars than one with smaller widths. If the distribution is bimodal, the two separate peaks will disappear if the observations they represent are merged into the same class.

50.

48.

52. Stemplots and histograms are similar in that they both present observations within classes. If the leaf of each observation in a stemplot is the ones digit, the width of each class of a similar histogram would be the difference in the stems (tens digit). Since the width defines each class, by subtracting one stem from another, which is the succession of one class to another, it is possible to determine the width. If the leaf of each observation in a split stem is the ones digit, the width of each class of a similar histogram is the difference between successive, unique stems.

54. The distribution of the age of a car (in use) would be skewed right. The reason is most people drive cars that are fairly new or moderately new, but only a few drive cars that are very old.

56. The distribution of a year on a penny in circulation is skewed left. The reason is most pennies in circulation are older. Newer pennies take time to circulate, so there are fewer.

58. The distribution of the price (in dollars) of a gallon of gasoline at the gas station is symmetric. The reason is the price of different grades of fuel is usually priced accordingly. So the lowest grade is the cheapest, the highest grade the most expensive, with the middle grade priced in between.

60. A relative frequency bar graph or a pie chart would be appropriate for a single categorical variable.

62. Because the data vary over the years from 2000 to 2019, a time series chart is appropriate.

64. There are two categorical variables involved–gender and favorite type of music–so a two-way table or a multiple bar graph would be appropriate.

66. a. Answers may vary. b. Answers may vary. c. Answers may vary.

d. Answers may vary. e. Answers may vary.

68. a.

Royal Caribbean, Carnival, and Princess own 0.1456  0.1392  0.1076  0.3924 of the cruise ships. They 3  0.15 of the companies. Since the proportion of cruise ships owned is more than twice the are 20 proportion of all the companies, these three companies are likely three of the largest cruise ship companies.

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64 ISM: A Pathway to Introductory Statistics 68. (continued) b.

The distribution of the lengths is unimodal and skewed left. c.

The two outliers are Royal Caribbean’s Oasis, with a crew size of 2100, and Carnival’s Conquest, with a crew size of 1910. d. The distribution of the crew sizes is unimodal and skewed left, which is the same as the distribution of the lengths. This makes sense because larger ships will tend to have larger crew sizes. e. The 50th percentile of crew size is between 800 and 1000 crew members. The 50th percentile of ship length is between 800 and 900 feet. A cruise ship will need about 1 crew member per foot.

Homework 3.5 2. True. If the vertical axis of a time-series plot does not start at 0, the changes in the variable described by that axis are being emphasized.

4. Graphs in three dimensions can be misleading.

6. a. The histogram with class widths of 3 percent is bimodal and skewed left. b. The histogram with class widths of 4 percent is unimodal. c. The histogram with class widths of 3 percent suggests the difference between the Western states and the rest of the country, with one mound centered around the class 64–67 and another centered around the class 70–73.

49  52  50.5%.

2 e. The histogram with class widths of 3 percent makes it easy to estimate the number of states where the percentage of adults who exercise is between 70% and 73%; the frequency of that class is 11 states.

d. For the smallest class, an estimate of the single value is

8. a. Boston’s mayor’s office would use the bar graph with the vertical axis starting at 11%. Because the scaling on the vertical axis increases by a smaller amount than in the other graph, this bar graph emphasizes the differences in percentages and seems to show that the percentage of commuters who walk or bike to work in Boston is greater than that of the other cities (especially San Francisco, Seattle, and Portland). b. San Francisco’s mayor’s office would use the bar graph with the vertical axis starting at 0%. Because the scaling on the vertical axis increases by a larger amount than in the other graph, this bar graph deemphasizes the differences in percentages and seems to show San Francisco’s percentage of walking or biking commuters to be similar to Boston’s.

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Chapter 3: Constructing Graphical and Tabular Displays of Data 65 8. (continued) c. The bar graph with the vertical axis starting at 11% gives a better estimate of the percentage of San Francisco’s walking or biking commuters. The graph is more “zoomed in” on the vertical axis. The estimate percentage is 13.9%. d. We estimate the number of commuters in Seattle who walk or bike to work to be approximately 0.129  425.8  54.928 thousand. e. Boston has about 16.7% of commuters walking or biking for a total of 0.167  368,800  61,589.60 or 61,590 (61.590 thousand) walking or biking commuters. San Francisco has 13.9% of commuters walking or biking for a total of 0.139  511, 400  71, 084.6 or about 71,085 (71.085 thousand) walking or biking commuters. San Francisco’s mayor’s office would want to show the results that it has 71,085 (71.085 thousand) walking or biking commuters while Boston has 61,590 (61.590 thousand).

10. a. If the college wants to deemphasize how much its tuition has increased, it would use the time-series plot with vertical axis starting at $13.2 thousand. The differences between the tuitions are deemphasized because the scaling increases by a larger amount than in the other time-series plot. b. A better estimate of the tuition in 2016 comes from the time-series plot with the vertical axis starting at $13.2 thousand. It is easier to estimate the tuition because the scaling on the vertical axis increases by a smaller amount than in the other time-series plot. The 2016 tuition is approximately $13.850 thousand. c. The tuition increased the most from 2016 to 2017, about $0.35 thousand. d. The change in tuition from 2013 to 2017 was about $14, 200  $13,350  $850 or about $0.85 thousand. If the same change occurred from 2017 to 2021, the 2021 tuition would be $15.05 thousand. We are not at all sure that this would be the case. Many factors could influence the cost of tuition so we cannot assume the changes in tuition over time will be similar.

12. a. To de-emphasize how much McDonald’s revenue has decreased, it should display the time-series plot with vertical axis starting at $0 billion. The differences in revenue are de-emphasized because the scaling on the vertical axis increases by a larger amount than in the other time-series plot. b. To emphasize how McDonald’s revenue has decreased, Burger King should display the time-series plot with vertical axis starting at $21 billion. The differences in revenue are emphasized because the scaling on the vertical axis increases by a smaller amount than in the other time-series plot. c. The bar graph with the vertical axis starting at $21 billion makes it easier to estimate the revenue in 2014.

The graph is more “zoomed in” on the vertical axis. The estimated revenue in 2014 is $28.2 billion. d. The change in revenue from 2015 to 2016 was about $25.1  $26.0  $0.9 billion. e. The change in revenue from 2014 to 2018 was about $21.6  $28.2  $6.6 billion. If the revenue were to change by the same amount from 2018 to 2022, the revenue in 2022 would be $21.6  $6.6  $15 billion.

We are not at all sure that this would be the case. Many factors could influence the revenue over time.

14. a. The bar graph is misleading because the scaling on the horizontal axis is not uniform. b.

This graph is not misleading because the scaling on the horizontal axis is uniform. c. If someone does not look carefully, the bar graph makes it seem like the annual revenue is increasing by greater and greater amounts. The fact that the horizontal axis is not uniform gives the impression that the revenue is increasing by greater and greater amounts. d. The revenue in 2013 is about $1.5 billion.

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66 ISM: A Pathway to Introductory Statistics 14. (continued) e. The annual revenue from 2007 to 2017 increased about $2.48  $0.4  $2.08 billion. If it increased by the same amount from 2017 to 2027, the annual revenue in 2027 would be about $2.48  $2.08  $4.56 billion. No, we should not have much faith; we cannot assume the increase in revenue from 2017 to 2027 will be the same as from 2007 to 2017. Many factors could influence the revenue over time.

16. a. The confusing thing about the bar graph is that it is difficult to tell how the tops of the bars line up with the scaling on the vertical axis. b. A (two-dimensional) bar graph would be more straightforward.

c. The wine consumption per person for the United States is 861  328  2.63 gallons per person. The wine consumption per person for France is 713  65  10.97 gallons per person. France has the larger consumption per person. d. Italy consumes 597  59  10.12 gallons per person; Germany, 534  82  6.51 gallons per person; China, 473  1410  0.34 gallons per person. Of these five countries, France has the largest wine consumption per person; China has the least. e. It is possible for some other country to have the largest wine consumption per person because the ratio is not dependent on population. For example, a country consisting of ten people, each of whom consumes 12 gallons of wine in a year, will have a larger consumption per person than France. Since it would still have to consume fewer than 473 million gallons (China’s consumption), its population would have to be less than 473  12  39 million people.

18. If the numbers on the vertical axis do not increase by the same amount and are not equally spaced, the visual information given by the bar graph or time-series plot will not reflect the true differences in values of the variable measured on that axis.

20. If the categories for a bar graph are various years and the years do not increase by the same amount, the graph will not give an accurate picture of how much the observations change from year to year. A time-series plot is a better choice.

22. Answers may vary.

Chapter 3 Review Exercises 1. The distance a student drives to school is numerical. There is a possible value between two different possible values.

2. A person’s favorite flavor of ice cream is a categorical variable. The variable assigns labels to different flavors.

3. a. The proportion who say they are late every day is 0.05. b. The proportion who say they are NOT late every day is 1  0.05  0.95. c. The proportion who say they are late less than once a month OR never is 0.21  0.47  0.68. d. The proportion who say they are late at least once a week is 0.05  0.11  0.04  0.20. e. Response bias is probably present because some surveyed adults might underestimate how often they are late to work because they are in denial. Also, if the survey is not anonymous, some survey adults might try to impress the data collector. Without response bias, the “Never” category would probably be smaller.

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Chapter 3: Constructing Graphical and Tabular Displays of Data 67 4. a.

Category

Frequency

Alternative

2

Country

1

Dance

1

Hip-hop

10

Pop

5

R&B/Soul

1

Total

20

Relative Frequency 1 10

1 20

1 20 1 2 1

4 1 20

 0.10  0.05  0.05  0.50  0.25  0.05

20 20

1

c.

b.

d. The proportion that are NOT Pop is 1  0.25  0.75. e. The proportion that are hip-hop OR alternative is 0.5  0.1  0.60.

5. a. The proportion who chose Wendy’s is 0.07. b. The proportion who said the restaurant with best fries was NOT Wendy’s is 1  0.07  0.93. c. The proportion who chose Arby’s OR McDonald’s is 0.07  0.34  0.41. d. Since the “Other” category has 24% of the votes, there must be at least 24%  2%  12 other restaurants, making at least 12  8  20 restaurants.

6. a. The proportion who think the government should take no action is 9489  28, 468  0.333. b. The proportion who are Independents is 10, 240  28, 468  0.360. c. The proportion of the Democrats who think the government should reduce the income difference is 5895  10,313  0.572. d. The proportion of the Republicans who think the government should reduce the income difference is 2344  7513  0.312. The proportion of Republicans who think the government should reduce the income difference is less than the proportion of Democrats who feel that way. e. The proportion of those who think the government should reduce the income difference who are Republicans is 2344  13, 257  0.177. f. The student is incorrect because in part (d) we are treating all Republicans as a whole and finding a certain fraction of that whole, but in part (e) we are treating the surveyed adults who think the government should reduce the income difference as the whole, and finding a certain fraction of that whole.

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68 ISM: A Pathway to Introductory Statistics 7. a. The proportion who are NOT Republicans is 10,313  10, 240  402  28, 468  0.736. b. The proportion who are Independents AND think the government should not take action is 3212  28, 468  0.113. c. The proportion who are Independents OR think the government should not take action is 10, 240  9489  3212  28, 468  0.580. d. The proportion of all Americans who think the government should reduce the income difference may not equal 47% due to sampling bias and sampling error. e. It would be stratified sampling.

8. The length is a continuous variable.

9. The number of musicians is a discrete variable.

10. a. The distribution is symmetric. b. The 50th percentile is in the class 4–4.99 surface-wave magnitude. The size of a typical earthquake is between 4 and 4.99 surface-wave magnitudes. c. The number with surface-wave magnitudes that were at least 3 and less than 5 is approximately 800  1190  1990 earthquakes. d. The number with surface-wave magnitudes greater than or equal to 4 is approximately 1200  390  50  1640 earthquakes; round down to 1600 earthquakes. e. The proportion of U.S. earthquakes occurring in California in 1985 is approximately 862  2550  0.338 or 34%.

11. a. Reading from the density histogram, 0.15 or 15% of the islands have between 50 and 59 species, inclusive. b. The percentage of the islands that have between 30 and 49 species, inclusive, is 0.08  0.13  100%  21%. c. The labeled relative frequencies add up to 0.98, so the unlabeled bar has relative frequency 0.02. The proportion with at least 60 species is thus 0.17  0.06  0.02  0.25. d. 0.12  0.19  0.08  0.08  0.13  0.60 of the islands have no more than 49 species.

12. a. Since there are two peaks, the distribution is bimodal. b. The 50th percentile is in the class 40–49 species. c. Adding the relative frequencies for classes containing at most 69 species gives 0.92, so Malaita Island is at the 92nd percentile. d. Since the proportion of islands having at most 19 species is 0.12  0.19  0.31, Samarai Island has 19 species.

13. a.

b. If there were one outlier, it would be 60 miles. This means that a government employee commutes 60 miles to work. c. The proportion of the commute times between 10 and 20 miles inclusive is d. The proportion of commute times that are at most 10 miles is

15  0.6.

25

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5  0.2.

25

Chapter 3: Constructing Graphical and Tabular Displays of Data 69 14. a.

4  0.4.

10 c. The 30th percentile is $15 million; the earnings $15 million is greater than or equal to approximately 30% of the top-10 annual earnings of DJs in 2018. d. Since half of the values are less than or equal to $22 million, and the other half is greater than $22 million, $22 million is the 50th percentile. e. $46 million is at the 90th percentile. The earnings $46 million is greater than or equal to approximately 90% of the top-10 annual earnings of DJs in 2018.

b. The proportion of the observations that are no more than $20 million is

15. a.

b. Annual per-person wine consumption has generally increased. From the graph we see increases from 1995-2015 at which point the consumption stays the same. c. The change in annual per-person wine consumption from 1995 to 2000 is 2.0  1.8  0.2 gallons per person. This means that annual per-person wine consumption increased by 0.2 gallons from 1995 to 2000. d. The change in annual per-person wine consumption from 2010 to 2017 is 2.9  2.6  0.3. If annual perperson wine consumption changes by the same amount from 2017 to 2024, the per-person wine consumption in 2024 will be 2.9  0.3  3.2 gallons. No, we cannot assume the change in annual perperson wine consumption from 2017 to 2024 will be the same as from 2010 to 2017.

16. a. Class (numbers of species) Frequency Relative Frequency 13  0.481 0  49 13 27 7  0.259 50  99 7 27 5  0.185 100  149 5 27 1  0.037 150  199 1 27 0 0 200  250 0 27 0 0 250  300 0 27 1  0.037 300  350 1 27 27  0.999 Total 27 27

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70 ISM: A Pathway to Introductory Statistics 16. (continued) b.

c.

d. The distribution is skewed right. e. The 50th percentile is in the class 50–99 threatened species. f. The outlier would be 310 threatened species.

17. a. The salesperson would refer to the bar graph with the vertical axis starting at 25 thousand cars; the differences between the numbers of stolen cars are emphasized because the scaling on the vertical axis increases by a smaller amount than in the other bar graph. b. The salesperson would refer to the bar graph with the vertical axis starting at 0 thousand cars; the differences between the numbers of stolen cars are de-emphasized because the scaling on the vertical axis increases by a larger amount than in the other bar graph. c. The better estimate comes from the bar graph with the vertical axis starting at 25 thousand cars; it is easier to make the estimation because the scaling on the vertical axis increases by a smaller amount than in the other bar graph. We estimate the number of Honda Accords stolen to be 44 thousand. d. We estimate that 44  35  9 thousand more Honda Accords were stolen than full-size Ford Pickups.

18. a. The graph is misleading because the scaling on the horizontal axis is not uniform. b. This graph is not misleading because the scaling on the horizontal axis is uniform.

c. The bar graph makes it seem like the enrollment is increasing by greater and greater amounts because the scaling on the horizontal axis is not uniform. d. We estimate the charter-school enrollment in 2013 to be about 2.3 million students. e. The increase from 2007 to 2017 is approximately 3.2  1.3  1.9 million students. If it increases by the same amount from 2017 to 2027, there will be approximately 3.2  1.9  5.1 million students enrolled. No, we should not have much faith; we cannot assume the change in the charter-school enrollment from 2017 to 2027 will be the same as from 2007 to 2017.

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Chapter 3: Constructing Graphical and Tabular Displays of Data 71 19. a. The graph is confusing because it is difficult to line up the tops of the boxes with the scaling on the vertical axis. b. A relative frequency bar graph would give a straightforward description.

c. The proportion who were not married couples is 1  0.55  0.45. d. The proportion who were single women or single men is 0.19  0.11  0.3.

Chapter 3 Test 1. The cost of a car is a numerical variable because it describes a measurable quantity.

2. a.

b.

Category

Frequency

Democrat

6

Independent

2

Libertarian

1

None

2

Republican

4

Total

15

Relative Frequency 6 15 2 15 1 15 2 15 4 15

 0.4

 0.133  0.067  0.133  0.267

15 15

1

c.

d. The proportion of the observations who are NOT Republican is 1  0.267  0.733. e. The proportion who are Independent OR Democrat is approximately 0.4  0.133  0.533.

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72 ISM: A Pathway to Introductory Statistics 3. a. The proportion of surveyed Democrats who think the wall should be built is 34  347  0.098. b. The proportion of surveyed Republicans who think the wall should be built is 244  287  0.850. c. The proportion of those surveyed who are Republican and think the wall should be built is 244  1062  0.230. d. The results in parts (b) and (c) won’t be equal because part (c) takes into account all those who think the wall should be built, including Democrats and Independents, whereas part (b) only takes Republicans into account. e. The proportion of those surveyed who are Republican or think the wall should be built is (287  34  195)  1062  0.486.

4. The high temperature is a continuous variable because it can take on any value between two possible values.

5. a. The shapes of the distributions are skewed right. b. The city-gas-mileage distribution is slightly narrower than the highway gas mileage distribution. This means 2019 cars get slightly better gas mileage on the highway versus the city. c. For city gas mileages, the class that contains the 50th percentile is 20-24 miles per gallon. For highway gas mileages, the class that contains the 50th percentile is 25-29 miles per gallon. d. We estimate the difference of the 50th percentile for highway gas mileages and the 50th percentile for city gas mileages to be: (25  30)  2  27.5 (20  25)  2  22.5 5.0 miles per gallon.

This means that the highway gas mileage is about 5 miles per gallon greater than the city gas mileage for a typical car. e. The distribution is unimodal because the city-gas-mileage distribution and the highway-gas-mileage distribution are both fairly wide and they overlap. f. Even though the distribution of city and highway gas mileages together is unimodal, the city-gas-mileage distribution is different than the highway-gas-mileage distribution.

6. a. The shape of the distribution is skewed right. b. Yes, $8018.80 is an outlier. A fare of $8018.80 is much larger than at least 99% of the other fares.

Response bias likely occurred. A respondent likely reported an exaggerated fare. c. The proportion of the fares between $10 and $29.99, inclusive, is approximately 0.32  0.08  0.40. d. The proportion of fares that were at most $29.99 is 0.53  0.32  0.08  0.93. e. Since the proportion of the fares that were at least $30 is approximately 1  0.93  0.07, the number of fares is approximately 0.07 1, 048,574  73, 400 fares.

7. a. We estimate the percentile of a $19.99 fare to be the 85th percentile (0.53  0.32  0.85). b. We estimate the fare at the 93rd (0.53  0.32  0.08  0.93) percentile to be approximately $29.99. c. Estimating each class with upper limit under $30 by its middle value gives the total of fares less than $30 as 0.53 1, 048,574  $5  0.32 1, 048,574  $15  0.08 1, 048,574  $25  $9,909, 024.30.

8. a.

b. The distribution is skewed right. c. The 50th percentile is 64 years. d. The proportion of justices who are at least 80 years in age is 2  9  0.222. e. Since 7 of the 9 justices’ ages are less than or equal to 70, this observation is at the 78th percentile.

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Chapter 3: Constructing Graphical and Tabular Displays of Data 73 9. a. Class (billions of dollars) Frequency Relative Frequency 25 0  24 25  0.781 32 3 25  49 3  0.094 32 2 50  74 2  0.063 32 0 75  99 0 0 32 0 100  124 0 0 32 1 125  149 1  0.031 32 1 150  174 1  0.031 32 32 Total 32 1 32 b.

c.

d. If there were four outliers, they would be $50 billion, $70 billion, $125 billion, and $160 billion.

10. a. If Sony Group wants to de-emphasize the decline of its number of employees, the time-series plot that it should display is the first, where the vertical axis starts at 0 thousand employees. The large increases in the scaling of the vertical axis in the first graph de-emphasizes the changes in the number of employees. b. The time-series plot that best helps to estimate the number of employees in 2014 is the second, where the vertical axis starts at 110 thousand employees. It is easier to make the estimation because the scaling on the vertical axis increases by a smaller amount than in the other time-series plot. We estimate the number of employees to be 141 thousand. c. The number of employees decreased the most from 2017 to 2018. The difference is about 129, 000  117, 000  12, 000 employees. d. The change in the number of employees from 2013 to 2018 is 117, 000  146, 000  29, 000 employees. If the number of employees changed by the same amount from 2018 to 2023, the number of employees to 2023 would be about 117, 000  29, 000  88, 000 employees. No, we cannot expect that the change in the number of employees from 2018 to 2023 will be the same as the change in the number of employees from 2013 to 2018. Because so many different factors impact employment (e.g., the economy), employee numbers can fluctuate quite a bit within several years.

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Chapter 4: Summarizing Data Numerically 75

Chapter 4: Summarizing Data Numerically Homework 4.1 2. The mean is a measure of the center.

4. False. If a distribution is skewed, the median measures the center better than the mean.

6. The symbol M stands for the population median.

8. Mean: x for  ; Median: M̂ for M ; Proportion: p̂ for p 10. Arrange the observations in ascending order:13, 25, 29, 29, 36, 42, 47, 48, 48, 50, 51. The median is the middle value, 42.

12. Arrange the observations in ascending order:0.3, 0.6, 0.8, 0.9, 1.4, 2.3, 3.5, 9.6, 10.4, 12.4. The median is 1.4  2.3  1.85.

2

14. a. xi  x1  x2  x3  x4  x5  2  6  5  2  3  18 b. xi2  x12  x22  x32  x42  x52  22  62  52  22  32  78 16. a. x1  105.8, x2  45.6, x3  27.8, x4  20.6, x5  19.8, x6  15.8 b. xi  105.8  45.6  27.8  20.6  19.8  15.8  235.4; The total cost of the six costliest U.S. hurricanes is $235.5 billion. c. xi2  105.82  45.62  27.82  20.62  19.82  15.82  15,111.88 18. a. x1  103.2, x2  102.2, x3  70.6, x4  68.4, x5  67.2 b. xi  103.2  102.2  70.6  68.4  67.2  411.6; The total compensation of these five CEOs in 2018 was $411.6 million. c. xi2  103.22  102.22  70.62  68.42  67.22  35, 273.84 20. x 

xi 4  9  10  12  13  14  15   11 7 n

22.  

xi 97  114  118  119  149  180  238  254  315  429   $201.3 million 10 n

24. x 

xi 10  17  21  22  23  24  26  27  31   22.3; Mˆ  23 9 n

xi 1715  1821  2675  2675  3000  3000  3750   2662.3 milliampere hours; Mˆ  2675 7 n milliampere hours

26. x 

28. x 

xi 68  68  68  68  68  73  73  73  73  93  93  93 73  73   $75.9; Mˆ   $73.0 12 n 2

30. x 

xi 10  11  12  15  16  20  24   15.43 hours 7 n

32. Arrange the observations in ascending order: 0, 1, 5, 40, 50, 90, 300. The median is 32 texts.

34. x 

xi 11  16  24   17 hours 3 n

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76 ISM: A Pathway to Introductory Statistics xi 7.25  7.25  7.25  7.25  7.25  8.25  8.55  8.60   $7.71 per hour 8 n 7.25  7.25  $7.25 per hour.

The median is 2 b. The total income would be 12  . Since  is the total of the minimum wages divided by 12, 12  is the total.

36. a.  

P working in each state. For example, if all 12 P work in the state with minimum wage $7.25, their total income would be less than either P  or PM .

c. Neither would be the total income, unless there are exactly

38. a.

The distribution is the distribution is skewed right. b. Since the distribution is the distribution is skewed right, the median should be used. c. M  40.0 texts d.

StatCrunch

TI-84

Random Number

Value

Random Number

Value

13 32 29 14 22 1 9 19 31 5 28

10 40 50 200 40 35 50 200 10 20 40

21 19 3 26 18 24 25 7 5 22 28

10 200 100 40 3 100 100 4 20 40 40

Mean

63.18

Mean

59.73

e. The number of sample medians that were less than the population median would be approximately equal to the number of sample medians that were greater than the population mean; the chances of a sample median being less than the population median is equal to the chances of a sample median being greater than the population median because the samples are randomly selected.

40. a. The median household income of the 3.54 million randomly selected U.S. households is $60,336 (Mˆ  60,336). b. The median household income of all U.S. households, M , is unknown.

42. a. The mean purchase of lottery tickets by the 1000 randomly selected American adults is $17 weekly ( x  17). b. The mean purchase of lottery tickets by all American adults,  , is unknown.

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Chapter 4: Summarizing Data Numerically 77 44. The mean is less than the median, because the distribution is skewed left.

46. The mean is greater than the median, because the outlier affects the mean but does not affect the median.

48. (b) because the mean is greater than the median.

50. (a) since the mean and median are equal, the distribution must be symmetric and centered at 75.

52. a. The distribution is skewed right. The mean is greater than the median; the median is a better measure of the center because of the skewed-right distribution. b. The median is in the class 200–249 pounds; 47% of the observations are less than 199 pounds and 18% are greater than or equal to 250 pounds. c. For some skewed distributions, the mean and median are not in the class with the largest density. d. The proportion of observations less than 250 pounds is 0.04  0.43  0.35  0.82, so 249 pounds is at about the 82nd percentile. e. The proportion of observations less than 300 pounds is 0.04  0.43  0.35  0.13  0.95, so the 95th percentile is about 299 pounds.

54. a. x  163.1 minutes b. Mˆ  172.0 minutes c.

The distribution is skewed left. d. The mean is not a typical finish time. The median is a typical finish time, because the distribution is skewed left. e. The difference in the median finish times for men and women is 167.0  141.5  25.5 minutes; a typical woman’s finish time is 25.5 minutes greater than a typical man’s finish time.

56. a. The distribution is skewed right. b. Because of the right skew, the mean is larger than the median. The median is a better measure of the center; for skewed distributions, it is a better measure of the center than the mean. c. The median distance of the flat-stage distribution is in the class 200–250 kilometers.

The median distance of the mountain-stage distribution is in the class 200–250 kilometers. d. This comparison is not true. Median distance of the flat-stage distribution is in the same class as the median distance of the mountain-stage distribution, so the medians are relatively close to each other. e. The distribution of the flat-stage distances has a greater skew to the right than the distribution of the mountain-stage distance. The left half of each distribution are very similar.

58. a. The distribution is unimodal and symmetric.

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78 ISM: A Pathway to Introductory Statistics 58. (continued) b. The distribution is bimodal and skewed to the right.

c. Because of the skew of the taxi fares, the median should be used. The median taxi fare is $18.54 and the median Uber fare is $23.62. d. the typical taxi average fare is less than the typical Uber average fare. e. The result in part (d) takes all the data into account.

60. a. Values in the third column: 7, –4, 18, 17, 3, –5, 1, 10, 11, –2. b. Values in the sixth column: –1, 12, –1, –3, 3, –5, 5, –2, –11, –1, –3. c. The mean difference for the calcium group is 5. On average, the systolic blood pressure of the treatment group decreased by 5 mmHg. d. The mean difference for the placebo group is about –0.6. On average, the systolic blood pressure of the control group increased very slightly, by 0.6 mmHg. e. Yes, calcium tended to lower the systolic blood pressure. We can conclude causality because this is an experiment.

62. a. The study is an experiment. Participants were randomly assigned to the treatment and control groups. b. The explanatory variable is the calcium supplement. The response variable is systolic blood pressure. c. Random selection was used to determine which persons were in the treatment group and which persons were in the control group; the researchers could construct a frame of the 21 participants and randomly select 10 of the participants to be in the treatment group. The other 11 participants would be in the control group. d. The sample is the 21 African American men. The population is all African American men. e. There is causality; we can determine whether there is causality when conducting experiments.

64. a. It is not true for the first week of August 2011. In fact, there are no observations that are exactly one hour (60 minutes) apart. b. It is probably better to view the distribution as unimodal with outliers. The outlier would be consistent with a small number of short eruptions that have a relatively short time before the next eruption. c. The observations could be divided into the relatively long times (80 minutes or more), and the relatively short times (65 to 69 minutes). The median for the long times would be in the class 95–99 minutes, and the median for the short times in the class 65–69 minutes. d. The number of eruptions is the sum of all frequencies, approximately 107 eruptions.

66. x 

xi 2  2  3  3  6  6  6  8  8  9 66   53; The median is  6 and the mode is 6. n 10 2

68. The mode is fastball because it occurs in the list 11 times, more often than any other observation.

70. a. x 

xi 7  8  8  8  8  8  9  9  10  12   8.7 nominations n 10

88  8 nominations.

2 c. The mode is 8 nominations because it occurs in the list 6 times, more often than any other observation. d. The largest observation is 12 nominations.

b. The median is

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Chapter 4: Summarizing Data Numerically 79 70. (continued) e. The mean is larger than the median; this is consistent with the finding in (d) because the mean is sensitive to outliers, and the median is resistant to outliers.

72. a. The median salary of respondents who earned a bachelor’s degree is in the class $30,000–$39,999. b. The median salary of respondents who earned a master’s degree is in the class $40,000–$49,999. c. The median salary of respondents who earned a degree beyond a master’s is in the class $60,000–$69,999. d. It makes sense that the medians should increase from part (a) to part (c) because we expect that more education would lead to a higher salary. e. Respondents may not have revealed their salaries because they were embarrassed that they were small, or because their salaries were very large and they didn’t want the University to approach them for money.

Either way, there would be nonresponse bias. Assuming that the proportions of degrees for the nonrespondents is roughly the same as those for the respondents, the results to parts (a), (b), and (c) should not change very much, since the median is resistant to outliers. f. Three problems with the conclusion are: the median for respondents who earned a bachelor’s degree is not in the class $20,000–$29,999; the conclusion may not apply to all alumni since there was a large nonresponse rate to the survey; the conclusion may not even apply to the respondents because a relatively large number of respondents did not reveal their salary.

74. Answers may vary.

76. If each person loses 10 pounds, the new mean weight is 157  10  147 pounds.

78. The median salary will stay the same. 50% of the salaries will still be less than $60 thousand and the other 50% will be at least $60 thousand.

80. a. x  574.8 friends; Mˆ  475 friends. b.

The distribution is skewed right. We would expect this since the mean is greater than the median. c. The 10% trimmed mean is 503.1 friends. d. The 10% trimmed mean is slightly larger than the median in part (a). e. The 10% trimmed mean is more resistant to outliers than the mean, because the calculation ignores the outliers.

33 33  3.  3. iii. The median is 2 2 ii. The median is 3. iv. The median is 3. b. The student is incorrect. The median measures the center of the data, not the number of observations.

82. a. i. The median is

84. The mean height of 10 randomly selected professional NBA basketball players would tend to be larger. The mean measures the center and is unaffected by the number of observations.

86. The student is incorrect. 50% of observations are below the median and 50% of observations are above the median.

88. Answers will vary.

90. Answers may vary.

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80 ISM: A Pathway to Introductory Statistics 92. a.

Almost half the volcanic eruptions occurred in Indonesia, Japan, and the Philippines. b.

c. The death distribution is skewed right. d. The median is a better measure of the center. The median measures the center better than the mean when a distribution is skewed. e. x  246.0 deaths; M  1 death. The mean is much larger than the median because the mean is sensitive to outliers, and the median is resistant to outliers. f. In 50% of the volcanic eruptions, the number of deaths was less than 1. In the other 50% of eruptions, the number of deaths was at least 5.

Homework 4.2 2. If a distribution is unimodal and symmetric, then 68% of the observations lie within one standard deviation of the mean.

4. The variance is the square of the standard deviation.

6. The symbol  2 stands for the population variance.

8. Standard deviation: s for  ; Median: M̂ for M ; Range: R̂ for R; Proportion: p̂ for p; Mean: x for  ; Variance: s 2 for  2 10. Rˆ  94  25  69 12. R  429  97  332 14. s 

 ( xi  x ) 2  n 1

20. Rˆ  8  1  7; s 

64 4 4

16. s 

 ( xi  x ) 2  n 1

2922  24.2 5

18. s 

 ( xi  x ) 2  n 1

24.8943  2.0 years 6

 ( xi  x ) 2  ( xi  x ) 2 52 52   2.9; s 2    8.7 n 1 6 n 1 6

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Chapter 4: Summarizing Data Numerically 81

22. Rˆ  3750  1715  2035 milliampere hours; s  s2 

24. s 2 

 ( xi  x ) 2 3,116, 454   667.2 milliampere hours; n 1 7

 ( xi  x ) 2 3,116, 454   445, 208 milliampere hours2 n 1 7  ( xi  x ) 2 2.8571428571   0.48 hours 2 n 1 6

26. R  21  19  2 years 28. s 2 

 ( xi  x ) 2 0   0 years 2 n 1 3

30. Graph (d). Almost all the data must be within three standard deviations of the mean; in this case, the data must be between approximately 2.8 and 77.2.

32. Graph (a). The median of this distribution is about 70, and the range is approximately 80  40  40.

34. a. The areas of the bars, from left to right, are 0.10, 0.30, 0.20, 0.30, 0.10. b. The total area of the five bars is 0.10  0.30  0.20  0.30  0.10  1. c. The area of a bar is equal to the relative frequency of the bar’s class, so the total area of the bars is equal to the total of the relative frequencies, which is always equal to 1. d. Answers may vary.

36. a. x  b. s 

xi 7500  8040  8114  8260  8876  9220  9366  10, 470  10, 696  12, 040   $9258.2 n 10  ( xi  x ) 2 183,14,171.6   $1426.5 n 1 9

c. The 10 private colleges have the higher typical tuition; the mean tuition for the sample of private colleges is greater than the mean tuition for the sample of public colleges. d. The 10 private colleges have tuitions with more spread; the standard deviation of tuitions for the sample of private colleges is greater than the standard deviation of tuitions for the sample of public colleges. e. The student cannot draw such a conclusion because the spread of all 4-year, private colleges might be different than the randomly selected 4-year, private colleges due to sampling error. Also, the spread of all 4-year, public colleges might be different than the randomly selected 4-year, public colleges due to sampling error.

38. a.

b. The DAL distribution has more spread. c. The MCO standard deviation is 9.8 percent; the DAL standard deviation is 19.4 percent.

The DAL distribution has more spread. d. The conclusions are the same.

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82 ISM: A Pathway to Introductory Statistics 40. a. The histogram for the sample shows a left skewed distribution.

b. The histogram for the sample shows a left skewed distribution.

c. Women: The mean is 60.22 minutes; the median is 60.00 minutes.

Men: The mean is 57.94 minutes; the median is 58.60 minutes.

Men have better finish times. d. Women: The range is 32.20 minutes; the standard deviation is 8.67 minutes.

Men: The range is 24.70 minutes; the standard deviation is 7.38 minutes.

The distributions of men’s time have less spread. e. A woman had the best finish time. Extreme values are more likely in the distribution of women’s times since they have more spread.

42. a.

Both distributions are skewed to the right. b. The range should be used for both distributions because both distributions are skewed rather than symmetric. c. Pfizer range: 4.6  1.1  3.5 AstraZeneca range: 5.6  0.6  5 d. AstraZeneca has the larger range of sales. This means that AstraZeneca’s sales tend to be less predictable. e. Answers may vary; 5.6 billion dollars 44. a. The student would probably get higher test scores with Professor B; all of Professor B’s scores will probably be no less than 71%, but Professor A’s scores could be as low as 50%. b. Yes; the scores on Professor A’s tests might have larger spread due to very high scores, not low scores, so all the students in Professor A’s class might have scored higher than the bottom student in Professor B’s class. Another possibility is that the students in Professor B’s class might have been weaker students due to what they learned in previous math courses, and Professor A might actually be the more effective professor.

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Chapter 4: Summarizing Data Numerically 83 46. a. The standard deviations for both were 5 minutes on the commute to home; almost all the data are within 14  4.7 minutes. about 14 minutes from the center, so the standard deviation must be close to 3 b. The standard deviations for both were 1 minute on the commute to work; separating the two mounds at 27 minutes results in two distributions, each of which has almost all the data within 3 minutes of the center. c. The distribution for the commute to work is bimodal because the spreads of the Route A times and the Route B times are relatively small and, hence, do not intermix much. The distribution for the commute home is unimodal because the spreads of the Route A times and the Route B times are relatively large and, hence, their times are intermixed quite a bit. d. The standard deviations should also be compared to help determine how much the times intermix.

48. a. The variance of the highway gas mileages of the fifty randomly selected 2019 cars is 35.9 (mpg) 2 ( s 2  35.9).

b. The variance of the highway gas mileages, in (mpg) 2 , of all 2019 cars,  2 , is unknown.

50. a. The proportion of the 3000 American adults that drink a cup of coffee every day is 64%  pˆ  0.64 . b. The proportion of all American adults that drink a cup of coffee every day, p, is unknown.

52. a. Since 51.0  3(9.4)  22.8 and 51.0  3(9.4)  79.2, 99.7% of the death-row inmates have ages between 22.8 years and 79.2 years. b. Since 51.0  1(9.4)  41.6 and 51.0  1(9.4)  60.4 68% the death-row inmates have ages between 41.6 years and 60.4 years. c. 95% of the observations lie between 51.0  2(9.4)  32.2 years and 51.0  2(9.4)  69.8 years, so the interval is (32.2, 69.8).

54. a. The Empirical Rule can be applied because the distribution is unimodal and approximately symmetric. b. Approximately 95% of the song lengths are between 253.1  2(74.6) and 253.1  2(74.6), or in the interval

103.9, 402.3. c. No. The interval found in part (b) only describes the lengths of alternative rock songs. d. The mean and standard deviation of songs played on Alt 105.3 are both smaller than the respective mean and standard deviation found in part (b). This suggests that a typical song played on Alt 105.3 is shorter (by about 20 seconds), and that the spread of song lengths is smaller. With a smaller spread, Alt 105 must not play very short or very long songs very often. e. The interval 141.5,313.9 is narrower than the interval found in part (b). It would be unusual for Alt 105.3 to play songs that are longer than 313.9 seconds or shorter than 141.5 seconds.

56. a.

The distribution is roughly symmetric, though it is a bit skewed to the left. The Empirical Rule can be applied, but use caution. b. Mean: 13.0 units; standard deviation: 2.9 units. c. Approximately 95% of the observations should lie within two standard deviations of the mean.

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84 ISM: A Pathway to Introductory Statistics 56. (continued) d. The actual percentage within two standard deviations of the mean is

40 100%  95.2%. This is very close 42

to the 95% determined in part (c).

58. a. The proportion of scores between 15 and 25 points is 0.34  0.34  0.68. b. The proportion of scores between 10 and 30 points is .14  0.34  0.34  0.14  0.96. c. Since the total of the relative frequencies must be 1, the proportion of scores between 5 and 35 points is 1. d. Using the Empirical Rule, 15 points is one standard deviation less than the mean and 25 points is one 25  15  5 points. standard deviation greater than the mean. So,   2 60. a. Because the distribution is symmetric, the median must be approximately 64 years. b. Because the distribution is symmetric, the mean must be approximately equal to the median, 64 years. c. Using the Empirical Rule, 68 percent of the distribution lies between 60 and 68 years, or within 4 years of the mean;   4 years. d.  2  42  16 years 2 e. The sum of proportions in the unlabeled bars must be 1 – 0.96 = 0.04, so each unlabeled bar must have relative frequency 0.01. Adding the proportions for the bars left of 62 gives 0.30, so the 30th percentile is approximately 62 years.

62. 51.3  2(12.0)  75.3; Yes, the observation 77 years is more than two standard deviations away from the mean.

64. 740  2(129)  482; No; the observation 485 mg/cm 2 is not more than two standard deviations away from the mean.

66. a. People following the Zone diet had the largest mean decrease in weight after 2 months, –3.8 pounds. b. People following the Ornish diet had the largest mean decrease in weight after 12 months, –3.3 pounds. c. Atkins: 2.1   3.6  1.5; Zone: 3.2   3.8  0.6; Weight Watchers: 3.0   3.5  0.5; Ornish: 3.3   3.6  0.3. For each of the four diets, the average change in weight between 2 months and 12 months was positive; a typical participant gained back some of the initial weight loss. d. It may be that participants were not following their diet as carefully after the first 2 months. Other possibilities: they may have decreased their physical activity, or the diets might have become less effective after the initial period, as the participants’ bodies adapted to the diets. e. The standard deviations of both diets are larger than the difference in mean weight gain between the two diets. It is likely that some Zone participants lost more weight than some of the Ornish participants.

68. Answers will vary.

70. The standard deviation will stay the same. If each person loses the same amount of weight, the spread will be unaffected.

72. The variance will increase. The mean will be unaffected, but the difference of these two observations from the mean will increase.

74. The standard deviation will decrease. The mean will be unaffected and the sum of squared differences from the mean will be unaffected, but that sum will be divided by a larger number.

76. Answers may vary.

78. The standard deviation of the prices of 7 randomly selected new homes would be larger; the standard deviation measures the spread and is unaffected by the number of observations.

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Chapter 4: Summarizing Data Numerically 85 80. Lowest: Student B; All the values are the same, so the standard deviation will be 0.

Middle: Student C; The scores are uniformly distributed around the middle of the distribution.

Highest: Student A; The scores are farthest from the middle of the distribution.

82. The range measures the difference between the largest and smallest observations. If the distribution is unimodal and symmetric, then by the Empirical Rule, 99.7% (almost all) of the observations are within 3 standard deviations of the mean. That means that the range is approximately 6 times the standard deviation, so s  R / 6.

84. a. The range is R  22.3%  7.7%  14.6%.

14.6  2.4%.

6 c. The error is 2.4%  2.7%  0.3 percentage points.

b. The standard deviation is approximately

86. a. Answers may vary. b. Answers may vary. c. Answers may vary. d. Answers may vary.

88. a.

e. Answers may vary. f. Answers may vary. g. Answers may vary. b.

California had the most earthquakes. c. The magnitude distribution is roughly symmetric. d. The mean magnitude is 4.2; earthquakes in the U.S. typically have a magnitude around 4.2 Ms. e. 0.886 f. About 97% of the magnitudes are between 2.4 and 6.0 Ms. Homework 4.3 2. The interquartile range measures the spread of the middle 50% of the observations.

4. False. The IQR is resistant to outliers.

6. An outlier lies outside the fences.

8. List the data in ascending order: 1, 2, 2, 2, 3, 4, 4, 4, 5, 6, 8, 8, 12 Five-number summary: 1,

22 68  2, 4,  7, 12 2 2

10. List the data in ascending order: 10.1, 37.96, 38.75, 49.64, 53.19, 56.76, 68.42, 74.07, 86.52, 93.44, 100.13, 104.03 Five-number summary: $10.1,

38.75  49.64 56.76  68.42 86.52  93.44  $44.195,  $62.59,  $89.98, 2 2 2

$104.03

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86 ISM: A Pathway to Introductory Statistics 12. a. List the data in ascending order: 1, 65, 80, 259, 471, 1332, 1564, 2130, 2720, 6675, 8677, 12,201 Five-number summary: $1,

80  259 1332  1564 2720  6675  $169.5,  $1448,  4697.5, $12,201 2 2 2

b. Range: 12, 201  50  $12,151; The largest pledge is $12,151 more than the smallest pledge. c. Yes, the range would be only $3524 if the largest observation, $12,201, were decreased to the next largest observation, $8677. d. IQR: 4697.5  169.5  $4528; The range of the middle 50% of the observations is $4528. e. No, the IQR would still be $4528 if the largest observation, $12,201, were decreased to the next largest observation, $8677.

14. The matching boxplot is (d), because the distribution is unimodal and symmetric with observations between about 37 and 43.

16. The matching boxplot is (a), because the distribution is skewed right.

18. a. The longest hike is the largest observation, approximately 49 kilometers. b. Since the middle line in the box is at approximately 11.9, approximately 50% of the hikes were less than 11.9 kilometers. c. Since the middle line in the box is at approximately 11.9, approximately 50% of the hikes were no less than 11.9 kilometers. d. 50%  50%  100%; the sum of the parts is equal to the whole. e. The number of hikes that were between 8.6 kilometers and 16.0 kilometers is 50% of the total, or 500 hikes.

20. a. The distribution is skewed right. The smallest 25% of the observations are less spread out than the largest 25% of the observations. b. The 75th percentile is the loan amount represented by the right edge of the box, approximately $22 thousand. The loan amount $22 thousand is greater than or equal to approximately 75% of the loan amounts and less than approximately 25% of the loan amounts. c. Since $8.5 thousand is approximately at the left edge of the box, it represents the 25th percentile. The loan amount $8.5 thousand is greater than or equal to approximately 25% of the loan amounts and less than approximately 75% of the loan amounts. d. The largest loan amount is approximately $40 thousand. e. The IQR is approximately $22  $8.5  $13.5 thousand. The relatively small value of the IQR suggests that the middle 50% of the data has small spread.

22. a. The cutoff scores are A: 80 points; B: 70 points; C: 50 points. b. 25% of 40, or 10 students earned a C on the test. c. Each side of the box represents 25% of the scores, so there were the same number of Cs and Bs. d. Students would prefer the cutoffs described in part (a). Students who scored between 70 and 80 points would have earned Cs with the normal cutoffs, but earned Bs with the cutoffs described in part (a).

Students who scored between 50 and 70 points would have earned Ds with the normal cutoffs, but earned Cs with the cutoffs described in part (a).

24. a. The distribution is skewed right. The median is a better measure of center; the median is better than the mean at measuring the center of a skewed distribution. b. The median is about 31 thousand kilometers; approximately half of the planets have diameter less than or equal to 31 thousand kilometers and approximately half of the planets have diameter greater than 31 thousand kilometers. c. Q1  10 thousand kilometers; 25% of the planets have diameter less than or equal to 10 thousand kilometers and approximately 75% of the planets have diameter more than 10 thousand kilometers.

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Chapter 4: Summarizing Data Numerically 87 24. (continued) d. The student is incorrect. The left whisker together with the left side of the box represent a total of 50% of the observations, but the right part of the box represents only 25% of the observations. e. Since the diameter of Earth is greater than Q1 and less than M , we can be certain of 50% of 8, or 4 planets with diameters larger than Earth’s.

26. Answers may vary.

28. a. List the data in ascending order: 5.2, 5.5, 5.5, 5.5, 5.9, 6.0, 6.0, 6.0, 7.4, 10 5.9  6.0  6.0, 6.0, 16; all in millions of albums 2 b. IQR: 6.0  5.5  0.5 million albums

Five-number summary: 5.2, 5.5,

c. Lower fence: 5.5  1.5(0.5)  4.8 million albums; Upper fence: 6.0  1.5(0.5)  6.8 million albums d. Outliers:7.4 and 10 million albums 30.

32. a. List the data in ascending order: 1175, 1338, 1824, 2620, 2764, 3000, 3540, 4455, 5060, 5313 Five-number summary: $1175, $1824,

2764  3000  $2882, $4455, $5313 2

b.

c. The distribution is slightly skewed to the right. The left side of the box and the left whisker are both a little shorter than the right side of the box and the right whisker, respectively. d. The distribution is skewed right. The median is a better measure of center; the median is better than the mean at measuring the center of a skewed distribution. e. The IQR is $4455  $1824  $2631; it is a measure of the spread of the middle 50% of tuitions.

34. a. List the data in ascending order: 2, 3, 3, 3, 3, 3, 4, 4, 4, 6, 7, 10 Five-number summary: 2,

33 3 4 46  3,  3.5,  5, 10; all in numbers of people 2 2 2

b. The right fence is 5  1.5(5  3)  8 people, so 10 people is an outlier. c.

d. The distribution is skewed right; the right side of the box is longer than the left side, and the right whisker is longer than the left whisker. e. The median is better; the median is a better measure of the center than the mean for a skewed distribution.

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88 ISM: A Pathway to Introductory Statistics 36. a.

b. The median is 77 aired episodes; the value 77 episodes is greater than or equal to approximately 50% of the number of the aired episodes of cooking shows of all time and less than approximately 50% of aired episodes of cooking shows of all time. c. The IQR is 181 aired episodes; it is a measure of the spread of the middle 50% of the aired episodes of cooking shows of all time. d. (Interpreting “top” to mean the largest values and “bottom” to mean the smallest values.) The number of aired episodes in the top 25% of the sample have the greater range; the right whisker is longer than the left whisker. e. The student is definitely incorrect. On the basis of the boxplot, at least 50% of the observations lie in the class 0–99 but less than 25% of the observations lie in the class 100–199. In fact, 34 of the observations lie in the class 0–99 episodes, and 10 observations lie in the class 100–199 episodes.

38.

40. a. The highway distribution has the larger median; a typical highway gas mileage is greater than a typical city gas mileage. b. The highway gas mileage has the larger Q3 ; the 75th percentile of the highway distribution is greater than the 75th percentile of the city distribution. c. The Hyundai Ioniq Blue has a city gas mileage of approximately 58 miles per gallon. It is surprising that it has a higher city than highway mileage, because the typical mileage is greater for the highway distribution than the city distribution. d. The standard deviation is sensitive to outliers, such as the one described in part (c). e. Answers may vary.

42. a. The boxplot would not have a right whisker if Q3 equals the maximum value; Q3  100%. The longest 25% of the matches took 100% of the maximum allowed rounds. b. The boxplot would not have a vertical line to indicate the median if the median and Q3 are equal; in this case, M  Q3  100%. The longest 50% of the matches took 100% of the maximum allowed rounds. c. The median for the last 28 is greater than the median number of rounds for the first 28 matches. It is possible that the level of competition changed, but it is more likely that Ali’s age and physical condition affected the number of rounds it took to win his matches. d. The median is approximately 50%; the median number of rounds would be 50% of 10 is 5 rounds. This is slightly less than the actual median. e. The losses should be analyzed separately because the percentages of maximum allowed rounds may not follow the same distribution as with the wins.

44. If the IQR is equal to the range, then Q3 must be the maximum value and Q1 must be the minimum. The smallest observation is 7.

46. The student is incorrect. For any sample, the IQR is less than or equal to the range.

48. The IQR is still 43; Q3  60  5  55, Q1  55  43  12, and Q3  Q1  55  12  43.

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Chapter 4: Summarizing Data Numerically 89 50. a. The range of the temperatures will increase by 3F, since the largest temperature increases by 2F and the coldest temperature decreases by 1F. b. The IQR stayed the same. Changing the values of the smallest and largest numbers would not affect the values of Q1 and Q3 .

52. The student may be incorrect. Because the IQR is resistant to outliers and the standard deviation is sensitive to outliers, the standard deviation of the salaries at company A could be smaller than the standard deviation of the salaries at company B. 54. a. The student is incorrect; answers may vary. b. The student is correct. Since the largest 25% of observations must be greater than Q3 and the smallest 25% of the observations must be less than or equal to Q1 , the range must be greater than or equal to the IQR. 56. a. s  $1.3 thousand

c. s  $288.6 thousand

b. IQR = $6 thousand d. IQR = $6 thousand e. The outlier $715 greatly increased the standard deviation but did not affect the IQR. 58. a. Scenario C: None of the speeds can be below 66 mph, so all commuters speed. b. Scenario B: None of the speeds can be above 64 mph, so none of the commuters speed. c. Scenario A: 25% of speeds are below 64 mph and 25% of speeds are above 68 mph (speeding), but no claim can be made about whether the other 50% of commuters are speeding.

Scenario D: 25% of speeds are below 62 mph and 25% of speeds are above 66 mph (speeding), but no claim can be made about whether the other 50% of commuters are speeding.

60. A dotplot, stemplot, relative frequency histogram, or boxplot would be appropriate.

62. A multiple bar graph or two-way table would be appropriate.

64. a.

b. 19.7  2.7  30.6  53.0% of the National League players are pitchers. c. The median base salaries, all in millions of dollars, are: starting pitchers, 0.57; relief pitchers, 0.55; relief pitchers/closers, 2.8; catchers, 0.56; first base, 0.85; second base, 0.83; third base, 0.84; shortstop, 0.57; left field, 0.56; center field, 0.58; right field, 0.61. Relief pitchers/closers have the highest median salary. d. For third basemen, the median salary is $0.84 million and the range $22.5 million; for catchers, the median is $0.56 million and the range $19.46 million. A player would tend to earn more as a third baseman, because the median is greater. e. The player would tend to earn a larger salary as a third baseman. At a minimum, the player could expect the median salary of $0.84 million, or as much as the maximum, $23 million. The median for catchers is $0.56 million and the maximum is $20 million.

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90 ISM: A Pathway to Introductory Statistics Chapter 4 Review Exercises

1. a. x1  460.4, x2  450.0, x3  421.0, x4  420.0, x5  417.5 b. xi  460.4  450.0  421.0  420.0  417.5  2168.9; The total dead lift of the top five dead lifts is 2168.9 kilograms. c. xi2  460.42  450.02  421.02  420.02  417.52  942, 415.41 2. Mean: x  3. a.

xi 25  25  40  56  57  62  80  93 56  57  56.5; Mode: 25   54.8; Median: Mˆ  n 8 2

xi 105  114  125  129  130  130  136  136  139  147  151  153   132.9 million gallons 12 n

130  136  133.0 million gallons 2 c. 130 million gallons and 136 million gallons

b.

d.

e. Yes; the mean is very close to the median, which is what we would expect for a distribution that is symmetric.

4. a. The distribution is unimodal and skewed right. b. The mean is larger than the median. Since the median is a better measure of the center than the mean for a skewed distribution, the median is the better measure of center. c. Since the smallest 50% of the observations belong to the class 0–9.99 billion dollars, that class contains the median. d. Since the range is a better measure of the spread than the standard deviation for a skewed distribution, the range is the better measure of the spread. e. The largest possible range is 79  0  79 billion dollars.

5. a. The proportion of observations less than $10 billion is 0.94, so Infor is at the 89th percentile. b. Since the proportion of observations less than $15 billion is .94  0.04  0.98, Stripe’s value is approximately $20 billion. c. Since the smallest 94% of the observations belong to the class 0–10 billion dollars, the largest possible value of Q3 is $9.99 billion. d. Since the smallest possible value for Q3 is 1, the largest possible value of the IQR is $9.99  $1  $8.99 billion. e. The largest possible value of the right fence is 10  1.5  9  $23.5 billion. Since these companies have a value greater than $23.5 billion, they are outliers.

6. The mean will stay the same. The sum of the numbers will be changed by 2  2  0, so the mean is unaffected.

7. The median will stay the same. Increasing the largest number and decreasing the smallest will not change the observations in the middle.

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Chapter 4: Summarizing Data Numerically 91 8. Mean:

xi 34  42  43  45  48  55  66   47.6; Median: 45; Range: 66  34  32; 7 n

Standard variation: s 

 ( xi  x ) 2  n 1

 ( xi  x ) 2 637.7143 637.7143  10.3; Variance: s 2    106.3 6 n 1 6

9. a. For the 20 channels, the standard deviation of the number of subscribers per channel is 2.748 million subscribers; s  2.748. b. The standard deviation of the number (in millions) of subscribers per channel for the top 5000 channels;  ; unknown 10.

 ( xi  x ) 2  n 1

20.83  2.0 units 5

11.

 ( xi  x ) 2 22   7.3 novels 2 n 1 3

12. 5  2  3 hours per day 13. (d), because 95% of the observations should be between 70  2(12.4)  45.2 and 70  2(12.4)  94.8.

14. (a) because 95% of the observations should be between 70  2(2.5)  65 and 70  2(2.5)  75.

15. (b) because the range should be approximately 42  34  8.

16. (c) because the range should be approximately 50  10  40.

17. a. The Empirical Rule can be applied because the distribution is unimodal and symmetric. b. Approximately 68% of the professors should have ages within one standard deviation of the mean. c. Since 27.3 years is two standard deviations below the mean and 75.3 is two standard deviations above, about 95% of the professors should have ages between 27.3 years and 75.3 years. d. Adding the relative frequencies gives 0.98; 98% of the professors have ages between 30 and 75 years. e. All of the observations between 30 and 75 years lie in the class 27.3–75.3 years, so the estimate of 95% for that class is too low.

18. a. 21 million sperm per milliliter b. 24 million sperm per milliliter; the range of the middle 50% of the observations is 24 million sperm per milliliter. c. 25% of 17 patients is approximately 4 patients. d. 50 million sperm per milliliter; yes; an increase of 50 million sperm per milliliter would raise any patient suffering from low sperm count into the normal range. e. Answers may vary. There must be a control group (and random assignment) for there to be chance to show causation.

19. a. 33.7, 36.3, 47.55, 76.3, and 105.9, all in millions of viewers b.

c. The distribution is skewed right; the upper 50% of the observations have more spread than the lower 50%. d. The median is the better measure of the center than the mean for a skewed distribution. e. The observation 47.55 million viewers is greater than or equal to approximately 50% of the observations and less than approximately 50% of the observations. f. The student is incorrect. Each whisker represents 25% of the observations.

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92 ISM: A Pathway to Introductory Statistics 20. a.

b. pop c. The median for pop is 911.5 million streams; the median for hip-hop/rap is 795 million streams. The median for the pop distribution is greater than the median for the hip-hop/rap distribution. The typical number of streams for a pop song is greater than the typical number of streams for a hip-hop/rap song. d. The IQR for pop is 261 million streams; the IQR for hip-hop/rap is 181.5 million streams. The IQR for the pop distribution is greater than the IQR for the hip-hop/rap distribution; the middle 50% of the observations of the pop distribution has more spread than the middle 50% of the observations of the hiphop/rap distribution. e. For pop, Q3 is 1060 million streams; for hip-hop/rap, Q3 is 954.5 million streams. Q3 for the pop distribution is greater than Q3 for the hip-hop/rap distribution; the 75th percentile for the pop distribution is greater than the 75th percentile for the hip-hop/rap distribution.

Chapter 4 Test 1. The mean is less than the median because the distribution is skewed left.

2. a. The distribution is skewed right. b. The mean is larger than the median; the median is the better measure. The median is a better measure of the center than the mean for a skewed distribution. c. The median is in the class 45–49 years. d. The typical age of a female tenured or tenure-eligible (TTE) professor is less than the typical age of a male TTE professor. e. A female professor who is 65 years in age is at the 96th percentile.

3. The new person’s income must be $40,000.

4. Mean: x 

xi 16  27  41  48  55  62  80  80  92   55.7; Median: 55; Mode: 80; 9 n

Range: 92  16  76; Standard variation: s  Variance: s 2 

 ( xi  x ) 2 5214   25.5; n 1 8

 ( xi  x ) 2 5214   651.7 n 1 8

5. a. Write the values in ascending order:17, 17, 20, 24, 30, 41, 41, 47, 49, 71. The median is years. xi 17  17  20  24  30  41  41  47  49  71   35.7 years 10 n c. Range: 71  17  54 years

b. x 

d. Standard variation: s  e. Variance: s 2 

 ( xi  x ) 2  n 1

2722.1  17.4 years 9

 ( xi  x ) 2 2722.1   302.5 years 2 n 1 9

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30  40  35 2

Chapter 4: Summarizing Data Numerically 93 6. a. Approximately 99.7% of the scores were within three standard deviations of the mean. b. Approximately 68% of the scores were between 68 points and 83 points (within one standard deviation of the mean). c. Since 59  75  2(8) and 91  75  2(8), approximately 95% of 40, or 38, students scored between 59 and 91 points. d. The center of the first-test scores is greater than the center of the second-test scores. The spread of the first-test scores is less than the spread of the second-test scores.

7. a. The distribution is skewed right; the upper 50% of the observations have more spread than the lower 50% of the observations. b. The 50th percentile is about 15 deaths per 1000 births. c. The largest infant mortality rate is approximately 88 deaths per 1000 births. Yes, it is an outlier; the boxplot shows that it is larger than the right fence. d. The rate 32.7 deaths per 1000 births is approximately at the 75th percentile. e. The estimated number of infant deaths is

6.3  3,853,472  24,277 deaths.

1000

8. a. The variable is numerical, because it describes something that can be measured (or counted). b. 0,

17  19 34  35 48  48  18,  34.5,  48, 86, all in numbers of tornados 2 2 2

c.

The distribution is skewed right; the right side of the box is longer than the left side, and the right whisker is longer than the left whisker. d. The typical number of tornados per state in 2018 for states in the Midwest and the south is 35. e. The third quartile is 48 tornados. f. The IQR is 48  18  30 tornados; it is a measure of the spread of the middle 50% of the observations.

9. a. Additional research should be performed to determine why the observations are negative. If the six pitches all hit the ground before reaching the batter, then each negative observation should be replaced with 0 feet.

If the negative observations were due to camera error, the observations should be removed. If the negative observations were due to data entry, the values should be corrected, or if this is not possible, they should be removed. If the observations are removed, then this should be stated in the report. b. The 75th percentile is 3 feet; 75% of the pitches were at a height of 3 feet or less, and 25% were at a height of more than 3 feet. c. Some of the outliers have the same value or are so close in value that they appear to be one dot. d. The smaller median is in the right-hand-hitter distribution; a typical pitch height to a right-hand hitter is less than a typical pitch height to a left-hand hitter. e. The right-hand-hitter distribution has the smaller IQR; the middle 50% of pitch heights to right-hand hitters has less spread than the middle 50% of pitch heights to left-hand hitters.

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94 ISM: A Pathway to Introductory Statistics 10. a.

b. Wetsuit: The mean is 1.507 meters per second.

Swimsuit: The mean is 1.429 meters per second.

Yes, the mean maximum speed for wetsuits appears to be at the center of the wetsuit distribution, and the mean maximum speed for swimsuits appears to be at the center of the swimsuit distribution. c. The mean maximum speed for wetsuits is greater than the mean maximum speed for swimsuits. d. The values for the fifth column are: 0.08, 0. 10, 0.07, 0.08, 0. 10, 0.11, 0.05, 0.05, 0.06, 0.08, 0.05, 0.10. e. The mean is 0.078 meter per second f. The difference in the mean wetsuit maximum speed and the mean swimsuit maximum speed is 1.507  1.429  0.078 meter per second. The result is equal to the result in part (e). The difference of a typical wetsuit maximum speed and a typical swimsuit maximum speed should equal a typical difference of a wetsuit maximum speed and a swimsuit maximum speed.

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Chapter 5: Computing Probabilities 95

Chapter 5: Computing Probabilities Homework 5.1 2. The probability of an impossible event is 0.

4. We can interpret the area of a density histogram as the probability of randomly selecting an observation that lies in the bar’s class.

6. a. A 4 occurred in the second roll, since the relative frequency of the outcome 4 is 0.5 for that roll and 0 for the first roll. b. After 5 rolls, the relative frequency of the outcome 4 is 0.2. No, this is not a good estimate of the probability of rolling a 4; the estimate 0.2 is not that close to the probability 0.167. c. After 10,005 rolls of the die, the relative frequency of the outcome 4 is 0.1676. Yes, this is a good 1 estimate; the estimate 0.1676 is very close to the probability  0.167.

6 8. P a cook  

3 1  12 4

10. P  President Abraham Lincoln  

0 0 12

1 12. Since there is 1 correct answer out of 5 possibilities, P correct   .

5

14. P  26 

1 38

16. P green  

2 1  38 19

18. Since there is no slot numbered 999, P 999  0.

20. He would most likely lose money overall. The probability of rolling 1 or 2 is less than the probability of rolling 3 or 4 or 5. (If the number 6 is rolled, the player neither wins nor loses.)

1 22. Since the area of the region for 2 is one eighth of the total, P  2  .

8 1 24. Since the area of the region for 4 is one fourth of the total, P  4  .

4

26. P an even number   P  2  P  4  28. P at least 3  P 3  P  4 

3 8

30. P at most 3  P 1  P  2  P 3 

3 4

32. The student made a mistake. The event is “1”, not “1/2”.

3 8

34. The student has made a mistake. The sum of the probabilities of all grades should equal 1.

36. The values –0.37 and 1.29 cannot be probabilities because they are not between 0 and 1, inclusive.

38. P Thursday  

5 1  30 6

40. P  weekday  

22 11  30 15

42. P in the fifth week  

5 1  30 6

44. a. P 3 times a week   15.9%  0.159 b. P  At most twice a week   0.174  0.282  0.246  0.702 c. P  More than twice a week   0.159  0.065  0.074  0.298 d. The student cannot draw such a conclusion. The data reflect the habits of adults who are at least 22 years of age; we cannot assume that they also reflect the behavior of teenagers.

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96 ISM: A Pathway to Introductory Statistics 46. P  X  5 

1 6

52. P  2  X  4 

3 1  6 2

48. P  X  4 

3 1  6 2

54. P  2  X  5 

2 1  6 3

50. P  X  5 

4 2  6 3

56. a. P  X  20  0.19  0.14  0.12  0.45 b. P  X  15  0.33  0.10  0.43 c. P 10  X  25  0.10  0.12  0.19  0.41 d. The student cannot draw such a conclusion because these data are based on a convenience sample. The sample may not be representative of all statistics students in the United States. e. The percentage who worked 25 hours or more per week is 14%  12%  26%.

58. a. The proportion of MLB stadiums with average ticket prices between 0 and 30 dollars is 0.03  0.17  0.27  0.47. b. The probability that a randomly selected MLB stadium has average ticket price between $0 and $30 is 0.73. c. P  no more than $25  0.03  0.17  0.20

d. P at least $35  0.07  0.03  0.07  0.17

e. P  between $20 and $40  0.17  0.27  0.20  0.13  0.77 60. a. Since the mean minus three standard deviations is 54  3(6)  36 and the mean plus three standard deviations is 54  3(6)  72, the probability that a veteran selected at random is between 36 and 72 years old is approximately 99.7% of 1, or 0.997. b. Since the mean minus one standard deviation is 54  1(6)  48 and the mean plus one standard deviation is 54  1(6)  60, the probability that a veteran selected at random is between 48 and 60 years old is approximately 68% of 1, or 0.68. c. Since the mean minus two standard deviations is 54  2(6)  42 and the mean plus two standard deviations is 54  2(6)  66, the probability that a veteran selected at random is between 42 and 66 years old is approximately 95% of 1, or 0.95.

62. a. Since the mean is 34.2 and the mean plus one standard deviation is 34.2  1(8.4)  42.6, the probability that a temperature reading selected at random is between 34.2 F and 42.6 F is approximately 34% of 1, or 0.34. b. Since the mean minus three standard deviations is 34.2  3(8.4)  9.0, the probability that a temperature reading selected at random is less than 9.0 F is approximately 0.15% of 1, or 0.0015. c. Since the mean plus two standard deviations is 34.2  2(8.4)  51.0, the probability that a temperature reading selected at random is greater than 51.0 F is approximately 2.5% of 1, or 0.025.

64. a. Since the mean plus one standard deviation is 63.8  1(4.2)  68.0, the probability that a height selected at random is greater than 68.0 inches is approximately 17% of 1, or 0.17. b. Since the mean minus two standard deviation is 63.8  2(4.2)  55.4 and the mean is 63.8, the probability that a height selected at random is between 55.4 inches and 63.8 inches is approximately 47.5% of 1, or 0.475. c. Since the mean plus three standard deviations is 63.8  3(4.2)  76.4, the probability that a height selected at random is less than 76.4 inches is approximately 0.15% of 1, or 0.0015.

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Chapter 5: Computing Probabilities 97

66. a. Since the mean minus two standard deviation is 8022  2(3427)  1168 and the mean is 8022, the probability that a tuition selected at random is between $1168 and $8022 is approximately 47.5% of 1, or 0.475. b. Since the mean minus one standard deviation is 8022  1(3427)  4595, the probability that a tuition selected at random is less than $4595 is approximately 16% of 1, or 0.16. c. Since the mean minus two standard deviation is 8022  2(3427)  1168, the probability that a tuition selected at random is greater than $1168 is approximately 97.5% of 1, or 0.975.

68. a. Since the mean plus two standard deviations is 194.7  2(52.1)  298.9, the probability that a query letter selected at random has a word count of less than 298.9 words is approximately 47.5% of 1, or 0.475. b. Since the mean plus one standard deviations is 194.7  1(52.1)  246.8 and the mean is 194.7, the probability that a query letter selected at random has a word count between 194.7 and 246.8 words is approximately 34% of 1, or 0.34. c. Since the mean minus two standard deviations is 194.7  2(52.1)  90.5, the probability that a query letter selected at random has a word count of no more than 90.5 words is approximately 2.5% of 1, or 0.025. d. Since the mean plus three standard deviations is 194.7  3(52.1)  351.0, the agent would not have read 0.0015(375)  0.5625, or about one letter. e. Since the mean minus two standard deviations is 194.7  2(52.1)  90.5, 100 words in not an unusual word count.

70. The probability that a randomly selected Harvard Business School MBA student in the class of 2016 is a minority is 0.26.

72. The proportion of American adults who use a tablet is 0.66.

74. The student is incorrect because the 50 states do not have equal numbers of residents.

76. We would expect that a 12% of all American households have a net worth (including real estate) of at least $1 million.

78. The student cannot draw such a conclusion. As more and more blackjack games are played, the proportion of wins would approach the true probability of winning, but the proportion from only 10 games is likely not a good estimate.

80. a. They did not run the experiment enough times to be able to tell whether the student is psychic. b. If the surgery is successful, that does not mean that exactly 80% of the first 20 surgeries will be successful. c. It would it be possible for the surgery to be successful 8000 times, but it is unlikely. We would expect the relative frequency of successful surgeries to be fairly close to or equal to 0.80.

82. The student must have made a mistake because a probability cannot be less than 0.

84. The researcher would need to determine the number of 60-year-old Americans who smoked last year and the number of those smokers who died last year. The researcher would then need to divide the number of those smokers who died by the number of 60-year-old Americans who smoked last year.

86. Answers may vary.

88. a. Answers may vary. b. Answers may vary. c. It would be close to the probability, 1 6.

90. All results should be close to 1 6. a. Answers may vary. b. Answers may vary. c. Answers may vary. d. Answers may vary.

Homework 5.2

2. P  NOT E   1  P  E 

6. P  NOT R   1  P  R   1  0.2  0.8

4. P  E OR F   P  E   P  F   P  E AND F 

8. P  M OR R   P  M   P  R   0.7  0.2  0.9

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98 ISM: A Pathway to Introductory Statistics 10. P C OR D   P C   P  D   P C AND D   0.5  0.4  0.2  0.7

12. P a backup singer  

3 5  8 8

3 8

16. P a stringed instrument player   P a backup singer   18. P  NOT black   1 

14. P  NOT a backup singer   1 

18 20 10   38 38 19

3 3 3   8 8 4

20. P  red OR black  

22. Since this event contains the outcomes 2, 4, and 6, P  black AND at most 7  

18 18 36 18    38 38 38 19

3 .

38

24. This event contains the outcomes 1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36 and also the 18 5 23   . outcomes 28, 29, 31, 33, and 35; P  red OR at least 28  38 38 38 26. Since there are 3 odd numbers as outcomes, P  NOT odd   1  28. P odd AND at most 3  P 1 OR 3 

2 1  6 3

30. P odd OR at most 3  P odd   P 1, 2, OR 3  P 1 OR 3  32. P less than 2 AND greater than 4  0 34. P  4 

1 4

36. P  NOT 4  1 

3 3 1   .

6 6 2

3 3 2 2    6 6 6 3

38. P  NOT odd   P  2  P  4 

1 1 3   8 4 8

40. P even OR at least 3  P  2  P  4  P 3 

1 3  4 4

1 1 1 7    8 4 2 8

42. a. P (Personal weapon)  0.046 b. P (NOT a personal weapon)  1  0.046  0.942 c. P (Personal weapon OR knife or other cutting instrument)  0.105  0.046  0.151 d. The second statement is correct. The number of murders is not the same as the number of weapons used in murders.

44. a. P ( S )  0.21 b. P (NOT S )  1  0.21  0.79 c. P ( S OR O)  0.21  0.45  0.66 d. Even if the survey was carried out well, the proportion for the population might be a bit different than the proportion for the sample due to sampling error.

46. a. P (between 30 and 69)  0.35  0.28  0.15  0.05  0.83 b. P (between 20 and 29 OR between 50 and 59)  0.14  0.15  0.29 c. P (younger than 30 OR older than 69)  P (NOT between 30 and 69)  1  0.29  0.71 d. The information provided relates to federal prisons, not state prisons, so the conclusion is not valid. e. 180,184(0.14  0.35)  180184(0.49)  88, 290 Copyright © 2021 Pearson Education, Inc.

Chapter 5: Computing Probabilities 99 48. a. A delay of –40 minutes means the plane departed 40 minutes early. b. P (between  25 and  1)  0.54 c. P (between 25 and 74)  0.07  0.03  0.10 d. Answers will vary.

50. P  NOT in the third week   1  52. P Tuesday OR Thursday  

7 23  30 30

7 4 1   30 30 30 10 1   30 3

56. P second week OR Sunday  

4 5 9 3    30 30 30 10

54. P  third week AND Saturday  

1 30

58. a. P (NOT lower class)  1  P (lower class)  1  b. P (lower class OR middle class) 

670  0.728 2464

670 1287   0.794 2464 2464

c. P (completely satisfied AND not completely satisfied)  0 d. P (completely satisfied AND middle class)  e. P (completely satisfied OR upper class)  60. a. P 18  29 OR 50  64 

399  0.162 2464

751 507 218    0.422 2464 2464 2464

4242 10, 484   0.424 34, 695 34, 695

b. P  Agreed OR disagreed   1, since “agree” and “disagree” are the only two options. c. P  Agreed AND disagreed   0, because the respondent cannot both “agree” and “disagree.” d. P  Disagreed   P 50  64  P  Disagreed AND 50  64  e. P  Disagreed AND 50  64  62. a. P  NOT C   1 

18,384 10, 484 5557    0.672 34, 695 34, 695 34, 695

5557  0.160 34, 695

461  0.426 803

b. P  N AND W   0 e. P  N OR C   P  N   P C   P  N AND C  

c. P  N   P W  

152 155   0.382 803 803

d. P  N AND C  

101  0.126 803

152 461 101    0.638 803 803 803

64. No; A and C may share an outcome that is not contained in B. 66. P ( E )  P (NOT E )  P ( E )  1  P ( E )   1 68. Answers may vary.

70. a. No; the number of outcomes in E OR F is at least the number of outcomes in E. b. Yes; all the outcomes in F could be contained in E.

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100 ISM: A Pathway to Introductory Statistics 72. P  E OR F OR G   P  E   P  F   P G    P  E AND F 

  P  F AND G    P  E AND G 

 2  P  E AND F AND G 

Homework 5.3

2. Two events E and F are independent if P  E | F   P  E  .

4. If E and F are disjoint events, then P  E AND F   0.

6. plays guitar | plays an instrument   8. P  4 | no less than 3 

number of guitarists 2 1   number who play an instrument 4 2

1 , since one outcome is 4, but there are four outcomes that are no more than 3.

4

10. Since the outcome “at most 5” has five possibilities, 1, 2, 3, 4, and 5, and three of those are odd, 3 P (odd | at most 5)  .

5 3 3 1 12. No, because P (odd | at most 5)  , but P (odd)   .

5 6 2

14. Since 18 outcomes are red, and one of them is 5, P 5 | red  

1 .

18

16. Since 18 outcomes are black, and ten of them are even, P even | black  

10 5  .

18 9

18. Since 18 outcomes are red and two of them are less than 5, P less than 5 | red  

2 1  .

18 9

20. Yes. The events red and black are disjoint because a slot cannot be both red and black; there are no outcomes that belong to both events.

1 22. Consider just the sectors that contain1 and 2. Since the areas of both sectors are equal, P 1 | at most 2  .

2

24. Consider just the sectors that contain 2 and 4. Since the area containing 4 is twice the area of the sector 2 containing 2, P  4 | even   .

3 1 26. There are 5 Thursdays, one of which is in the second week, so P (second week | Thursday)  .

5

28. No, they are not independent. For example, P (Wednesday) 

5 1 1  , but P (Wednesday | fifth week)  .

30 6 5

30. P  D | C   0.784

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Chapter 5: Computing Probabilities 101 32. a. P 18  29 | agreed  

2333  0.143 16,311

b. P 30  49 | agreed  

5793  0.355 16,311

2333 5793  0.550, but P agree | 30  49   0.490. Surveyed adults ages 4242 11, 823 30–49 years are less likely to agree with the statement than surveyed adults ages 18–29 years. d. The student is incorrect. The student has limited the space to adults ages 50–64 years instead of the space limited to adults who are not completely satisfied by their jobs.

c. No. P agree | 18  29 

247 243 b. P (fairly happy | man)   0.494  0.518 500 469 c. The result from part (b) is slightly larger; a surveyed adult who was a man was more likely to be fairly happy than a surveyed adult who was a woman. d. The student is incorrect. On the basis of the results from parts (a) and (b), we conclude that married men were more likely to be fairly happy than married women.

34. a. P (fairly happy | woman) 

134  0.200; The probability that a person randomly selected from the survey is completely 670 satisfied with their job, given the person is a member of the lower class, is 0.200.

36. a. P (C | L) 

218  0.430; The probability that a person randomly selected from the survey is completely 507 satisfied with their job, given the person is a member of the upper class, is 0.430. c. The result found in part (b) is larger; for adults in the survey, adults who are members of the upper class are more likely to completely satisfied with their jobs than adults who are members of the lower class.

b. P (C | U ) 

218  0.290; The probability that a person randomly selected from the survey is upper class, 751 given the person completely satisfied with their job, is 0.290.

d. P (U | C ) 

e. The student is incorrect. For P (C | L), the space is limited to adults who are members of the lower class, but for P (U | C ), the space is limited to adults who are completely satisfied by their jobs.

428  0.563; The probability that a person randomly selected from the survey wore a seat belt, 760 given the person had undergone a dental checkup within two years, is 0.563.

38. a. P ( S | D) 

90  0.457; The probability that a person randomly selected from the survey wore a seat belt, 197 given the person had undergone a dental checkup more than two years ago, is 0.457. c. The result from part (a) is slightly larger; a surveyed adult who had a dental checkup within the past two years was more likely to use a seat belt than a surveyed adult who did not have a dental checkup within the past two years. d. The student is incorrect. Causality cannot be concluded from an observational study. The extent to which people take care of themselves is a likely lurking variable. Age and income may also be lurking variables. e. Systematic sampling was used to select the households. f. Response bias, nonresponse bias, and sampling bias likely occurred. Response bias likely occurred because the behaviors were self-reported; nonresponse bias and sampling bias likely occurred because of those who refused to take part, and because the sampling method favored people who own a car.

b. P ( S | M ) 

40. Dependent; the probability of earning more than $100,000 per year is most likely different from the probability of earning more than $100,000 per year given that the person has a bachelor’s degree.

42. Independent; wearing contact lenses likely does not change a person’s preference for chocolate ice cream.

44. P  H AND Y   P  H   P Y   (0.9)  (0.5)  0.45

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102 ISM: A Pathway to Introductory Statistics 3

1 1 .

46. Since each roll is independent, P 5 AND 5 AND 5  P 5  P 5  P 5     6 216

48. Because the selection is random, we can assume the events are independent, the probability that all four have experienced street harassment is (0.57)  (0.57)  (0.57)  (0.57)  0.10556.

50. Since we can reasonably assume the guesses are independent, the probability of correctly guessing all six 6

1 1 answers is    . 2 64

52. a. The probability that all three say McDonald’s makes the best French fries is 0.34  0.03930.

3

b. The probability that at least one says McDonald’s makes the best French fries is 1 minus the probability that none do: 1  0.66  0.71250.

3

54. a. The probability that all four have hearing loss is 0.203  0.00170.

4

b. The probability that at least one has hearing loss is 1 minus the probability that none do: 1  0.797   0.59651.

4

56. a. P (agree) 

16,311  0.470 34, 695

b. P (agree | 18  29) 

2333  0.550 4242

c. No, the events are not independent; the probabilities found in parts (a) and (b) are not equal.

915 738  0.618 b. P  L | D    0.780 1481 946 c. No, the events are not independent; the results found in parts (a) and (b) are not equal. d. The student is incorrect. Causation cannot be concluded because this is an observational study. e. The events are not disjoint. It is possible that the person who thinks upper-income people pay too little in taxes could also be a Democrat.

58. a. P  L  

6

1 1 60. a. P THTHTH   P T   P  H   P T   P  H   P T   P  H      2 64 6

1 1 b. P all tails   P T   P T   P T   P T   P T   P T      2 64

c. The probabilities found in parts (a) and (b) are equal. d. The student is incorrect because in both parts (a) and (b), exactly one of two outcomes is specified for each coin flip.

5 1 1 1 b. P 3 | first four rolls were 3s   P 3  62. a. P (all 3s)     6 6 7776 c. The result found in part (a) is much smaller than the result found in part (b). d. The student is incorrect. In part (a), the probability describes getting 3 on all five rolls, but in part (b), the probability describes getting 3 on only the fifth roll.

64. Since E and F have no outcomes in common, P  E | F   0.

66. The student might be incorrect. For P  E | F  , the sample space is narrowed to F, but for P  E AND F  , the sample space may contain a larger number of outcomes.

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Chapter 5: Computing Probabilities 103 68. Assuming the die is fair (each of the six outcomes is equally likely), the probability of guessing correctly is 1 still .

6 70. Answers may vary.

72. Answers may vary.

Homework 5.4

2. For the probability distribution of a discrete random variable, 0  P ( xi )  1.

4. For the probability distribution of a discrete random variable, the standard deviation can be computed using the formula    ( xi   ) 2 P ( xi ) 6. continuous random variable

10. neither

8. discrete random variable

12. discrete random variable

14. No, one of the probability values of X is negative.

16. Yes, the probabilities are all between 0 and 1, inclusive, and the sum of all the probabilities is 0.3  0.1  0.2  0.4  0  1.

18.   xi P ( xi )  30(0.04)  35(0.23)  40(0.45)  45(0.21)  50(0.07)  40.2 20. a. Let X be the number of days per week a student randomly selected from the survey exercised. b. The probabilities are all between 0 and 1, inclusive, and the sum of all the probabilities is 0.157  0.075  0.097  0.111  0.097  0.132  0.066  0.265  1. c.

bimodal and skewed left

d.   xi P ( xi )  0(0.157)  1(0.075)  2(0.097)  3(0.111)  4(0.097)  5(0.132)  6(0.066)  7(0.265)  3.9 e. P ( X  7)  1  P (7)  1  0.265  0.735 22. a.

skewed left

b.   xi P ( xi )  1(0.009)  2(0.039)  3(0.065)  4(0.06)  5(0.039)  6(0.086)  7(0.138)  8(0.134)  9(0.25)  10(0.073)  11(0.091)  12(0.009)  13(0.009)  7.5

c. The median is 8 since P ( X  7)  0.436 and P ( X  8)  0.570. d. The median is slightly larger than the mean. This makes sense since the distribution is skewed left. e. P ( X  6)  P (1)  P (2)  P (3)  P(4)  P(5)  P (6)  0.009  0.039  0.065  0.06  0.039  0.086  0.298

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104 ISM: A Pathway to Introductory Statistics 24. a. P ( X  1)  1  P ( X  0.5)  1  0.184  0.132  0.684 b.   xi P ( xi )  0(0.184)  0.5(0.132)  1(0.107)  2(0.147)  3(0.129)  4(0.092)  5(0.209)  2.3 hours c. The actual mean would be greater. The actual distribution is probably even more skewed right, and the mean is sensitive to extreme values. d. The median is 2 hours, since P ( X  1)  0.423 and P ( X  2)  0.570. e. The actual median would be equal. Even though the actual distribution is probably even more skewed right, the median is resistant to extreme values.

26.   xi P ( xi )  20(0.05)  30(0.27)  40(0.39)  50(0.21)  60(0.08)  40

   ( xi   ) 2 P( xi )  (20) 2 (0.05)  (10)2 (0.27)  (0) 2 (0.39)  (10) 2 (0.21)  (20) 2 (0.08)  10  2  (10) 2  100 skewed right

28. a

b.   xi P ( xi )  0(0.784)  1(0.091)  2(0.053)  3(0.029)  4(0.016)  5(0.01)  6(0.008)  7(0.005)  8(0.004)  0.5 triples

(0.513) 2 (0.784)  (0.487) 2 (0.091)  (1.487) 2 (0.053)

c.

  ( xi   ) 2 P( xi )  (2.487) 2 (0.029)  (3.487) 2 (0.016)  (4.487) 2 (0.01)  1.3 triples (5.487) 2 (0.008)  (6.487) 2 (0.005)  (7.487) 2 (0.004)

 2  (1.31)2  1.7 triples 2 d. P ( X  5)  P (5)  P (6)  P(7)  P(8)  0.01  0.008  0.005  0.004  0.027 e. P ( X  3)  P (0)  P(1)  P(2)  0.784  0.091  0.053  0.928 30. a.

skewed left

b.   xi P ( xi )  1(0.002)  1.5(0.006)  1.75(0.002)  2(0.018)  2.25(0.008)  2.5(0.071)  2.75(0.144)  3(0.19)  3.25(0.169)  3.5(0.217)  3.75(0.117)  4(0.055)  5(0.001)  3.185

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Chapter 5: Computing Probabilities 105 30. (continued)

  ( xi   ) 2 P( xi ) (2.1845) 2 (0.002)  (1.6845) 2 (0.006)  (1.4345)2 (0.002)  (1.1845) 2 (0.018) 

(0.9345) 2 (0.008)  (0.6845) 2 (0.071)  ( 0.4345) 2 (0.144)  ( 0.1845) 2 (0.19) (0.0655) 2 (0.169)  (0.3155) 2 (0.217)  (0.5655) 2 (0.117)  (0.8155) 2 (0.055)

 0.479

(1.8155) 2 (0.001)

c. P (3.2  0.5  X  3.2  0.5)  P(2.7  X  3.7)  P(2.75)  P(3)  P(3.25)  P(3.5)  0.144  0.19  0.169  0.217  0.72 The actual probability is greater than what the Empirical Rule predicts (0.68). The two values are 0.04 apart. d. P (3.2  0.5(2)  X  3.2  0.5(2))  P (2.2  X  4.2)  P(2.25)  P(2.5)  P (2.75)  P(3)  P(3.25)  P(3.5)  P(3.75)  P(4)  0.008  0.071  0.144  0.19  0.169  0.217  0.117  0.055  0.971 The actual probability is greater than what the Empirical Rule predicts (0.95), but the two values are only 0.021 apart. e. P (3.2  0.5(2)  X  3.2  0.5(2))  P (1.7  X  4.7)  1  P (0)  P(1.5)  P (5)  1  0.002  0.006  0.001  0.991 The actual probability is smaller than what the Empirical Rule predicts (0.997), but the two values are only 0.006 apart.

32.   xi P ( xi )  372(0.009)  249(0.012)  224(0.012)  124(0.039)  233(0.221)  35(0.707)  $90.10 For a large number of families of four, the expected cost of attending one of the given events is $90.10.

34. a. x

P( x)

$1500  $250, 000  $248,500

1  0.99499  0.00501

$1500

0.99499

b.   xi P ( xi )  248,500(0.00501)  1500(0.99499)  $247.50; If a large number of men who are 50 years of age buy the policy, the company can expect to earn about $247.50 in profit per policy. c. 1300(247.50)  $321, 750 36. a. x

P ( x)

x

P( x)

$1  $4  $3

1 6

$4  $4  $0

1 6

$2  $4  $2

1 6

$5  $4  $1

1 6

$3  $4  $1

1 6

$6  $4  $2

1 6

1 1 1 1 1 1 b.   xi P ( xi )  3    (2)    (1)    0    1    2    $0.50; If the person plays the 6 6 6 6 6 6 game a large number of times, the person can expect to lose about $0.50 per roll of the die.

c. 1200(0.50)  $600

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106 ISM: A Pathway to Introductory Statistics 38. a. Let X be the player’s winnings (in dollars) from playing the game once. x

P( x)

20

20 38 13 38

20

 20   18  b.   xi P ( xi )  20    20    $1.05; If the player plays a large number of times, she can expect  38   38  to lose about $1.05 per spin.

c. 2000(1.05)  $2100 d. The expected winnings would be equal, since each result is equally likely.

40. Strategy A: Each student is expected to earn 10(2)  10(0)  20 points, for a total team score of 5(20)  100 points.

Strategy B: Each student is would earn 10(2)  10(0)  20 points, for a total team score of 5(20)  100 points.

Both strategies would result in the same team score.

42. a. P (NOT pass algebra)  1  0.6  0.4 b. P (pass algebra AND NOT pass trigonometry)  (0.6)(1  0.4)  0.36 c. P (pass algebra AND pass trigonometry AND NOT pass precalculus)  (0.6)(0.4)(1  0.5)  0.12 d. P (pass algebra AND pass trigonometry AND pass precalculus)  (0.6)(0.4)(0.5)  0.12 e. x

P ( x)

x

P( x)

0 5

0.4 0.36

8 13

0.12 0.12

f.   xi P ( xi )  0 0.4  5 0.36  8 0.12  13 0.12  4.32 units 44. The mean of 2, 3, 4, 5, and 6 is 4, but the student did not take into account the probabilities related to those values.

46. a. The mean will decrease; the standard deviation will remain the same. b. The mean will decrease; the standard deviation will remain the same. c. The mean and standard deviation will both decrease.

48. Answers will vary.

50. a. Answers will vary. b. Answers will vary.

52. City A has a higher standard deviation. Since it is more likely to have numbers of snowstorms farther from the mean, the record number of snowstorms would likely be higher.

54. Answers may vary.

56. Answers may vary.

Homework 5.5 2. If a distribution is normal, then the probability that a randomly selected observation lies within 3 standard deviations of the mean is approximately 0.997.

4. If an observation lies to the left of the mean, then its z-score is negative.

6. Since it is unimodal and approximately symmetric, the distribution is approximately normal.

8. Since it is skewed right, the distribution is not approximately normal.

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Chapter 5: Computing Probabilities 107 10. (a). Almost all the observations are between 50 and 80.

12. (b). Almost all the observations are between 10 and 60.

14. (a). Centered on 6 and almost all the observations are between 3 and 9.

16. (b). Centered on 50 and almost all the observations are between 20 and 80.

18. M is the value at the center of the observations, 50 points.

20. Q3 is between 55 and 60 points.

22.

26.

24.

28.

30. a. The areas of the bars, from left to right, are 0.04, 0.12, 0.20, 0.24, 0.20, 0.12, 0.06, and 0.02. b. The sum of the areas is 1. The total area of the bars of a density histogram represent the probabilities of all possible events, which sum to 100%, or 1. c.

d. The total area of the regions where the curve undercounted is equal to the total area of the regions where the curve overcounted. e. The total area of the bars would have been 1; the total area under the normal curve would have been 1.

32. z 

210  100  7.33; Kim Ung-yong’s IQ is 7.33 standard deviations higher than the mean.

15

34. z 

32  51.0  2.02; The inmate’s age is 2.02 standard deviations less than the mean.

9.4

36. P ( Z  2.38)  0.0087

42. P ( Z  1.96)  0.0250

38. P ( Z  0.40)  0.6554

44. P (3.10  Z  1.80)  0.0350

40. P ( Z  2.62)  0.9956

46. P (0.25  Z  2.74)  0.3982

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108 ISM: A Pathway to Introductory Statistics 48. a. Let X be the lifetime (in hours) of a randomly selected Activair battery.

Let Y be the lifetime (in hours) of a randomly selected Rayovac battery. b. P ( X  285)  0.0382

c. P (Y  285)  0.0382

d. The standard deviation of the tested Activair batteries is greater than the standard deviation of the tested 285  272.1  2.39 standard deviations larger than the mean life of the Rayovac Rayovac batteries. 285 is 5.4 285  252.4  1.77 standard deviations larger than the mean life of the Activair batteries, but only 18.4 batteries. This means there is a greater probability that the Activair batteries will last past 285 hours.

50. a. Let X be the scores on the Wechsler IQ test. b. P ( X  70)  0.0228

c. P ( X  75)  0.0478

d. If the probability that a randomly selected person has an IQ less than 75 is 0.0478, we would expect that 0.0478  392  19 prisoners would have an IQ less than 75. So, 19  9  10 additional prisoners could be removed from death row. e. The estimate in part (c) depends on the assumption that the IQ of prisoners on death row has the same distribution as the general population.

52. a. z 

17  20.5  0.56; The result should be negative because 17 points is less than the mean.

6.3

b. P (ACT  16)  0.2375; About 24% of students who took the ACT scored at most 16 points. c. The student might be incorrect. The distribution of ACT scores for all incoming Thomas More College students may have varied from the distribution of all 2018 ACT scores.

54. a. 15 is in the 19th percentile; 25 is in the 76th percentile; 35 is in the 99th percentile. b.

Percentile English Score (Assuming Normality) Actual Percentile 10 5 1 15 19 13 20 47 40 25 76 75 30 93 92 35 99 99 c. The percentiles assuming normality are close to the actual percentiles; this suggests that the ACT English scores are approximately normally distributed.

56. a. P (HDL  40)  0.2877

c. P (40  HDL  60)  0.5631

b. P (HDL  60)  0.1492 d. The number of men in the study with HDL between 40 and 60 mg/dl is 0.5631(1541)  868.

58. a. P (BMD  908)  0.0964

c. P ( z  1.5)  0.0668

b. P (572  BMD  636)  0.1137 60. a. The probability that a randomly selected veteran from the study has age less than 48 years is 0.13. b. The proportion of veterans from the study with age less than 48 years is 0.13. c. A veteran from the study with age less than 48 years is in the 13th percentile.

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Chapter 5: Computing Probabilities 109 62. a. The probability for African American girls is 0.0711. b. The probability for Caucasian girls is 0.0589. c. The standard deviation of salt intake is larger for the African American girls than for the Caucasian girls.

64. a. (i)

P (50  time  70)  0.5927

(ii)

P (40  time  80)  0.9026

(iii)

P (30  time  90)  0.9871

b. The distribution is unimodal and roughly symmetric. c. (i)

P (50  time  70)  0.32  0.31  0.63

(ii)

P (40  time  80)  0.12  0.32  0.31  0.16  0.91

(iii)

P (30  time  90)  0.02  0.12  0.32  0.31  0.16  0.05  0.98

d. The results in parts (c) and (a) are approximately equal; the distribution is approximately normal.

66. The probability of randomly selecting a distribution that is greater than the mean is 0.5 because the distribution is symmetric; the mean and the median are the same.

68. Almost all the area under the normal curve is within two standard deviations of the center; answers may vary.

70. Empirical rule: P (1  Z  1)  0.68; StatCrunch: P (1  Z  1)  0.683; The results are very close.

72. P ( Z  1.8)  P( Z  1.8); Sketches will vary.

74. Approximately 1.54% of young men have HDL levels below 20 ml/dl.

76. The z-score for a value is computed as the value minus the mean, divided by the standard deviation. The z-score is the number of standard deviations that the value is above the mean (or below the mean, if the z-score is negative.)

78. a. If the distribution were normal, then a wait time of 0 minutes would have a z-score of –1.02; 15.47% of the patients would have a negative wait time. Since that is impossible, the distribution cannot be normal. b. The distribution must be skewed right. The difference from the mean to the minimum value is 63.7 minutes, but assuming that almost all the values are within two standard deviations of the mean, the difference from the mean to the maximum value may be as much as 63.7  2(62.4)  188.5 minutes.

80. a.

b. The distribution is roughly unimodal and symmetric. c. Mean: 253.1; standard deviation: 74.6.

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110 ISM: A Pathway to Introductory Statistics 80. (continued) d. P (200  length  300)  0.4969 220  0.5528. This 398 proportion is greater than the one found in part (c); it suggests that there are too many observations near the mean for the distribution to be normal. f. Assuming the distribution is approximately normal, the estimated proportion of songs with lengths between 100 and 400 seconds is 0.9555. The actual proportion of songs with lengths between 100 and 400 376  0.9447. The estimated and actual proportions are almost equal. seconds is 398

e. The actual proportion of songs with lengths between 200 and 300 seconds is

Homework 5.6 2. True; a z-score gives the size of a value relative to its group.

4. If an observation’s z-score is greater than the number 1.96, then it is unusual.

6. 73.8  0.71(2.14)  75 bpm

8. 72.8  2.591(3.4)  64 inches

10. a. 3.45% of the students would get As. b. A cutoff value of 83 points would give As to 7.29% of the students.

12. The 88th percentile would allow 12% of the students to get As, but the 88-point cutoff would allow 15.87% of the students to get As. 14. a. The 30th percentile for women’s heights is 61.6 inches. b. The 30th percentile for men’s heights is 66.9 inches.

63.8  69.4  66.6 inches.

2 d. Since there are more women than men, the estimate in part (c) is an overestimate of the height of American adults.

c. The mean height of all American adults would be

16. a. The 10th percentile is 35 years. b. The 90th percentile is 60 years. c. The percentage under 18 years of age is 0.21%. d. No such inmate would be on death row: 0.0021(152)  0.3192  0.

18. a. It would be unwise because only half of the loaves would meet the fiber requirement. b. 97% of the loaves contain at least 5.6 g of fiber.

20. a. Since the mean of a normal distribution is also the median, 151 words is at the 50th percentile. b. The percentage with WCPM scores between 141 and 161 words is about 17.59%. c. The percentage of slow readers would equal the percentage of fast readers:

100  17.59  41.21%.

2

Answers may vary. d. The middle 80% would have a WCPM score between 93 and 209 words.

22. a. For Student A, z 

89  74  2.14; Student A’s score was 2.14 standard deviations larger than the mean.

7

78  66  3.00; Student B’s score was 3.00 standard deviations larger than the mean.

4 c. Student B did relatively better; Student B’s z-score is larger than Student A’s z-score.

b. For Student B, z 

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Chapter 5: Computing Probabilities 111 24. The student’s ACT score is relatively better; the z-score for the student’s ACT score z  larger than the z-score for the student’s SAT score, z  26. a. z 

15  20.5  0.87, is 6.3

410  531  1.06.

114

60  54  0.40; This woman’s HDL is 0.40 standard deviations above the mean for her age group.

15.0

62  57  0.29; This woman’s HDL is 0.29 standard deviations above the mean for her age group.

17.2 c. The relatively better score is that of the woman in her thirties. d. The woman with the higher reading has the lower z-score because the mean HDL for women in their fifties is larger than the mean HDL for women in their thirties. It is also because the standard deviation of the HDL readings for women in their fifties is greater than the standard deviation of the HDL readings for women in their thirties.

b. z 

893  920  0.20; This woman’s BMD is 0.20 standard deviations below the mean for her age 136 group.

28. a. z 

485  920  3.20; This woman’s BMD is 3.20 standard deviations below the mean for her age 136 group. c. The woman with BMD 485 mg/cm2 has eight times the risk of a hip fracture than the woman with BMD 893 mg/cm2.

b. z 

30. a. An IQ in the 87th percentile is 117 points. No, it is not unusual; the probability of a score that large or larger is 0.130, which is greater than 0.025.

160  100  4.00; Yes, it is unusual, because the z-score is greater than 1.96.

15 c. Both scores are in the 99th percentile. No, this result does not mean that Shakira is as intelligent as Newton was. Assuming that an IQ score is a good measure of intelligence, Newton’s score is 190  100 z  6.00 standard deviations above than the mean, while Shakira’s score is 15 140  100 z  2.67 standard deviations above than the mean.

15

b. z 

32. a. z 

31  63.8  7.81; De Treaux’s height was 7.80 standard deviations less than the mean.

4.2

32  69.4  7.96; Troyer’s height is 7.96 standard deviations less than the mean.

4.7 c. Since his z-score is the smaller, Troyer has the relatively smaller height. d. Yes, De Treaux had an unusual height for women and Troyer has an unusual height for men. Both people have z-scores greater than 1.96.

b

z

34. a. No, 25 points is not unusual; z 

25  20.5  0.71, so the score is not more than 1.96 standard deviations 6.3

away from the mean. b. Yes, 35 points is unusual; z 

35  20.5  2.301, so the score is more than 1.96 standard deviations away 6.3

from the mean. c. The cutoff for an unusually high score is 20.5  1.96(6.3)  33 points.

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112 ISM: A Pathway to Introductory Statistics 36. a. P (at least 3.5 g of fat per cup)  0.0062 b. Yes; the cup containing 3.5 g of fat per cup is an unusual event. Its probability is less than 0.025. c. It would not be an unusual event if the mean fat level per cup is 3.5 g. d. No, the assumptions made in part (a) do not seem reasonable; a more thorough investigation should occur.

38. a. P (mileage is at most 38 mpg)  0.0004 b. Yes, a gas mileage of 38 mpg is an unusual event; its probability is less than 0.025. c. It would not be an unusual event if the mean gas mileage is 38 mpg. d. No, the assumptions made in part (a) do not seem reasonable; a more thorough investigation should occur.

40. a. P (28.6 g of protein)  0.0026 b. Yes, a package containing 28.6 g of protein per serving is an unusual event; its probability is less than 0.025. c. It would not be an unusual event if the mean amount of protein per package is 28.6 g. d. No, the assumptions made in part (a) do not seem reasonable; a more thorough investigation should occur.

42. There is not enough information to determine who has the lower relative test score.

44. c will be negative, since the value will be below the mean.

46. Yes. Finding z-scores of observations preserves the order of the observations because a z-score measures the number of standard deviations an observation is from the mean.

48. An observation is unusual if its z-score is less than –1.96 or greater than 1.96; or, it is unusual if the probability of randomly selecting the observation, or one on the same side of the mean, but even more extreme, is less than 0.025. Answers may vary.

Chapter 5 Review Exercises 1. The student is incorrect because the outcomes winning and losing are not equally likely events.

2. P 9 

1 38

3. Since there are 18 odd numbered slots out of 38, P odd   4. Since there are 18 red slots out of 38, P  NOT red   1  5. Since there are no blue slots, P  blue   0.

18 9  .

38 19

18 10  .

38 19

6. P  red OR green  

7. Since there are four black slots numbered at most 9, P  black AND at most 9 

18 2 20 10    38 38 38 19

4 2  .

38 19

8. Since there are 18 red slots and 6 black slots numbered at least 26, P  red or at least 26

18 6 24 12    .

38 38 38 19

9. Yes, since no slot is simultaneously green and black.

10. No; P odd  

18 9 8 4  , but P odd | black    .

38 19 18 9

11. a. P coffee | 18  29 years  

67  0.411 163

c. P coffee | 45  64 years  

174  0.602 289

b. P coffee | 30  44 years  

121  0.55 220

d. P coffee | over 64 years  

97  0.752 129

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Chapter 5: Computing Probabilities 113 11. (continued) e. The probabilities from parts (a) through (d) increase; as the classes increase in age, the probability that a randomly selected individual drinks coffee increases. No, causation cannot be assumed; the study is observational.

242  0.302 801 c. No; the results of parts (a) and (b) are not equal.

12. a. P  tea  

d. P coffee AND 45  64  e. P  tea OR over 64 

b. P  tea | 18  29 

67  0.411 163

174  0.217 801

242 129 24    0.433 801 801 801

13. a. The probability is P all five  (0.68)5  0.145. b. P at least one  1  P  none  (1  0.32)5  0.997 14. discrete random variable

15. continuous random variable

16. a.

b.   xi P ( xi )  6(0.1)  7(0.3)  8(0.4)  9(0.1)  10(0.1)  7.8 points; The center of the distribution is 7.8 points. c.

   ( xi   ) 2 P( xi )  (2) 2 (0.1)  (1) 2 (0.3)  (0) 2 (0.4)  (1) 2 (0.1)  (2) 2 (0.1)  1.1 points  2  (1.1) 2  1.2 points 2

d. P ( X  8)  P (6)  P(7)  0.1  0.3  0.4

e. P ( X  9)  P(9)  P (10)  0.1  0.1  0.2

17. a. x

P( x)

$1500  $200, 000  $198,500

1  0.993152  0.006848

$1500

0.993152

b.   xi P ( xi )  $198,500(0.006848)  $1500(0.993152)  130.40; If a large number of women who are 60 years of age buy the policy, the company can expect to earn about $135.41 in profit per policy. c. 1000($135.40)  $130, 400 18.

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114 ISM: A Pathway to Introductory Statistics 19. z 

35  47.4  1.28; The inmate’s age is 1.28 standard deviation less than the mean age.

9.7

20. P  1.7  Z  2.5  0.9492 21. a. Since the mean plus two standard deviations is 6.5  2(1.4)  9.3, the probability that a student slept at least 9.3 hours is approximately 2.5% of 1, or 0.025. b. Since the mean is 6.5 and the mean plus three standard deviations is 6.5  3(1.4)  10.7, the probability that a student slept between 6.5 hours and 10.7 hours is approximately 49.85% of 1, or 0.4985. c. Since the mean is 6.5 and the mean plus three standard deviations is 6.5  1(1.4)  7.9, the probability that a student slept no more than 7.9 hours is approximately 84% of 1, or 0.84.

22. a. Let X be the age (in years) of a randomly selected American who died of alcohol poisoning. b. P ( X  21)  0.0203 c. P (50  X  60)  0.2510 d. P ( X  65)  0.0705 23. 21.6  2.24(2.5)  16 miles per gallon 24. More students would get As if the cutoff is the 90th percentile; P (at least 90 points)  5.48%, less than 10%.

25. a. Yes, it is unusual; z 

765  531  2.05, so the score is more than 1.96 standard deviations away from the 114

mean. b. Yes, it is unusual; z 

33  20.5  1.98, so the score is more than 1.96 standard deviations away from the 6.3

mean. c. The student from Texas performed better; the z-score for the student from Texas is greater than the z-score for the student from Florida.

26. a. P (at least 3.8 g of fat)  0.0062 b. Yes, a package containing 3.8 g of fat per serving is an unusual event; its probability is less than 0.025. c. It would not be an unusual event if the mean fat level per serving is 3.8 g. d. No, the assumptions made in part (a) do not seem reasonable; a more thorough investigation should occur.

Chapter 5 Test

1. P 3 

1 6

2. P greater than 4 

4. P odd OR at least 4  P 1,3, 4,5, OR 6  2 1  6 3

3. P even AND at most 3  P  2  6. Yes. P less than 5 

5. P odd | less than 4  1 6

4 2 2  and P less than 5 | even   .

6 3 3

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2 3

5 6

Chapter 5: Computing Probabilities 115 7. a. P  nonfatal heart-related episode 

33  0.0545 605

b. P  nonfatal heart-related episode OR died   c. P cancer OR Mediterranean diet   d. P cancer | prudent AHA diet  

33 38   0.1174 605 605

24 302 7    0.5273 605 605 605

17  0.0561 303

e. No; P  healthy | Mediterranean diet  

273 510  0.9040, but P (healthy)   0.8430.

302 605

8. a. The probabilities are all between 0 and 1, inclusive, and the sum of all the probabilities is 0.200  0.510  0.178  0.090  0.022  1. b.

c.   xi P ( xi )  0(0.20)  1(0.510)  2(0.178)  3(0.090)  4(0.022)  1.22 d.

   ( xi   ) 2 P( xi )  (1.224) 2 (0.200)  ( 0.224) 2 (0.510)  (0.776) 2 (0.178)  (1.776) 2 (0.090)  (2.776) 2 (0.022)  0.94

 2  (0.94)2  0.88 e. P ( X  3)  P (0)  P(1)  P(2)  P(3)  0.200  0.510  0.178  0.090  0.978 9. a. x

P ( x)

x

P( x)

$1  $3.80  $2.80

1 6

$4  $3.80  $0.20

1 6

$2  $3.80  $1.80

1 6

$5  $3.80  $1.20

1 6

$3  $3.80  $0.80

1 6

$6  $3.80  $2.20

1 6

1 1 1 1 1 1 b.   xi P ( xi )  2.8    1.8    0.8    0.2    1.2    2.2    $0.30; If the person plays the 6 6 6 6 6 6 game a large number of times, the person can expect to lose about $0.30 per roll of the die.

c. 1300(247.50)  $321, 750 10. a. Let X be the tuition (in dollars) in the academic year 2016–2017 of a randomly selected 4-year, public college. b. P (3000  X  4000)  0.0489

d. P ( X  5000)  0.8111

c. P ( X  4500)  0.1520 11. The 90th percentile is age 51.3  1.28(12.0)  67 years.

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116 ISM: A Pathway to Introductory Statistics 12. a. No, it is not unusual; the z-score is

508  454.1  0.73, which is less than 1.96 standard deviations away 74.2

from the mean. b. No, it is not unusual; the z-score is

541  490.5  0.70, which is less than 1.96 standard deviations away 72.0

from the mean. c. The better relative performance was in 2011 because the z-score is larger.

13. a. P (at least 93 mg of sodium)  0.0038 b. Yes, a package containing 93 mg of sodium per serving is an unusual event; its probability is less than 0.025. c. It would not be an unusual event if the mean sodium level per serving is 93 mg. d. No, the assumptions made in part (a) do not seem reasonable; a more thorough investigation should occur.

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Chapter 6: Describing Associations of Two Variables Graphically 117

Chapter 6: Describing Associations of Two Variables Graphically Homework 6.1 2. For a coordinate system, the values of the response variable are described by the vertical axis.

4. If the response variable tends to decrease as the explanatory variable increases, we say there is a negative association.

6. The explanatory variable is n, the number of office parties, and the response variable is p, the percentage of employees that enjoy working at the company. n should be on the horizontal axis, and p should be on the vertical axis.

8. The explanatory variable is h, the player’s height, and the response variable is P, the player’s mean points scored per game. h should be on the horizontal axis, and P should be on the vertical axis.

10. The explanatory variable is a, age, and the response variable is p, the percentage of men at that age with gray hair. a should be on the horizontal axis, and p should be on the vertical axis.

12. The explanatory variable is G, the student’s GPA, and the response variable is p, the percentage of colleges that would accept the student. G should be on the horizontal axis, and p should be on the vertical axis.

14. The ordered pair is (3,167). b should be on the horizontal axis, and p should be on the vertical axis.

16. The ordered pair is (10,160). s should be on the horizontal axis, and p should be on the vertical axis.

18. The ordered pair is (17, 42.1). t should be on the horizontal axis, and w should be on the vertical axis.

20. A television salesperson that compliments people an average of 72 times per day sells, on average, 39 televisions per month.

22. After ingesting 30 milligrams of an experimental acne medication daily for three months, a teenager has 2 pimples on his face.

24. In 2108, Starbucks’s annual revenue was $25 billion dollars.

26. In 2013, 61% of Americans were satisfied with the size and power of major corporations.

28. Negative; As the number of hours a college student spends gaming per week increases, one would expect the number of hours the student spends studying per week to decrease.

30. Positive; As the temperature (in degrees Fahrenheit) at a beach increases, one would expect the number of people at the beach to increase.

32. Negative; As the price (in dollars) of a bottle of wine increases, one would expect the annual sales (in hundreds of bottles) of the wine to decrease.

34. The association is negative.

36. The association is positive.

38. a. The explanatory variable is rating of a given wine in points. The response variable is price of that wine in dollars. b. The association is positive. The higher a wine’s rating, the higher its cost. c. The student in incorrect, there are wines with a rating of 94 points that are more expensive that wines with a rating of 96 points. d. The prices of wines with a rating of 94 points are more spread out than the prices of wines with a rating of 90 points. e. The highest point is approximately (97,310). The wine represented had a rating of 97 points and cost $310. The point does not fit the overall pattern of the data.

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118 ISM: A Pathway to Introductory Statistics 40. a. At least one of the dots represents more than one data pair. b. The explanatory variable is the temperature in degrees Fahrenheit. The response variable is the relative humidity as a percent. c. The association is negative. As the temperature decreases, the relative humidity tends to decrease. d. The humidity tends to decrease as the temperature increases; the scatterplot does not show any information about time of day. e. The humidity would most likely have increased; since the association is negative, if the temperature decreased, then the humidity tends to increase.

42. a. The explanatory variable is d, the value of the bill in dollars. The response variable is L, the bill’s life-span in years. b. d should be on the horizontal axis since it is the explanatory variable. L should be on the vertical axis since it is the response variable. c.

d. The association is positive. As the denomination of the bill increases, the life-span of the bill will tend to increase. This makes sense because smaller denominations are used more often, and thus will wear out more quickly. e. Answers may vary; $1 bills have a shorter life-span than $50 bills, and $1 bills are used more commonly for small transactions than $50 bills.

44. a.

b. The association is negative. As people get older, they tend to be less likely to order more takeout food than they did two years ago. c. The highest point is (21,34), which means that 34% of 21-year-olds ordered more takeout food than they did two years ago. d. The lowest point is (70, 7), which means that 7% of 70-year-olds ordered more takeout food than they did two years ago.

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Chapter 6: Describing Associations of Two Variables Graphically 119 46. a. The explanatory variable is t, the number of years since 2010, and the response variable is C, the total airline fuel cost, in billions of dollars. b.

c. Total airline fuel cost was the least in 2015, with a cost of $181 billion dollars. d. Total airline fuel cost was the greatest in 2013, with a cost of $230 billion dollars. e. The mean price per barrel of crude oil was the greatest in 2012, which is not the same year that the total airline fuel cost was the greatest. It is possible that the number of flights was greater in 2013 than in 2012, offsetting the lower price of crude oil.

48. a.

b. The association is positive. Mean hourly manufacturing pay tended to increase over time. $18.49 40 hours   50 weeks  $36,980; Answers may vary.

1 hour 1 week d. The mean hourly manufacturing pay grew faster than the increase in the mean price of goods and services.

21.92 The mean hourly pay increased by a factor of  6.91 from 1970 to 2019, which is more than 6.05.

3.17

c.

50. a. The explanatory variable is d, the depth of the ocean (in meters), and the response variable is s, the salinity (in grams per kilogram). b.

c. The association is positive. As the depth in the ocean increases, the salinity tends to increase. d. Saltier water sinks in the ocean. This makes sense since saltier water has higher density. e. The lowest point is (0,32.6). The salinity of the surface of the ocean is 23.6 grams per kilogram.

52. a.

b. The association is positive. The more nominees in a given genre, the more winners in that genre. c. The point farthest from the others is (456,80). Drama had 456 nominees and 80 winners. d. The percentages are 11 58  19% for adventure, 6 34  18% for family, 36 217  17% for romance, and 8 456  18% for drama. e. The point for drama is in line with the direction of the other points in the scatterplot. It also has a percentage of winners that is consistent with the other points from part (d).

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120 ISM: A Pathway to Introductory Statistics 54. a. The explanatory variable is GDP (in dollars per person). The response variable is percentage of smartphone penetration. b.

c. The association is positive. As GDP per person increases, smartphone usage will tend to increase. This makes sense as greater GDP per person would increase the chances that a person could afford a smartphone. d. Yes, Taiwan’s GDP per person is relatively large, as is its smartphone penetration. However, the point for smartphone penetration is a slightly below the overall pattern. e. The highest point is (62.6, 77.0). The United States’ GDP per person is 62.6 thousand dollars, and its smartphone penetration is 77.0%, which is the highest percent of all the countries described by the data.

56. a. The explanatory variable is the number of friends in high school. The response variable is the number of those high school friends that they still consider to be friends. b.

c. The association is positive. The more friends a student had in high school, the more high school friends they still consider to be friends. d. The points farthest from the data is (60, 30). A student still considers 30 of their high school friends as friends. e. The percentages for the student from part (d) and those listed in blue are 30 60  50%, 8 20  40%, 8 15  53%, and 5 10  50%, respectively. The percentage is for the student listed in red is 5 5  100%. This student is still friends with all of his or her high school friends.

58. a.

b. That home’s list price is equal to its sales price. c. The line comes close to most of the data points on the scatterplot. If a point lies near the line, the list price is approximately equal to the sale price. d. The house that sold for the most over its list price was on S Desplaines St. By sizing up the scatterplot, we can limit the search to data points that are high up and off to the left. e. The house that sold for the most under its list price was on N Surrey Ct. By sizing up the scatterplot, we can limit the search to data points that are low and off to the right.

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Chapter 6: Describing Associations of Two Variables Graphically 121 60. a.

b. The association is positive. As the number of bedrooms increases, the sale price tends to increase. c. 2 bedrooms:

292  287  158  230  149  223.2 thousand dollars 5

3 bedrooms:

475  235  178  190  306  240  220  236.4 thousand dollars 7

4 bedrooms:

563  412  1338  771 thousand dollars 3

5 bedrooms:

2425  725  1575 thousand dollars 2

d.

The association of number of beds and the mean sales price is much stronger than the association of the number of beds and the sale price. e. There are five houses with 2 bedrooms with sales prices of 292, 287, 258, 230, and 149 thousand dollars.

Answers will vary, but location and overall condition of the home are two possible reasons.

62. No. If the association between number of milligrams ingested per day and blood pressure is negative, then it is appropriate to say that blood pressure tends to decrease as the medication increases, but this does not mean a causal relationship since there may be other variables.

64. a. This is not possible. Each axis of a scatterplot describes a numerical variable. b. This is not possible. Each axis of a scatterplot describes a numerical variable, so both variables must be numerical. c. This is possible.

66. This data would be appropriately described by a frequency and relative frequency table, dotplot, stemplot, relative frequency histogram, or boxplot.

68. This data would be appropriately described by a frequency and relative frequency table, relative frequency bar graph, or pie chart.

70. Yes; since the two variables are negatively associated, as x decreases, y tends to increase. Thus, as y increases, x would tend to decrease.

72. Answers may vary.

76. Answers may vary.

74. Answers may vary.

Homework 6.2 2. If r is positive, then there is a positive association.

4. False. Association does not imply causation.

6. A lurking variable is a variable that causes both the explanatory and response variables to change during the study.

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122 ISM: A Pathway to Introductory Statistics 8. No association. The scatterplot shows no pattern.

10. All of the points lie close to a curve that is a line, so there is a linear association.

12. (c); if r  0, then there is no association between the variables, and this scatterplot shows no association.

14. (e); if r  0.9, then there is a strong negative association.

16. (a); this scatterplot shows an exact positive linear association.

18. a.

b. There is a linear association. c. r  0.96 d. Because r  0.96 and the scatterplot shows the points lying near the same line, the association is strong and negative.

20. a. There are no outliers. The association is fairly strong, linear, and positive. b. The scatterplot has a dot near (69,185), so Isaiah Thom’s weight is about 185 pounds. c. Two players may have the same weight. The estimated range is about 55 pounds. d. Answers may vary.

22. a. This point is not close to most of the other points; this state has a lower obesity rate than expected, given the percentage of residents that exercise. This makes sense because Puerto Rico may have different cultural norms for exercise and eating. b. The association is fairly strong, negative, and linear. c. These 5 states have obesity rates of approximately 19%, 22%, 22.5%, 24.5%, and 25%. The range is 6%.

This is consistent with a fairly strong association. d. This scatterplot only shows a negative association for exercise and obesity in the United States; the association may not be the same in all countries.

24. a.

b. There are no outliers. The association is strong, positive, and nonlinear. c.

11.2 km 0.621 mi 3600 s    25, 039 miles per hour 1s 1 km 1 hr Copyright © 2021 Pearson Education, Inc.

Chapter 6: Describing Associations of Two Variables Graphically 123 24. (continued) d. The mass of Venus is between the masses of Mars and Earth; the estimated escape velocity for Venus is between 5.0 km/s and 11.2 km/s. e. The ratio of masses is 1898 102  18.61; the ratio of escape velocities is 59.5 23.5  2.53. The ratio of masses is greater than the ratio of escape velocities. Since the escape velocity increases rapidly as mass increases before leveling off, this makes sense.

26. a. There are no outliers. The association is strong, positive, and linear. b. There are no outliers. The association is fairly strong, positive, and linear. c. The correlation coefficients suggest that the association for Saurischians is stronger than the association for Ornithischians, which is consistent with the results from (a) and (b). d. No; while the values of the individual variables are outliers, this point is consistent with a strong association between the two variables. e. The values of the correlation coefficients would be closer to 0, since the spread of the observations would increase.

28. No. The scatterplot shows no pattern, and the value of r is 0.14, indicating either no association or a weak negative association.

30. a.

b. There is a linear association. c. r  0.99; the association is strong and negative.

105.8  0.514 105.8  100 e. Parents may try to influence the gender of the child to be a boy.

d.

32. a.

b. There is a linear association. c. r  0.973; the association is strong and negative. d. The association is negative; the record time has decreased over time. e. The change from 1900 to 1950 is 47.8  45.8  2.0 seconds; the change from 1950 to 1999 is 45.8  43.18  2.62 seconds. This is surprising because we would expect improvements to be harder to come by as the record decreases.

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124 ISM: A Pathway to Introductory Statistics 34. a.

b. There is a linear association. c. r  0.90; The association is fairly strong. d. The association is positive; Halloween spending tends to increase over time. e. An association does not guarantee causation. It is more likely that increased Halloween spending reflects an increase in population.

36. a.

b. Underestimation: There is a linear association.

Overestimation: There is a linear association. c. Underestimation: r  0.98; The association is strong.

Overestimation: r  0.96; The association is strong. d. Underestimation: The association is positive.

Overestimation: The association is negative. e. Yes, the percentages of younger women that overestimated their BMI are higher than those of the same age that underestimated their BMI. 38. a. There is a positive association; there have been more fires per year over time. b. There is a positive association; there have been more acres burned per year over time. c. There is most likely a positive association; both the number of fires and acres burned have increased over time. d. Yes; the scatterplot shows a positive association, which is consistent with the conclusion in part (c). e. There may have been fewer fires, but some of those fires were very large in terms of acreage burned.

40. a.

r  0.90

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Chapter 6: Describing Associations of Two Variables Graphically 125 40. (continued) b.

r  0.90, which is identical to the result from part (a). The shape of the scatterplot is the same as in part (a). This makes sense since all of the data points were changed by the same amount and direction.

c.

r  0.90, which is identical to the result from part (a). The shape of the scatterplot is the same as in part (a). This makes sense since all of the data points were changed by the same amount and direction.

42. Answers may vary.

44. The correlation coefficient does not measure causation. A lurking variable may be responsible for the association.

46. The outliers are points that do not fit the overall pattern on the scatterplot. If an outlier is due to error in measurement or recording, the coordinate(s) should be corrected, if possible, or the point should be excluded from the study.

The shape of the association is the pattern formed by the points on the scatterplot, if a pattern exists. This pattern can be linear or nonlinear. If no pattern exists, then we say there is no association.

The strength of the association is based on how well a line or curve fits the points on a scatterplot.

The direction of the association indicates whether the response variable increases, decreases, or does not change with changes in the explanatory variable.

48. Answers will vary.

50. A lurking variable is a variable that causes both the explanatory and response variables to change during the study.

Homework 6.3 2. An input is a permitted value of the explanatory variable that leads to at least one output.

4. If the residual of a data point is negative, then the data point lies below the linear model.

6. When a model gives a prediction that does not make sense or an estimate that is not a good approximation, we say model breakdown has occurred.

8. y  3

14. y  3

10. x  2

16. x  3

12. The y-intercept is (0, 4).

18. The x-intercept is (3, 0).

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126 ISM: A Pathway to Introductory Statistics 20. a.

b. There is a linear association. c. Answers may vary. One possible line is drawn on the scatterplot in part (a). d. Answers may vary, but they should be close to (6,8.7). e. Answers may vary, but they should be close to (8.9,12).

22. a.

b. There is a linear association. c. Answers may vary. One possible line is drawn on the scatterplot in part (a). d. Answers may vary, but they should be close to (0, 3.7). e. Answers may vary, but they should be close to (5.2, 0).

24. a. According to the linear model, a DBP of about 95 mm Hg corresponds to a SBP of 156 mm Hg. b. The actual DBP for point B is about 89 mm Hg. c. According to the linear model, point B has a DBP of 95 mm Hg. The residual is 89  95  6 mg Hg. The model overestimates of the actual DBP. This can be seen on the graph because the line drawn is above point B. 26. a. The two models are quite similar since the lines for each model on the graph are very close. It is reasonable to compare the models for lengths of up to about 16 meters since there are no Ornithischians longer than 16 meters. b. According to the linear model, the height of the new Saurischian will be about 20 meters. Since this is obtained using interpolation, we can place a lot of faith in this estimate. c. According to the linear model, the height of the new Ornithischian will be about 19 meters. Since this is obtained using extrapolation, we should place little or no faith in this estimate. d. According to the linear model, Saurischians of length 10 meters, 20 meters, and 30 meters will have heights of about 4.5 meters, 10 meters, and 15 meters respectively. e. 4.5 10  0.5; 10 20  0.5; 15 30  0.5; All three ratios are about the same. A paleontologist can estimate the height of a Saurischian by multiplying the length by 0.5.

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Chapter 6: Describing Associations of Two Variables Graphically 127 28. a. According to the linear model, a 30-year-old player will have about 7 years of experience. b. According to the linear model, a player with 5 years of experience will be about 27.5 years old. c. According to the linear model, a 35-year-old player will have about 12 years of experience. The leastexperienced 35-year-old player has 10 years of experience. The residual is –3 years. The observed age is 2 years less than the predicted age. d. According to the linear model, a 25-year-old player will have about 2.5 years of experience. The mostexperienced 25-year-old player has 6 years of experience. The residual is 3.5 years. The observed age is 3.5 years more than the predicted age.

30. a.

b. There is a linear association between year and annual revenue. c. Answers may vary. One possible model is drawn on the scatterplot in part (a). d. Answers may vary, but they should be close to $45 million. The residual is $42.3  $45  $2.7 million.

The model overestimates the revenue by $2.7 million. d. Answers may vary, but they should be close to 2012.

32. The model with explanatory variable arm span tends to better predict height. The scatterplot displays a stronger association.

34. A student with an arm span of 155 centimeters is predicted to have a height of 160 centimeters and a student with a height of 160 centimeters corresponds to a student with a foot length of 20 centimeters.

36. The data point that has a negative residual with largest absolute value is (165,168). The residual is approximately –10 centimeters, which means the model underpredicted the height of a student with an arm span of 165 centimeters by 10 centimeters.

38. a.

b. Answers may vary. One possible model is drawn on the scatterplot in part (a). c. Answers may vary, but they should be close to 90 years. This is interpolation because it is between the ages of two of the data points listed. d. Answers may vary, but they should be close to (0, 215). This means that for newborns, there are 215 males for every 100 females. There is little or no faith in the result because this result is beyond the scope of the data points listed. e. Answers may vary, but they should be close to (113, 0). This means that there are 0 men for every 100 women for people aged 113 year. There is little or no faith in the result because this result is beyond the scope of the data points listed.

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128 ISM: A Pathway to Introductory Statistics 40. a.

b. Answers may vary. One possible model is drawn on the scatterplot in part (a). c. Answers may vary, but they should be close to 159°F. This is interpolation because it is between the years of two of the data points listed. d. Answers may vary, but they should be close to 19.1 thousand meters. This is extrapolation because it is beyond the scope of the data points listed. e. Yes, the result from part (d) of 19.1 thousand meters is in this interval.

42. a.

b. One possible model is drawn on the scatterplot in part (a). c. Answers may vary, but they should be close to 2.6 million. This is extrapolation because it is beyond the scope of the data points listed. d.

e. The residual is about 0.5  2.6  2.1. The absolute value of the residual is so large because it was assumed that the linear model would continue to hold.

44. a.

b. There are no outliers. The association is linear, strong, and negative; r  0.91. c. The model is drawn on the scatterplot in part (a). d. If the mean difficulty rating of a professor is 3.0, the mean overall rating of that professor will be about 3.3. e. Self-response is used. We should not have much faith that all the ratings entered at the website represent how all students in the country would rate their professors Copyright © 2021 Pearson Education, Inc.

Chapter 6: Describing Associations of Two Variables Graphically 129 46. a.

b. There are no outliers. The association is linear, strong, and positive; r  0.98. c. The model is drawn on the scatterplot in part (a). d. According to the linear model, a wooden roller coaster with a height of 40 feet will have a length of about 1406 feet.

48. a.

b. There are no outliers. The association is linear, fairly strong, and positive; r  0.86. c. The model is drawn on the scatterplot in part (a). d. According to the linear model, the bicyclist fatality rate for Oklahoma City will be about 11 fatalities per 10,000 bicycling commuters. e. Answers may vary.

50. a.

b. The model is drawn on the scatterplot in part (a). c. According to the linear model, the firearm death rate for South Dakota will be about 11 deaths per 100,000 people. d. The residual is 10  11  1 firearm deaths per 100,000 people. e. Answers may vary, but they should be close to (3, 0). This means that there are 3 firearm deaths for every 100,000 people in states with no households that have firearms. There is little or no faith in the result because this result is beyond the scope of the data points listed.

52. Answers may vary.

54. The residual is the observed value of y minus the predicted value of y. A data point below a linear model has a negative residual since a smaller number minus a larger number is negative.

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130 ISM: A Pathway to Introductory Statistics 56. The student reversed the subtraction. The residual is $4700  $4750  $50; the opposite of the student’s answer.

58. Answers may vary.

60. A time of t  2 years means that the year is 2013. This is not an example of model breakdown, unless the model is measuring characteristics of something that did not exist prior to 2015.

62. The student is incorrect. An ordered pair does not have an x-intercept unless it lies on the x-axis.

64. The y-intercept is where a line crosses the y-axis. Any point on the y-axis has an x-coordinate of 0.

66. Answers may vary.

70

68. Answers may vary.

a.

b. There are some outliers, including several homes with square footages of 0 feet and one home with a square footage of about 12,000 feet and a sale price of $30 million. The association is linear, fairly strong, and positive; r  0.80. c. The model is drawn on the scatterplot in part (a). d. According to the linear model, a home with 2000 square feet will have a sale price of about $1.3 million. e. According to the linear model, a home with a sale price of $2.9 million will have about 4000 square feet.

Chapter 6 Review Exercises 1. The explanatory variable is t, the number of years of education. The response variable is s, the salary (in dollars). The number of years of education should be on the horizontal axis, and the salary in dollars should be on the vertical axis.

2. The explanatory variable is n, the number of pizzas that a person consumed last year. The response variable is w, the person’s weight in pounds. The number of pizzas consumed should be described by the horizontal axis, and the weight in pounds should be described by the vertical axis.

3. The situation can be described using the ordered pair (110,139). The number of miles the person runs weekly, M, should be described by the horizontal axis, and the person’s best marathon time, T, should be described by the vertical axis.

4. The situation can be described using the ordered pair (8, 29.3). The number of years since 2010, t, should be described by the horizontal axis, and the number of Starbucks stores, s, should be described by the vertical axis.

5. When a car is driven at 64 mph, its gas mileage is 42 miles per gallon.

6. In the year 2019, there were 2153 billionaires worldwide.

7. Negative; As the price (in dollars) of a bottle of wine increases, one would expect the annual sales (in thousands of bottles) of the wine to decrease.

8. Positive; As the sound level (in decibels) of a concert increases, one would expect the percentage of the concertgoers whose ears ring after the concert to also increase.

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Chapter 6: Describing Associations of Two Variables Graphically 131 9.

10. a.

b. The association is neither positive nor negative. c. The lowest point in the scatterplot is the point (40, 27). This means that a total of 27 sepsis patients were discharged when their stay reached 40 days. d. The highest point in the scatterplot is the point (20,931). This means that a total of 931 sepsis patients were discharged when their stay reached 20 days. e. There was an extremely large jump in discharges of sepsis patients when their stay reached 20 days.

11. a. The length in feet of the extension cord, L, is the explanatory variable. The safe maximum rate of electric current in amperes, A, is the response variable. b. The length in feet should be described by the horizontal axis, and the safe maximum rate of electric current in amperes should be described by the vertical axis. c.

d. The association is negative. This means that as the length of the extension cord increases, the safe maximum rate of electric current tends to decrease. e. The only length of 16-gauge extension cords that can be safely used with the screwdriver is 25 feet.

12. r  0.6 corresponds to scatterplot (c).

14. r  1 corresponds to scatterplot (b).

13. r  0.9 corresponds to scatterplot (a).

15. a.

b. The model is drawn on the scatterplot in part (a).

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132 ISM: A Pathway to Introductory Statistics 15. (continued) c. The predicted value is about 10%. The residual is 3%  10%  7%. The observed percentage of pneumonia cases that are penicillin resistant is 7 percentage points less than the predicted percentage of pneumonia cases that are penicillin-resistant. d. The predicted value is about 13%. The residual is 22%  13%  9%. The observed percentage of pneumonia cases that are penicillin-resistant is 9 percentage points more than the predicted percentage of pneumonia cases that are penicillin-resistant. e. The student is wrong. Because random assignment was not performed, we cannot assume causation.

16. y  1

18. The x-intercept is (4, 0).

17. x  6

19. The y-intercept is (0, 2).

20. a.

b. There is a linear association between the variables x and y. c. The model is shown on the scatterplot in part (a). d. y  8.8 e. x  5.8 21. a.

b. The association is linear, very strong, positive, and there are no outliers. The value of r is 0.999, which confirms that the association is very strong. c. The model is shown on the scatterplot in part (a). d. According to the linear model, the salary at step 5 would be about $80,000. Interpolation was performed to get this because step 5 is between two steps that are used to create the model. Since interpolation was used, we have good faith in the result. e. According to the linear model, the salary at step 8 would be about $92,000. Extrapolation was performed to get this because step 8 is beyond the steps that are used to create the model. Since extrapolation was used, we do not have faith in the result.

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Chapter 6: Describing Associations of Two Variables Graphically 133 22. a.

b. The model is shown on the scatterplot in part (a). c. According to the linear model, Mays stole about 28 bases in 1959. This is interpolation because 1959 is between two years that are used to create the model. Since interpolation was used, we can have good faith in the result. The residual is about 27  28  1. d. The n-intercept is (0, 45). This intercept means that Mays stole 45 bases in 1955. Model breakdown has occurred here. e. The t-intercept is (10.4, 0). This intercept means that Mays did not steal any bases in 1965. Model breakdown has occurred here.

23. The correlation coefficient r is not a probability; it is a measure of the direction and strength of a linear relationship between two quantitative variables.

Chapter 6 Test 1. The explanatory variable is a, the age of the Americans. The response variable is p, the percentage of the Americans who own a home. The age of the Americans should be described by the horizontal axis, while the percentage of the Americans who own a home should be described by the vertical axis.

2. The ordered pair is (6,390). The number of tickets should be described by the horizontal axis, and the total cost of the tickets should be described by the vertical axis.

3. The ordered pair means that in 2018, LeBron James made $35.65 million.

4. a. The explanatory variable is a, the age of the person. The response variable is s, the mean amount of savings (in millions of dollars) the person believes is enough to have at retirement. b.

c. The highest point in the scatterplot is (25, 4.3). This means that the mean amount of savings that a 25-year-old person believes is enough to have at retirement is $4.3 million. d. The lowest point in the scatterplot is (35, 2.4). This means that the mean amount of savings that a 35-year-old person believes is enough to have at retirement is $2.4 million. e. The 20–29.99 age group does not seem to fit the pattern of the others. For the other age groups, the older the age group, the more savings the group believes will be necessary to have at retirement.

5. There is no association between the variables.

6. There is a nonlinear association between the variables.

7. There is a linear association between the variables.

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134 ISM: A Pathway to Introductory Statistics 8. a.

b. The association is linear, very strong, positive, and there are no outliers. The value of r is 0.993, which confirms that the association is very strong. c. The association is linear, very strong, positive, and there are no outliers. The value of r is 0.997, which confirms that the association is very strong. d. A very strong association does not guarantee causation. e. Women are just as likely as men to have high blood pressure at age 59.5. For people younger than that age, men are more likely to have high blood pressure. For people older than that age, women are more likely to have high blood pressure.

9. a.

b. There is a linear association between the two variables. c. The model is shown on the scatterplot in part (a). d. The x-intercept is (11.2, 0). e. The y-intercept is (0, 24.3).

10. a.

b. The model is shown on the scatterplot in part (a). c. According to the linear model, there are about 5.0 thousand space debris in 1980. This is interpolation because the year is between two years used to create the model. Since interpolation was used, there is good faith in the estimate. The residual is 5.5  5.0  0.5 thousand. the observed number of space debris is 0.5 thousand debris more than the predicted number of space debris. d. According to the linear model, there are about 11.9 thousand space debris in 2010. This is extrapolation because the year is outside of the years used to create the model. Since extrapolation was used, there is little or no faith in the estimate. e. The estimated amount of space debris in 2010 caused by the collisions is 16  11.9  4.1 thousand debris.

To state this is to assume that the linear model would have predicted the exact number of debris in 2010, which it was determined there is little or no faith that this is true.

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 135

Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change Homework 7.1 2. The graph of an equation in two variables is the set of points that correspond to all solutions of the equation.

4. The graph of x  c, where c is a constant, is a vertical line.

6. 20  12  4(2) 20  12  8 20  20

8  12  4(1) 8  12  4 8  8

0  12  4(3) 0  12  12 00

The ordered pairs (2, 20), (1, 8), and (3, 0) satisfy the equation y  12  4 x.

8. 3  8  5(2) 3  8  10 3  18

8  8  5(0) 8  80

7  8  5(3) 7  8  15

88

7  7

The ordered pairs (0,8) and (3, 7) satisfy the equation y  8  5 x.

10. The y-intercept of the equation is (0, 4).

20. The y-intercept of the equation is (0, 0).

12. The y-intercept of the equation is (0, 6).

22. The y-intercept of the equation is (0, 0).

14. The y-intercept of the equation is (0, 0).

24. The y-intercept of the equation is (0, 2).

16. The y-intercept of the equation is (0, 0).

26. The y-intercept of the equation is (0, 1).

18. The y-intercept of the equation is (0, 0).

28. The y-intercept of the equation is (0, 4).

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136 ISM: A Pathway to Introductory Statistics 30. The y-intercept of the equation is (0, 2).

38.

40.

32. The y-intercept of the equation is (0, 2).

42.

34. The y-intercept of the equation is (0,5).

44.

36.

46. Points A, B, G, and H 48. Points E and F 50. Points A and B 52. a. The explanatory variable is t, the number of years since 1980. The response variable is n, the number of priests in thousands. b.

c. There are no outliers; the association is linear, strong, and negative; r  0.998, which confirms that the association is strong and negative. d. The model is shown in part (b). The model comes close to the data points. e. n  57.96  0.58(38)  35.92; According to the model, there are about 35.9 thousand priests in 2018. The residual is 36.6  35.9  0.7. The actual number of priests was about 700 higher than the number predicted by the model.

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 137 54. a. The explanatory variable is F, the temperature in degrees Fahrenheit. The response variable is R, the chirp rate of the crickets. b.

c. The association is positive. That means that, in general, as the temperature increases, the chirp rate of the crickets also increases. d. The model is shown in part (b). The model comes close to the data points. e. n  38.03  1.02(73)  36.43; According to the model, when the temperature is 73F, the chirp rate of the crickets will be about 36.43 chirps every 13 seconds. The residual is 36  36.43  0.43. The actual chirp rate was about 0.43 lower than the rate predicted by the model.

56. a. The explanatory variable is n, the number of pitches thrown in a game. The response variable is L, the length (in minutes) of the game.

b.

c. The association is positive. That means that, in general, as the number of pitches increases, the length of the game also increases. d. The model is shown in part (b). The model comes close to the data points. e. L  16.29  0.57(287)  179.88; According to the model, in a game with 287 pitches, the game will last about 180 minutes, or 3 hours. This is interpolation because 287 is between the lowest and highest number of pitches used to create the model.

58. a. Answers may vary; x y 0 5 1 1 2 3

b.

c. For each solution, the y-coordinate is 5 more than negative 4 times the x-coordinate.

60. a.

b. The arrows are included in part (a). There is only one output for the input x  3. c. For any single x value, only one arrow can be drawn directly vertical from that x value to the line, and then directly horizontal from that line to the corresponding y value.

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138 ISM: A Pathway to Introductory Statistics 60. (continued) d. Answers may vary. e. Answers may vary. There is only one output for the input x  2. f. For any single x value, only one arrow can be drawn directly vertical from that x value to the line, and then directly horizontal from that line to the corresponding y value. g. For any x value, only one arrow can be drawn directly vertical from that x value to any line of the form y  a  bx, and then directly horizontal from the line to the corresponding y value.

62. 4 x  5 x 9x  0 x0 y  4(0)  0

78. y  3 80. y  1  x 82. y   x 84. a. Answers may vary. b. y  2 x

The intersection point is (0, 0).

64. Answers may vary.

86.

66. y  1 68. y  2 70. x  2 72. x  4

88. The ordered pair (1, 2) satisfies both equations.

74. a. The points A and E satisfy equation 1. b. The points E and F satisfy equation 2. c. The point E satisfies both equations. d. The points B, C, and D do not satisfy either equation.

76. Answers may vary. There are infinitely many solutions.

90. The graph of y  a is a horizontal line because, for any value of x, the y value will be equal to a.

92. Answers may vary.

94. Answers may vary.

Homework 7.2 2. The slope of a line is the rate of change of y with respect to x.

4. The absolute value of the slope of a line measures the steepness of the line.

6.

48  4; The rate of change of the ti plant’s height is 4 inches per year.

12

8.

15  5; The rate of change of the temperature is 5F per hour.

3

10.

55  29  0.93; The rate of change of the price of a stamp for a 1-ounce, first-class letter is $0.93 per year.

2019  1991

12.

473  862  55.57; The rate of change of the number of bankrupt borrowers who asked a judge to cancel 2017  2010 their student loans is –55.57 people per year.

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 139 14.

16.

18.

20.

62  79  3.40; The rate of change of the percentage of Americans who say improving the job situation 2018  2013 should be a top priority for the President and Congress is –3.40 percentage points per year.

511, 250  249,900  $130, 675; The approximate rate of change of median price with respect to the number of 42 bedrooms is $130,675 per bedroom.

549  246  8.42; The rate of change of the number of calories with respect to the number of carbohydrates is 64  28 about 8.42.

62.75  47.07  1.57; The approximate rate of change of best viewing distance with respect to screen size is 40  30 1.57 inches per inch.

22. a. The explanatory variable is t, the number of hours the person has driven. The response variable is G, the volume of gasoline in gallons. c. b. t G 0

12

1

10

2

8

3

6

4

4

5

2

d. The association is exactly linear, negative, and there are no outliers. The value r  1 supports the conclusion that the association is exactly linear and negative. e. Answers may vary.

24. a.

b. There are no outliers; the association is exactly linear and positive; r  1, which confirms that the association is exactly linear and positive. c.

2350  1950  100 millimeters per inch 66  62

d.

3550  2750  100 millimeters per inch 78  70

3550  1550  100 millimeters per inch 78  58 The three results are equal. f. Answers may vary.

e.

26. a. The association is positive. This means that as the year increases, the total wealth of Americans also increases. b. The rate of change of the total wealth of Americans is about 5 trillion dollars per year.

28. a. The association is negative. This means that as the year increases, the 200-meter run record time decreases. b. The slope of the line is about –0.0275. This means that the men’s 200-meter run record time decreases by about 0.0275 seconds per year.

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140 ISM: A Pathway to Introductory Statistics 28. (continued) c. 25(0.0275)  0.6875; According to the model, the men’s record time decreased by about 0.6875 seconds over 25 years. d. Extrapolating the slope of the model into the future does not produce a reliable result.

30. a. The association is positive. This means that as the number of firefighters increases, the number of helicopters deployed also increases. b. The rate of change of the number of helicopters deployed with respect to the number of firefighters is around 2 helicopters per 6 firefighters. c. An association between variables does not guarantee that one of the variables causes the other.

32. a. The association is negative. This means that as the amount of herbicides used increases, and the number of bee colonies decreases. b. The slope of the model is about 0.005. This means that a 1-million-pound increase in the amount of herbicides used results in a about a 5000 decrease in the number of bee colonies. c. 300(0.005)  1.5; According to the model, a 300-million-pound increase in the amount of herbicides used results in an approximately 1.5 million decrease in the number of bee colonies. d. An association between variables does not guarantee that one of the variables causes the other.

34.

12  3  3; The slope is positive, so the line is increasing.

52

36.

28  3; The slope is negative, so the line is decreasing.

75

38.

57 1   ; The slope is negative, so the line is decreasing.

9 1 4

40.

2  6  2; The slope is negative, so the line is decreasing.

1  (3)

42.

6  (3) 1   ; The slope is negative, so the line is decreasing.

82 2

44.

2  (8)  5; The slope is positive, so the line is increasing. 4  (6)

46.

3  (2) 1   ; The slope is negative, so the line is decreasing. 1  (5) 4

48.

10  (5) 5   ; The slope is negative, so the line is decreasing. 2  4 2

50.

9  0 3  ; The slope is positive, so the line is increasing. 6  0 2

52.

6  (6)  0; The slope is 0, so the line is horizontal.

3  (4)

54.

72 5  ; The slope is undefined, so the line is vertical.

44 0

56.

5.5  4.8  0.58; The slope is negative, so the line is decreasing. 3.1  (1.9)

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 141 58.

2.7  (6.8)  0.79; The slope is negative, so the line is decreasing.

4.5  9.7

60.

6.13  (8.54)  0.59; The slope is negative, so the line is decreasing. 7.29  (3.22)

62. The slope of the line is 2.

70. Answers may vary.

5 64. The slope of the line is  .

3

72. Answers may vary.

66. The slope of the line is

74. Answers may vary.

40 4  .

0  (3) 3

68. Answers may vary.

76. The student was not consistent in subtracting the x and y values of one point from the other. The correct slope 94 5  . is 3 6 3 78. The student forgot to simplify the fraction. The correct slope is

1 .

2

80. Answers may vary.

82. a. Answers may vary. b. No, the lines have the same steepness. c.

2  2  2; The absolute values of the slope of each line are equivalent.

84. Answers may vary.

88. Answers may vary.

86. Answers may vary.

90. Answers may vary.

92. a.

b.

c.

d. A larger positive slope leads to a steeper positive line on the graph. e.

94. Answers may vary.

96. a. Answers may vary. b. Answers may vary. c. Answers may vary.

d. Answers may vary. e. Answers may vary.

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142 ISM: A Pathway to Introductory Statistics 98. a.

b. There are only a certain number of airports that the flights can go to, and the distance to these airports does not change. However, the time it takes to get there can vary based on a variety of factors. c. Answers may vary. It is impossible to be sure because there is no way to know how many dots are overlapping each other. d. The outliers tend to be at a time above the other data points. This means that for a given distance, flights with unusual times tend to arrive later than expected. This makes sense because planes very rarely arrive at their destination faster than they are scheduled to, but delays can happen for a variety of reasons. e. LIT: Bill and Hillary Clinton National Airport; Little Rock, Arkansas. f. Answers may vary. g. Answers may vary.

Homework 7.3 2. For a linear equation of the form y  a  bx, the y-intercept of the line is a.

4. It is true that one way to graph an equation of the form y  a  bx is to first plot the y-intercept and then use the slope to plot another point.

6.

14.

8.

16.

10.

18.

12.

1 .

5 The y-intercept of the equation is (0, 2).

20. The slope of the equation is

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 143 3 22. The slope of the equation is  .

2 The y-intercept of the equation is (0, 1).

5 .

2 The y-intercept of the equation is (0, 3).

24. The slope of the equation is

32. The slope of the equation is 1.

The y-intercept of the equation is (0, 2).

34. The slope of the equation is 4.

The y-intercept of the equation is (0, 0).

36. The slope of the equation is 1.

The y-intercept of the equation is (0, 0).

4 26. The slope of the equation is  .

5 The y-intercept of the equation is (0, 0).

38. The slope of the equation is 0.

The y-intercept of the equation is (0, 2).

28. The slope of the equation is 2.

The y-intercept of the equation is (0, 4).

40. The slope of the equation is 0.

The y-intercept of the equation is (0,1).

30. The slope of the equation is 3.

The y-intercept of the equation is (0,5).

42. a. The sign of b is negative. The sign of a is negative. b. The sign of b is positive. The sign of a is positive. c. The value of b is 0. The sign of a is positive. d. The sign of b is negative. The sign of a is negative.

44. Answers may vary. The graph will slope upward and cross the y-axis below 0.

46. Answers may vary. The graph will slope downward and cross the y-axis at 0.

48. The graph will be horizontal and cross the y-axis at 0. An equation for this graph is y  0.

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144 ISM: A Pathway to Introductory Statistics 50. y  5  2 x 52. y  2 

3 x 4

54. y  1 56. y  1 

58. x  6 60. y  1 1 62. y  1  x 3

2 x 3

64. a.

b.

p  23.57  0.82(7)  29.3; According to the model, in 2017, 29.3% of individuals said they experienced a lot of worry and stress yesterday.

66. a.

b. 1.0  0.7 H  4 0.7 H  3 H  4.30 According to the model, the price of a hot dog at an MLB stadium where a soft drink costs $4.00 will be $4.30.

68. a. The slope of the model is 5.75. This means that, according to the model, an increase in height of 1 inch results in an increase in weight of 5.75 pounds. b. The w-intercept is (0, 236.88). This means that an NBA player that is 0 inches tall would weigh 236.88 pounds. c.

d. w  236.88  5.75(72)  177; According to the model, Chris Paul should weigh about 177 pounds.

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 145 70. a. The explanatory variable is the number of weeks since February 1, t. The response variable is the number of songs that the band knows, n. b. The slope of the linear model is 2. The y-intercept of the linear model is (0,5). c. n  5  2t d.

e. n  5  2(7)  5  14  19; The band will know 19 songs on March 22.

72. a. The person always sets the burner to the same setting, which uses a fixed amount of propane. b. The slope of the linear model is 0.13. This means that for every minute of cooking, 0.13 ounce of propane is used. c. y  16.4  0.13 x d.

e. The arrows are shown on the graph in part (d). There will be 6 ounces of propane remaining in the tank 1 after 80 minutes or 1 hours.

3 74. a. The explanatory variable is u, the number of units of courses the student is taking. The response variable is T, the total cost of enrollment and health fees. b. The slope of the linear model is 46. This means that for every unit of course, the total cost increases by $46. c. T  20  46u d.

e. T  20  46(15)  710; The total one semester cost of tuition plus health fee for 15 units of classes is $710.

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146 ISM: A Pathway to Introductory Statistics 76. a. The rate of change in the conversion from Celsius to Fahrenheit is constant. b. The F-intercept of the model is (0,32). This means that a temperature in degrees Celsius of 0 is equivalent to a temperature of 32 degrees Fahrenheit. d. c. F  32  1.8C

e. F  32  1.8(30)  32  54  86; A temperature of 30C converts to 86F. 78. a.

b. The association is positive. This means the number of countries that require picture warnings on cigarette packages increased over time. c. The model is shown in part (a). The line comes close to the data points. d. The n-intercept of the linear model is (0, 11.53). This means that in the year 2005, there were 11.53 countries that required picture warnings on cigarette packages, which is model breakdown. e. n  11.53  9.57(14)  122.45; According to the model, about 122 countries required warnings on cigarette packages in 2019. The residual is 118  122.45  4.45, which means the model overestimated the number of countries in 2019 by 4.45.

80. a.

b. The association is linear, strong, positive, and there are no outliers, as supported by r  0.985. c. The model is shown in part (a). The line comes close to the data points. d. The slope of the linear model is 2.02. This means that, according to the model, a 1 year increase in age results in about a 2.2% increase in the percentage who have mild cognitive defects. e.

p  25  2.02(43)  61.86; According to the model, 61.86% of 43-year-old childhood brain cancer survivors have mild cognitive defects.

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 147 82. a.

b. The association is linear, strong, negative, and there are no outliers, as supported by r  0.844. c. The model is shown in part (a). The line comes close to the data points. d. The slope of the linear model is 2028. This means that, according to the model, a 1 year increase in age of a Subaru Outback results in about a $2028 decrease in the mean price. e. p  31,980  2028(6)  19,812; According to the model, the mean price of a 6-year-old Subaru Outback is about $19,812.

84. a. The graph is shown below. The line comes close to the data points.

b. The slope of the model is 6.92. This means that every year after 2005, Costco’s annual net sales will increase by about 6.92 million dollars.

99.1  60.2  6.48 7 1 141.6  99.1 2012–2018:  7.08 13  7 141.6  60.2 2006–2018:  6.78 13  1 The slope found in part (b) is higher than the rate of change of the values for 2006–2012 and 2006–2018, and lower than the rate of change for 2012–2018.

c. 2006–2012:

d. n  49.26  6.92(19)  180.7; According to the model, Costco’s net sales in 2024 will be about 180.7 million dollars. Extrapolation was used to find this value because it is beyond the scope of the data used to create the model. There is little to no faith in this prediction because extrapolation was used.

86. The student is incorrect because the slope is a negative value.

88. Answers may vary.

x

y

3

1

0 1 3 3

90. a.

c. For each solution, the y-coordinate is one less 2 than  times the x-coordinate.

3 92. The student has reversed rise and run.

b. Answers may vary.

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148 ISM: A Pathway to Introductory Statistics b. The equation of the line is y  5  3 x.

94. a.

96. a. The slope of each line is undefined. b. The slope of the graph of any equation of the form x  c, where c is a constant, is undefined.

98. The x-coordinate of the x-intercept is negative. Answers may vary.

100. Answers may vary.

Homework 7.4 2. A function is a relation in which each input leads to exactly one output.

4. A linear function is a relation whose equation can be put into the form y  a  bx.

6. a. Relation 3 and Relation 4 can be functions because there is only one y value for each x value given in the relation. Relation 1 cannot be a function since an x-value of 3 corresponds to y-values of 5 and 6. Relation 2 cannot be a function since an x-value of 5 corresponds five different y-values. b. Relation 3 could be a linear function because the increase in the y values is constant for each increase in the x values given.

8. Yes, the relation could be a function as long as there is only one output for every input.

10. No, because the input 4 leads to two different outputs, 5 and 9.

12. The graph is not a function because there are values of x that correspond to multiple values of y.

14. The graph is a function because there is only one value of y corresponding to each value of x.

16. The graph is a function because there is only one value of y corresponding to each value of x.

18. The graph is not a function because the value of x corresponds to infinitely many values of y.

20. Yes; the equation y  8  3 x is of the form y  a  bx, so the relation is a (linear) function.

22. Yes; the equation y  1 is of the form y  a  bx, so the relation is a (linear) function.

24. No; a vertical line intersects the graph of x  0 at more than one point.

26. a. Answers may vary. b.

c. For each input-output pair, the output is 2 more 3 than  times the input.

2

28. The domain of the function is 2  x  3 or [2, 3]. The range of the function is 2  y  4 or [2, 4].

30. The domain of the function is 4  x  5 or [4,5]. The range of the function is 3  y  4 or [3, 4].

32. The domain of the function is 3  x  5 or [3,5]. The range of the function is 0  y  4 or [0, 4].

34. The domain of the function is all real numbers or (, ). The range of the function is y  1 or [1, ).

36. The domain and range of the function are both all real numbers or (, ).

38. P (8) 

1 1 ; The input of 8 leads to the output of .

10 10

40. f (2)  4  6(2)  4  12  16

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 149 52. h(9)  4

5 5 42. f    4  6    4  15  11 2 2

44. h(4) 

54. f (17.28)  183.22  5.95(17.28)  80.40

4  3(4) 4  12 16 8    2  5(4) 2  20 18 9

58. f (2)  4

46. g (3)  2(3)  3(3) 2  6  17  21

60. f (0)  2

48. f (4)  7  2(4)  7  8  15 50. g (1)  2(1)  3(1) 2  2  3  5 66. a.

56. f (4)  0

62. f (3)  1 64. The range is all real numbers or (, ).

f (t )  431.27  5.11t

b.

f (17)  431.27  5.11(17)  344; There were about 344 drive-in movie sites in 2017.

c.

f (0)  431.27  5.11(0)  431.27; There were about 431 drive-in movie sites in 2000.

68. a.

f ( E )  212  5.9 E

b.

f (5)  212  5.9(5)  212  29.5  182.5; This means that, according to the model, the boiling point at an elevation of 5000 meters is 182.5F.

c.

f (0)  212  5.9(0)  212  0  212; This means that, according to the model, the boiling point at an elevation of 0 meters is 212F.

70. a.

b. The model is shown in part (a). The model comes close to the data points. c.

f (t )  55.79  0.41t

f (22)  55.79  0.41(22)  46.8; This means that, according to the model, about 46.8% of teachers were union members in the year 2017. e. According to the model, the percentage of teachers who were in unions decreased by about 0.41% per year.

d.

72. a.

b. The association is linear, strong, positive, and there are no outliers, as supported by r  0.997. c. The model is shown in part (a). The model comes close to the data points. d.

f (23.5)  581.49  2.17(23.5)  632.485; The residual is 637  632.485  4.515. The observed mean credit score is 4.515 greater than the predicted mean credit score.

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150 ISM: A Pathway to Introductory Statistics 72. (continued) e.

f (119)  581.49  2.17(119)  839.72; Extrapolation was used to find this result because the value found was beyond the age groups used to create the model. There is little to no faith in this prediction because extrapolation was used.

74. The scatterplot and graph of the model are shown below. The line comes close to the data points.

b. The association is positive. The longer the triple-hop distance, the higher the vertical-jump height. c.

f ( H )  0.88  0.089 H

e. The slope of the graph is 0.089. The vertical-jump height increases 0.089 centimeters for each increase of 1 centimeter in triple-hop distance. f.

f ( H )  0.88  0.089(553.2)  50.1; According to the model, a soccer player with a triple-hop distance of 553.2 cm will have a vertical-jump height of 50.1 cm. The residual is 37.7  50.1  12.4. The predicted value is 12.4 cm larger than the observed value.

76. a. The association is negative and exponential. b. The set of data points is not a function. At least two inputs have more than one output. c. The red model is a function since each input has only one output. d. The scatterplot describes the actual values of the variables. e. The model summarizes the situation more simply.

78. a. The graph comes close to the points in the scatterplot.

b. The graph comes close to the points in the scatterplot.

c. C (47)  38.61  1.15(47)  38.61  54.05  92.66; The value C (47) means that according to the model, the mean annual consumption of chicken in 2017 was 92.66 pounds.

R (47)  142.16  0.78(47)  142.16  36.66  105.5; The value R (47) means that according to the model, the mean annual consumption of red meat in 2017 was 105.5 pounds. d. The slopes indicate that the mean annual consumption of chicken is increasing, while the mean annual consumption of red meat is decreasing. e. 1.15  ( 0.78)  0.37; This means that the mean annual consumption of either red meat or chicken is increasing by 0.37 pounds per person each year.

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 151 80. a. The distance in miles is being treated as the explanatory variable. b. The association is linear, strong, positive, and there are no outliers.

1  7.14. This is an estimate of the mean 0.14 speed of all the airplanes because speed is the amount of distance traveled over a given amount of time.

This value is equivalent to 7.14(60)  428.4 miles per hour.

c. The slope of the model is 0.14. The reciprocal of the slope is

f (1267)  5.39  0.14(1267)  182.77; According to the model, the flight from Atlanta to Albuquerque will take about 183 minutes. e. 182.77  179  3.77; The velocity was a bit more than the result in part (c) in order to arrive sooner than predicted. The point that represents the flight is a bit below the graph of f because the time is lower than would be predicted by the function.

d.

82. a.

f (t )  6.1  5t

b. The slope of the function is 5. Annual U.S. retail sales of herbal supplements increase by $5 billion per year. c.

f (3)  f (2)  6.1  5(3)   6.1  5(2)   21.1  16.1  5; The result is equal to the slope. If the explanatory

variable increases by 1, then the response variable changes by the slope. d.

e.

84. a.

f (3)  6.1  5(3)  21.1; According to the model, annual U.S. retail sales of herbal supplements were $16.1 billion in 2016.

26.1  0.52; U.S. retail sales of herbal supplements were $0.52 billion, or $520 50 million, per state in 2017. Answers will vary.

f (4)  6.1  5(4)  26.1; f (t )  18.8  0.7t

b.

f (3)  18.8  0.7(3)  16.7; According to the model, there were $16.7 billion in retail sales of jeans in the United States in 2016.

c.

f (0)  18.8  0.7(0)  18.8; According to the model, there were $18.8 billion in retail sales of jeans in the United States in 2013.

86. a.

f (t )  8.6  0.86t

f (5)  8.6  0.86(5)  4.3; According to the model, the number of households that watched the Miss Universe Pageant was about 4.3 million in 2018. c. The slope of the function is 0.86. The number of households that watch the Miss Universe Pageant decreases by about 860,000 households per year.

b.

d. The n-intercept of the function is (0,8.6). 8.6 million households watched the Miss Universe Pageant in 2013.

88. a.

f (t )  12 

12 t 5

b.

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152 ISM: A Pathway to Introductory Statistics 88. (continued) c. The domain of the model is 0  t  5 or [0,5] because the person starts eating at t  0 and finishes eating at t  5. The range of the model is 0  f (t )  5 or [0,12] because when the person starts eating, there are 12 ounces of ice cream, and when the person finishes eating there are 0 ounces of ice cream remaining.

90. Answers may vary. The relation is not a function because there is a value of x that corresponds to multiple values of y.

92. Answers may vary.

94. Answers may vary.

96. It would make more sense to use the income amount as the explanatory variable and the percentage of adults as the response variable. Therefore, the function that best describes the association would be f (d )  p.

98. a.

f (5)  f (4)  [2  3(5)]  [2  3(4)]  17  14  3; This value is equal to the slope.

b.

f (7)  f (6)  [5  2(7)]  [5  2(6)]  19  17  2; This value is equal to the slope.

c.

f (3)  f (2)  [14(3)]  [1  4(2)]  13  9  4; This value is equal to the slope.

d. The results suggest that, for any linear function of the form y  a  bx, f (n)  f (n  1)  b.

100. The student forgot that the negative is part of the number being squared. The correct solution is g (5)  (5) 2  25.

102. The slope of the graph of the linear function is

5  1 6 3   . 2  (6) 4 2

104. The y-intercept of the graph is (0, 1); the x-intercept of the graph is (5, 0).

Chapter 7 Review Exercises

1. 9  3  2(3) 9  3 6 99

2  3  2(1)

5  3  2(4)

2  3 2

5  3  8

2 1

5  5

The ordered pairs (3,9) and (4, 5) satisfy the equation y  3  2 x.

2. The point (2,3) is on the graph, so y  3.

3. The y-intercept of the line is (0, 2), so y  2.

4. The point (4, 4) is on the graph, so x  4.

5. The x-intercept of the line is (4, 0), so x  4.

6. a.

b. The association is negative; the weekly time young adults spend watching traditional TV decreased over time. c. The model is drawn on the graph in part (a). The line comes close to the data points.

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 153 6. (continued) d. w  29.06  2.08(7)  14.5; According to the model, in 2017, young adults spent about 14.5 hours watching traditional TV. The residual is 14.5  14.5  0. The observed weekly time young adults spend watching traditional TV in 2017 is equal to the predicted one e. w  29.06  2.08(13)  2.02; According to the model, in 2023, young adults will spend 2.02 hours watching traditional TV. There is little to no faith in this prediction because extrapolation was used.

7.

6  1.5 degrees Fahrenheit per hour 4

8.

12.1  16.2  $1.03 billion per year 2019  2015

9. a. The explanatory variable is income. The response variable is the percentage of American adults who are confident they will retire ahead of their schedule. b. The association is negative. The greater the income that an American adult earns, the less confident the adult will be that he or she will retire ahead of schedule. c. 23.4% of American adults who earn $50 thousand annually are confident they will retire ahead of their schedule. The residual is 22  23.4  1.4%. The observed percentage of American adults who are confident they will retire ahead of their schedule is 1.4 percentage points less than the predicted percentage. d. The slope is 0.13. For American adults who earn $1 thousand more than American adults with a certain income, the percentage of them who are confident they will retire ahead of their schedule is 0.13 percentage point less than for Americans who earn $1 thousand less. e. 6.5 percentage points 10.

1  (3) 1  ; The slope is positive, so the line is increasing.

8 4 2

11.

5  (3) 1   ; The slope is negative, so the line is decreasing. 4  (10) 3

12.

3  7 10  ; The slope is undefined, so the line is vertical. 4  (4) 0

13.

22  0; The slope is zero, so the line is horizontal. 1  (5)

14.

4.77  (2.38)  0.94; The slope is positive, so the line is increasing. 1.16  ( 8.74)

15. Answers may vary.

16.

2 18. The slope is  ; the y-intercept is (0, 1).

5

17.

19. The slope is

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2 ; the y-intercept is (0, 0).

3

154 ISM: A Pathway to Introductory Statistics 20. The slope is 3; the y-intercept is (0,1).

22.

23.

21. The slope is 1; the y-intercept: (0, 2).

24. a. Answers will vary. b.

c. For each solution, the y-coordinate is 1 more than 2 times the x-coordinate.

25. The line is vertical with an x-intercept of (0,5), so the equation of the line is x  5.

26. y  4 

2 x 3

27. a. The explanatory variable is d, the length of the ride. The response variable is c, the charge for the taxi ride. b. The slope is 2. The charge increases by $2 per mile. c. The c-intercept (0,3.75). The charge for a ride of 0 miles is $3.75, which is model breakdown. Another explanation is that $3.75 is the base fare that added to the $2 per mile charge. d.

e. c  3.75  2(5.7)  15.15; According to the model, the charge for a 5.7 mile taxi ride is $15.15.

28. a. The rate of change in the person’s weight was constant. b. The slope is 4; the w-intercept is (0,195). c. w  195  4t d.

e. 195  4(5)  175; According to the model, the person will weigh 175 pounds after 5 months.

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 155 29. a.

b. There are no outliers; the association is linear, strong, and positive; r  0.997, which confirms that the association is strong and positive. c. The graph of the model is shown on the scatterplot in part (a). The line comes close to the data points. d. The slope is 1.26. The mean tuition increased by $1.26 thousand per year. e. L  0.82  1.26(37)  45.8; According to the model, in 2017, the mean tuition was $45,800. The residual is 46  45.8  0.2. In 2017, the observed mean tuition is $0.2 thousand greater than the predicted mean tuition.

30. Relation 2 is not a function since an input of 4 has two corresponding outputs. Relation 4 is not a function since an input of 2 has five different outputs. Relations 1 and 3 could be functions since in each relation one input results in exactly one output.

31. No; the vertical line x  0 intersects the graph of the function twice, so the vertical line test fails.

32. Yes; the relation is in the form y  a  bx, so the relation is a linear function.

33. No; the graph of a vertical line is not a function.

34. The domain of the function is all real numbers or (, ). The range of the function is y  4 or  , 4.

35. P (2) 

6 3 3  ; The input of 2 leads to the output of .

8 4 4

36. f (3)  3  10(3)  3  30  27

41. f (2)  0

37. g (2)  7  3(2) 2  7  12  5

42. f (4)  1

38. h(4) 

5  2(4) 5  8 13   6  3(4) 6  12 18

43. The domain is 5  x  6 or [5, 6].

44. The range is 2  y  4 or [2, 4].

39. f (0)  1

45. f (2)  4

40. f (3)  3.6

46. f (1)  2

47. a.

f (t )  186  6.25t

b. The slope is 6.25. The mean weekly cost of child care centers increased by $6.25 per year. c.

f (3)  f (2)  186  6.25(3)   186  6.25(2)   $6.25; The result is equal to the slope. If the explanatory

variable increases by 1, then the response variable changes by the slope. d.

f (4)  186  6.25(4)  $211; According to the model, in 2017, the mean weekly cost of child care centers was $211.

e.

f (5)  186  6.25(5)  217.25; According to the model, in 2018, the mean weekly cost was $217.25.

There is little to no faith in this prediction because extrapolation was used.

48. a. The association is positive. The larger a person’s 10-g threshold, the larger that person’s 50-g threshold will be. b.

f (1.5)  0.16  0.86(1.5)  1.13; 1.5 is larger than f (1.5); a person’s 10-g threshold of 1.5 mm is larger than their 50-g threshold of 1.13 mm.

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156 ISM: A Pathway to Introductory Statistics 48. (continued) c. 0.86(2)  0.16  1.56; 2 is larger than f (2); a person’s 10-g threshold of 2 mm is larger than their 50-g threshold of 1.56 mm. f (2)  0.16  0.86(3)  2.42; 3 is larger than f (3); a person’s 10-g threshold of 3 mm is larger than their 50-g threshold of 2.42 mm. e. Most of the data points lie below the line y  x. Most individuals have larger 10-g thresholds than 50-g thresholds; most individuals’ ability to feel gaps between grooves is better when more pressure is applied to their index finger.

d.

Chapter 7 Test

1. The point (3,3) is on the graph, so y  3.

3. The line crosses the y-axis at (0,1).

2. The point (3, 1) is on the graph, so x  1.

4. The line crosses the x-axis at (1.5, 0).

5. a. Explanatory variable: W, the weight in pounds of the turkey; response variable: T, the time it takes to cook the turkey in hours. b.

c. The association is positive; the heavier the turkey, the greater the cooking time. d. The graph of the model is shown on the scatterplot in part (b). The line comes close to the data points. e. T  1.64  0.24(20)  6.44; According to the model, a 20-pound turkey would take 6.44 hours to cook.

6.

75  64.61  $1.30 per year 2018  2010

7. a. There are no outliers; the association is linear, strong, and positive. b. The slope is approximately 23; the median square footage increased on average by about 23 square feet per year. c. 23  15  345 square feet d. We have little or no faith that the square footage will increase at the same rate as in the past. e. The median square footage in 1982 is about 1630; or about 1630 2.7  605 square feet per person. The median square footage in 2016 is about 2400; or about 2400 2.6  925 square feet per person. The square footage per person was lower in 1982 than in 2016. The results were found assuming that the mean household sizes for newly built homes and existing homes were equal in 1982 and in 2016.

8.

2  (8)  3; The slope is positive, so the line is increasing.

53

9.

4  (1) 1   ; The slope is negative, so the line is decreasing.

2  (4) 2

10.

44  0; The slope is zero, so the line is horizontal.

1  (5)

11.

3  (7) 10  ; The slope is undefined, so the line is vertical. 2  (2) 0

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Chapter 7: Graphing Equations of Lines and Linear Models; Rate of Change 157 12.

8.12  (3.27)  1.29; The slope is positive, so the line is increasing.

2.83  ( 5.99)

3 13. The slope is  ; The y-intercept is (0, 2).

2

15. The slope is 0; the y-intercept is (0, 2).

16. The slope is –2; the y-intercept is (0,3).

5 14. The slope is ; the y-intercept is (0, 0).

6

17. The y-intercept is (0,1) and the slope is 2. The equation of the line is y  1 

1 x.

2

18. a. b is positive; a is positive. b. b is zero; a is negative; c. b is negative; a is positive. d. b is negative; a is negative.

19. a. Yes, there is a linear association because the mean number of seats on U.S. domestic flights increased by an approximate constant rate. The slope is 3.2; the mean number of seats on U.S. domestic flights increased by about 3.2 seats per year. b. The n-intercept is (0,102). In 2013, The mean number of seats on U.S. domestic flights was around 102. c. n  102  3.2t d.

e. n  102  3.2(5)  118; In 2018, the mean number of seats on U.S. domestic flights was 118.

20. Yes; the relation is in the form y  a  bx, and any relation in that form is a linear function.

21. The relation is a function because there is no vertical line that intersects the graph of the function at more than one point. The domain of the function is 3  x  5 or [3,5]. The range of the function is 3  y  4 or [3, 4].

22. f (3)  2

25. The range is 3  y  1 or  3,1.

23. f (3)  0

26. f (3)  7  4(3)  7  12  19

24. The domain is 6  x  6 or  6, 6.

27. g (4)  1  5(4)  2(4) 2  1  20  32  11

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158 ISM: A Pathway to Introductory Statistics 28. a.

b. The model is shown on the scatterplot in part (a). The line comes close to the data points. c.

f (t )  30  0.56t

d.

f (48)  30  0.56(48)  3.12; According to the model, 3.12% of presidential-election donation boxes were checked off in 2018. The residual is 4  3.12  0.88. In 2018, the observed percentage of tax returns in which Presidential Election Campaign Fund boxes where checked off is 0.88 percentage point greater than the predicted percentage

e.

f (46)  30  0.56(46)  4.24; According to the model, 4.24% of tax returns had at least one Presidential

Campaign Fund check off in 2016. Therefore, 0.0424(136,500, 000)  5, 787, 600 tax returns had at least one box checked off. If half those returns had one box checked off and half had two boxes checked off, the  136,500, 000   136,500, 000  total money contributed would be   (3)    (6)  $26 million.  2 2

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 159

Chapter 8: Solving Linear Equations and Inequalities to Make Predictions Homework 8.1 2. Associative law for multiplication: a(bc)  (ab)c.

4. Like terms are either constant terms or variable terms that contain the same variable(s) raised to exactly the same power(s).

6. x  8  8  x

32. 6( x  y )  6 x  6 y

8. 6  8w  8w  6

34. ( x  8)(4)  4 x  32

10. pw  wp

36. 4( w  3)  4w  12

12. 7 n  12  n  7  12

38. 8(5  3 x)  40  24 x

14. (8  x)  y  8  ( x  y )

40. 6(2 x  4 y  5)  12 x  24 y  30

16. (7 p) w  7( pw)

42. (3x  7 y  4)  3 x  7 y  4

18. 9  (k  d )  (9  k )  d 20. k ( pw)  (kp) w

5 44.  (30  12 x)  25  10 x 6

22. 3(6 x)  18 x

46. 3x  6 x  9 x

24. (k  8)  1  k  9

48. 6 x  5 x  x 50. p  8 p  7 p

26. 1  (8m  4)  8m  5 2 28. (12 x)  8 x 3

52.

9 2 7 x x  x 5 5 5

54. 8 x  1  3 x  4  5 x  5

 x  6x 30. 6    5 5

56. 5  3x  7 y  6 x  6 y  5  3 x  y

58. 8.7  3.5 y  4.4 x  6.2 y  1.9 x  6.3 x  9.7 y  8.7 60. 5(k  8)  6k  5k  40  6k   k  40 62. 3.8(2.7 x  5.5)  8.4  10.26 x  20.9  8.4  10.26 x  12.5 64. 3a  2(3a  4)  5  3a  6a  8  5  3a  3 66. 8 x  (3x  7 y )  2 y  8 x  3x  7 y  2 y  5 x  9 y 68. 2(7 x  5 y )  5(3x  y )  14 x  10 y  15 x  5 y   x  15 y 70. (6 x  7)  (7  6 x)  6 x  7  7  6 x  0 72. 4 x  3 y  2(5 x  2 y  8)  4 x  3 y  10 x  4 y  16  6 x  y  16 74.

3 1 3 3 1 1 4 2 (t  1)  (t  1)  t   t   t  5 5 5 5 5 5 5 5

1 76. 7 x  (6 x  6)  7 x  2 x  2  5 x  2 3

78. The phrase translates to x  3 x, which simplifies to 2 x.

80. The phrase translates to 6( x  4), which simplifies to 6 x  24.

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160 ISM: A Pathway to Introductory Statistics 82. The phrase translates to x  5( x  2), which simplifies to 4 x  10.

84. The phrase translates to 2 x  9( x  4), which simplifies to 11x  36.

86. The expression translates to “3 times the number, minus 8 times the number.” The expression simplifies to 5 x.

88. The expression translates to “ 2 times the sum of the number and 4.” The expression simplifies to 2 x  8.

90. The expression translates to “the number, minus 8 times the difference of the number and 3.” The expression simplifies to 7 x  24.

92. The expression translates to “2 times the number, plus 6 times the difference of the number and 4.” The expression simplifies to 8 x  24.

94. (6 x  3)  (8 x  9)  14 x  12

96. (5 x  9)  (2 x  6)  3x  15

98. The student should have distributed 2 over the product of x and y, not over x and y individually. The correct simplification is 2 xy.

100. We have distributed 1 over x  5: ( x  5)  1( x  5)   x  5.

102. abc  (ab)c Multiply from left to right.  a (bc) associative law  a (cb) commutative law  (ac)b associative law  (ca)b commutative law  cab Multiply from left to right.

104. (a  b)c  c(a  b)  ca  cb  ac  bc

commutative law distributive law commutative law

106. a. Answers may vary. The result will be equal to the original number. b. Answers may vary. The result will be equal to the original number. c. 2( x  3)  x  6  2 x  6  x  6  x; The result will always be equal to the original number, regardless of the original number.

108. (2 x  3)  2 x  3 3  2 x  2 x  3 (3x  2)  3 x  2 2 x  3 (3x  2)  x  1  3 x  2  x  1  2 x  3 ( x  1)  ( x  2)   x  1  x  2  2 x  3

All the expressions except for (3 x  2) are equivalent.

110. Answers may vary.

112. y  3  2 x

114. y  4  x

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 161 116. Two or more expressions are equivalent expressions if, when each variable is evaluated for any real number for which all of the expressions are defined, the expressions all give equal results. If two expressions are equivalent, then associative, commutative, and distributive laws can be used to rewrite or simplify the expressions to be equal.

Homework 8.2 2. A number is a solution of an equation in one variable if the equation becomes a true statement when the number is substituted for the variable.

4. True. This is the statement of the Addition Property of Equality.

6.

2 x  5  4 2(2)  5  4 1 4 2 is not a solution.

8. 3( x  2)  0 3(2  2)  0 00 2 is a solution.

10. 5  x  3(2 x  1) 5  2  3(2(2)  1) 39 2 is not a solution.

12.

x49 x44  94 x  13

14.

x  1  5 x  1  1  5  1 x  6

16.

3  x  4

24. 24  2 x 24 2 x  2 2 x  12 26. 4 x  6 4x 6  4 4 3 x 2 28. 14 x  6 14 x 6  14 14 3 x 7 30. 5 x  0 5 x 0  5 5 x0 32.

3  4  x  4  4 x 1

18.

x5  0 x55  05 x5

34.

20. 4w  24 4w 24  4 4 w6 22. 5 x  20 5 x 20  5 5 x  4

36.

1 w4 2 1 2 w  24 2 w8 4  x  5 9 9  4  9     x     (5) 4  9  4 45 x 4 2x 5 5 5 2x  (8)   2 2 5 x  20 8 

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162 ISM: A Pathway to Introductory Statistics 38.

40.

44.

56. 7  4 x  1 6  4 x 3 x 2

5x 15  4 8 4  5  4  15     x      5  4  5  8 

x

42.

54. 2 x  7  23 2 x  16 x  8

4 5 b 7 21 7 4 7  5  b     4 7 4  21  5 b 12

58. 2  x  9  x  11 x  11

3 2

60. 5 x  2 x  3  6 3x  3  6 3x  9 x3

x  2 1x  2 1x 2  1 1 x  2 x  

62. 4 x  3  9 x  22 5 x  3  22 5 x  25 x5

3 4

64. 10  3x  6  7 x 10  4 x  6 16  4 x x  4

3 4  3 1( x)     (1)  4 x

46.

x  7.5  2.8 x  7.5  7.5  2.8  7.5 x  10.3

48.

5.27  x  28.85 5.27  28.85  x  28.85  28.85 x  34.12

66. 7 x  5  4 x  17 3 x  5  17 3 x  12 x4 6 w  3  4w  17

68.

10 w  3  17

50. 2.9 w  13.34 2.9 w 13.34  2.9 2.9 w  4.6

10w  20 w  2

70. 8  2 x  2  3 x  x 2 x  6  4 x 6x  6 x 1

52. 5 x  1  9 5 x  10 x2 271  1004; The total number of respondents was about 1004.

72.

0.27 74.

5645  13,128; About 13,128 students were surveyed.

0.43

76.

105  218.75; There were about 218,750 women diagnosed to have lung or bronchus cancer.

0.48

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 163 78.

4.19  11.0; Total revenue from all energy drinks was about $11 billion.

0.38

80. When y  2, x  2. The solution is x  2.

88. x  2

82. When y  2, x  6. The solution is x  6.

90. w  2

84. x  7

92. When y  7, x  2. The solution is x  2.

86. x  4

94. When y  8, x  1. The solution is x  1.

96. In the second step, the student did not use the Addition Property of Equality correctly, since they did not subtract 6 from both sides of the equation. The correct solution is as follows. x6  9 x66  96 x0  3 x3 98. 11  4 is equal to 7.

11  4  4 is equal to 7  4.

11  0 is equal to 11.

11 is equal to 11.

100. While the student’s solution is correct, the Multiplication Property of Equality could have been used to solve for x with one application instead of two.

3 x2 7 7 3 7  x  2 3 7 3 14 x 3 102. 2 x  10

x2 7

104. Answers may vary.

x5

x5

106. Answers may vary.

Yes, the equations are equivalent. x( x  1)  6

108.

2  x( x  1)  2  6 2 x( x  1)  12

Yes; the two equations are equivalent due to the Multiplication Property of Equality.

110. a.

2x  7

b.

2x 7  2 2 7 x 2

5x  9 5x 9  5 5 9 x 5

112. Answers may vary.

114. Multiplication by

1 is equivalent to dividing by c. c

Homework 8.3 2. False. The result of simplifying an expression in one variable is an expression.

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c.

cx  p cx p  c c p x c

164 ISM: A Pathway to Introductory Statistics 4. True. When making a prediction about the response variable of a linear model, we substitute a chosen value for the explanatory variable in the model. Then we solve for the response variable.

6. 3( x  4)  2 x  2 3 x  12  2 x  2 5 x  12  2 5 x  10 x2 8. 3  4(3 p  2)  7  (9 p  1) 12 p  5  9 p  8 3 p  13 13 p 3

18.

10  36 x  3x  2 33x  12 4 x 11

20.

10. 2(5 x  3)  (4 x  1)  5( x  2) 14 x  5  5 x  10 19 x  15 15 x 19 12.

14.

x 1 2   9 3 9 2  x 1 9     9  9 3 9 x3 2 x5

3x 1  1 8 2  3x 1  8      8 1  8 2

22.

24.

5( x  2)  4 x 3 5 x  10  12 x 17 x  10 10 x 17 3x  2 6 x  3  2 5  3x  2   6x  3  10    10    2   5  15 x  10  12 x  6 3 x  16 16 x 3

x4 3 5 1 t  t 8 6 4 3   5 1 24   t   24   t   8  6 4 9t  20t  6 11t  6 6 t 11

2 3 x  4  2x  5 4 2 3    20   x  4   20   2 x   5   4 8 x  80  40 x  15 32 x  65 65 x 32

3x  4  8 3x  12

16.

5 9 3 1  x  x 4 2 8 4 5 9 3 1    8   x  8  x   4 2  8 4

26.

2 p  4 5 p  7 11   3 6 12 2p  4 5p  7  11  12      12     3 6  12 8 p  16  10 p  14  11 2 p  30  11 2 p  19 19 p 2

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 165 28. 0.6 x  0.1  0.4 6x 1  4 6x  5 x  0.83

40. 3( x  2)  (7 x  2)  4(3x  1) 3 x  6  7 x  2  12 x  4 4 x  8  12 x  4 16 x  12 3 x 4

30. 8.25 x  17.56  4.38 x  25.86 12.63 x  17.56  25.86 12.63 x  43.42 x  3.44

42.

32. 3.2 x  0.5(7.3  x)  4.7  6.4( x  2.1) 2.7 x  3.65  6.4 x  18.14 9.1x  14.49 x  1.59 x  87.1 12.7 20.89  x  87.1 x  107.99

7 5 1 3 x   x 2 6 3 4 7 5   1 3  12   x    12    x  2 3 4  6 42 x  10  4  9 x 33 x  14 14 x 33

34. 1.645 

44.

x  50.92 8.39 19.52  x  50.92 x  31.40

36. 2.326 

7 5 1 3 x   x 2 6 3 4 14 5 2 3  x   x 4 6 6 4 17 7  x 4 6

38. 3( x  2)  (7 x  2)  4(3 x  1)  3x  6  7 x  2  12 x  4  16 x  12 46. The graph of y  2 

3 3 x contains the point (4, 4). The solution to 2  x  4 is x  4.

2 2

48. The graph of y  2  50. The graph of y  2 

1 1 x contains the point (2, 3). The solution to 2  x  3 is x  2.

2 2

3 3 x contains the point (0, 2). The solution to 2  x  2 is x  0.

2 2

52. x  2.67

56. x  1.58

54. x  2.48 58. For y  7  3 x, an output of 4 corresponds to an input of 1. The solution is x  1.

60. For y  15  5 x, an output of 25 corresponds to an input of 2. The solution is x  2.

62. f (2)  5  2(2)  5  4  1

64.

f ( x)  2 5  2x  2 2 x  3 3 3 x  2 2

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166 ISM: A Pathway to Introductory Statistics 66.

68. f (17.28)  183.22  5.85(17.28)  82.13

4 3 4 5  2x   3 4 2 x    5 3 19 2 x   3 19  19 x 3  2 6 f ( x)  

70.

f ( x)  72.06 183.22  5.85 x  72.06 5.85 x  111.16 x  19.00

72. a. n  273.6  15.3t b. n(3)  273.6  15.3(3)  319.5; In 2016, there were 319.5 million visits to national parks. c. 273.6  15.3t  335 15.3t  61.4 61.4 t 4 15.3 There will be 335 million visits to national parks in 2017.

74. a. R  5.74  0.21t ; R represents number of Radio Shack stores, in thousands; t represents the number of years since 2010. b. 5.74  0.21t  5.1 0.21t  0.64 0.64 t  3.05 0.21 There were 5.1 thousand Radio Shack stores in 2013. c. R (2)  5.74  0.21(2)  5.32; In 2012, there were 5.32 thousand Radio Shack stores. d. R (4)  5.74  0.21(4)  4.9; There were an additional 4.9  3.96  0.94 thousand, or 940 store closures.

76. a. b.

f  20  1.15t f (8)  20  1.15(8)  10.8; In 2018, 10.8% of adult Americans were uninsured.

c. 20  1.15t  13 1.15t  7 7 t 6 1.15 13% of adult Americans were uninsured in 2016.

78. a. F  3.91  2.31n; F represents the total fare, in dollars; n represents the miles traveled. b. F (3.5)  3.91  2.31(3.5)  $12.00; The fare was $12.00. c. 3.91  2.31n  40.87 2.31n  36.96 36.96 n  16 2.31 The ride was 16 miles.

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 167 80. Let x be the was the U.S. planned St. Patrick’s Day spending in 2016. x  0.227 x  5.3 1.227 x  5.3 5.3 x  4.3 1.227 The U.S. planned St. Patrick’s Day spending in 2016 was about $4.3 billion.

82. Let r be Facebook’s revenue in 2016. r  0.471r  40.6 1.471r  40.6 40.6 r  27.6 1.471 Facebook’s revenue in 2016 was about $27.6 billion.

84. Let n be the number of acres of farmland in the United States in 2000. n  0.037 n  910 0.963n  910 910 n  945 0.963 The number of acres of farmland in the United States in 2000 was about 945 million acres.

86. Let n be the number of Walmart store openings in the 2018 fiscal year. n  0.564n  24 0.436n  24 24 n  55 0.436 The number of Walmart store openings in the 2018 fiscal year was about 55.

88. a. The model comes close to the data points.

b. The p-intercept is (0, 20.84). The total annual revenue (in billions of dollars) of the pet industry in 2000 was $20.84 billion. This is extrapolation, so not much faith should be placed in this value. c.

p(17)  20.84  2.77(17)  $67.93 billion

d. 20.84  2.77t  50 2.77t  29.16 29.16 t  10.5 2.77 The total annual revenue of the pet industry will be $50 billion in 2011. e.

p(15)  20.84  2.77(15)  $62.4 billion; The residual is 60.3  62.4  2.1, the observed revenue is $2.1 billion less than the predicted revenue.

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168 ISM: A Pathway to Introductory Statistics 90. a.

b. The model in shown on the scatterplot in part (a). c. 0.78  0.86 L  2.92 0.86 L  2.14 L  2.48 d. The slope is 0.86. The median GPA of more-affluent high schools increases by 0.86 for each increase of 1 for the less-affluent high schools. e. The study is observational, which means that we cannot assume causation. Answers may vary.

92. a.

b. The association is positive; as American adults get older, they tend to be more likely to have earned over $200,000 at least once. This makes sense because once someone has earned over $200,000 at least once, they keep counting towards the percentage, so the percentage cannot go down. c. The model is shown on the graph in part (a). Yes, the model comes close to the data points. d. 23.46  0.93(30)  4.4% e. 23.46  0.93a  25 0.93a  48.46 a  52 About one-fourth of 52-year-old American adults have earned an annual salary over $200,000 at least once.

94. a.

b. The breed with a mean weight of 55 pounds and a mean life expectancy of 10 years may be an outlier.

The association is linear, fairly strong, and negative; r  0.87. c. The model is shown on the graph in part (a). Yes, the model comes close to the data points. d.

f (8)  14.52  0.035(8)  14.24; The mean life expectancy for a dog breed with a mean weight of 8 pounds is 14.24 years.

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 169 94. (continued) e. 14.52  0.035W  8 0.035W  6.52 W  186.3 pounds The mean weight of a dog breed with a life expectancy of 8 years is about 186.3 pounds.

96. The student did not multiply 3 by 2 when clearing the fractions.

1 5 x3  2 2 1 5 2 x  23  2 2 2 x6  5 x66  56 x  1

98. While both students solved the equation correctly, Student 1 applied the Addition Property of Equality and Multiplication Property of Equality twice each, when they only needed to apply it once each as shown in Student 2’s solution process.

100. The student cannot apply the Multiplication Property of Equality to an expression.

1 1 3 1 4 1 x x   x  x 4 3 3 4 4 3 3 4  x x 12 12 7  x 12

102. 3( x  1)  2  x  7 3x  3  2  x  7 3x  5  x  7 2x  2 x 1 On the left side, 3(1  1)  2  8, and on the right side, 1  7  8. The result satisfies both equations.

104. An equation that contains only integers is easier to solve than an equation containing fractions.

106. a.

b. The equation is shown in part (a). It comes close to the data points. f (30)  40.38  0.52(30)  24.8; According to the model, about 24.8% of adults are obese in a state where 30% of the adults exercise. d. 40.38  0.52 x  30 0.52 x  10.38 x  20

c.

According to the model, about 20% of adults exercise in a state where 30% of the adults are obese.

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170 ISM: A Pathway to Introductory Statistics 106. (continued) e. The slope is –0.52. For each increase of one percentage point in the number of adults that exercise in a state, the percent of adults in that state that are obese decreases by 0.52 percentage point.

Homework 8.4 2. False. In general, solving for a variable in a formula will not change the association between the variables in the formula.

4. If x is nonnegative, then

 x   x.

2

18.

6. P ( E AND F )  P( E ) P( F ) 0.14  P( E )  0.2 P ( E )  0.7 8.

pˆ 

x n

x 1040 x  676

0.65 

E t 2.95  t 

z0 

20.

10. P ( E OR F )  P( E )  P( F )  P( E AND F ) 0.63  0.39  P( F )  0.21 0.63  0.18  P( F ) P ( F )  0.45 12.

x y  1 a b 3 y  1 4 7 y 1  7 4 7 y 4

s n 4.83

r  19.162 2.943 1.14777  r  19.162 r  18.01

22. a. x  b.

x1  x2  x3  x4 4

87  92  86  x4 4 360  265  x4 90 

x4  95

t  1.83

24. a. z 

x    z 58.98  54.7  z (2.6) 4.28  2.6 z 4.28 z  1.65 2.6

x



180  100  5.33 15

 x    z

b.

180  100  z (15) 80  15 z 80 z  5.33 15

16. y  y1  m( x  x1 )

These results are identical.

4  7  m(1  5) 3  4m 3 m 4

26. a. P (C OR T )  P(C )  P(T ) b. P (C )  0.10, and P (C OR T )  0.52.

P (C OR T )  P(C )  P(T ) 0.52  0.10  P (T ) P (T )  0.42

r

0.39 

9 2.95  1.61t

14.

r  r

The probability that an individual randomly selected from the study said that he or she most wanted to be reimbursed for travel expenses is 0.42, or 42%.

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 171 28. yi  2  0  3  6  11

32. (yi ) 2  (2  0  3  6) 2  121

30. xi yi  (4  1  7  5)(2  0  3  6)  187

34. yi2  22  02  32  62  49

36.   0(0.240)  1(0.412)  2(0.265)  3(0.076)  4(0.008)  1.20 19(35  33) 2  24(39  33) 2  15(30  33) 2 19(4)  24(36)  15(9)  3 1 2 76  864  135  2  537.5

38. MST 

40.   [( xi   ) 2 P( xi )]  (10  12) 2 (0.6)  (12  12) 2 (0.1)  (16  12) 2 (0.3)  2.68 se

42. sb1 

 ( xi  x )

2



5 2

(2  4)  (7  4) 2  (3  4) 2

44.    t  t 

 1.34

   t 

46.  x 

60.

n

P ( E AND F ) P( F ) P ( E | F ) P ( F )  P ( E AND F )

48. P ( E | F ) 

62.

P( E OR F )  P ( E )  P ( F ) P ( E OR F )  P ( F )  P ( E )

64.

x    z

56.



y  y1  m( x  x1 ) y  y1  mx  mx1 y  y1  mx   mx1 mx  y  y1 x1  m

  npq

p

x    z x

x y  1 a a x y  a

 2  npq

P ( E AND F )  P( E )  P( F )  P( E OR F )

z

2n1n2 1 n 2n n ur  1  1 2 n n(ur  1)  n2 2n1 ur 

x  a y

P ( E OR F )  P ( E )  P( F )  P( E AND F )  P ( E AND F )  P ( E OR F )  P( E )  P( F )

54.

r

z0 r  r  r

 x n

52.

r  r

z0 r  r  r



50.

z0 

58.

66.

2 nq z0

rs 

n 1 z0 n 1  rs n 1 

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z02

z02

rs

rs2

, so n  2

1

172 ISM: A Pathway to Introductory Statistics P ( E OR F )  P ( E )  P( F )  P( E AND F )

68. a.

 P ( E AND F )  P ( E OR F )  P( E )  P( F ) P ( E AND F )  P( E )  P( F )  P( E OR F )

b. P ( E AND F )  0.3  0.4  0.6  0.1 70. a. x 

x1  x2  x3  x4  x5 5

b. 5 x  x1  x2  x3  x4  x5 x5  5 x  x1  x2  x3  x4

c. x5  5(80)  71  75  88  81  85 72. a. n  85  3.3t n  85  3.3t

b.

n  85  3.3t n  85 t 3.3 85  n t 3.3

85  30  16.67; 2017 3.3 85  40  13.64; 2014 40 executions: t  3.3 85  50  10.61; 2011 50 executions: t  3.3 85  60  7.58; 2008 60 executions: t  3.3 85  70  4.55; 2005 70 executions: t  3.3

c. 30 executions: t 

74. a. F  318  5.96t b. 318  5.96t  350 5.96t  32 32 t  5.34 5.96 The annual revenue from display fireworks was $350 million in 2017.

F  318  5.96t c.

F  318  5.96t 318  F t 5.96 318  350  5.34; The annual revenue from display fireworks was $350 million in 2017.

5.96 e. The results in parts (b) and (d) are the same. The equation in part (d) was easier to use.

d. t 

76. a. There are no outliers. The association is linear, strong, and positive. b. V  1.57  0.093(400)  35.6 cm; The error is about 49  35.6  14 cm. c.

V  1.57  0.093T V  1.57  0.093T T

V  1.57 0.093

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 173 76. (continued) d. The triple-hop distance for a vertical-jump height of 45 cm is

45  1.57  500.75 cm.

0.093

The triple-hop distance for a height of 50 cm is

50  1.57  554.52 cm.

0.093

The triple-hop distance for a height of 55 cm is

55  1.57  608.28 cm.

0.093

The triple-hop distance for a height of 60 cm is

60  1.57  662.04 cm.

0.093

The triple-hop distance for a height of 65 cm is

65  1.57  715.81 cm.

0.093

e. Answers may vary.

78. a.

b. There are no outliers. The association is linear, strong, and positive; r  0.94. c. Yes, the model comes close to the data points. d. The slope is 5.23; each year, mean per-person planned spending increases by $5.23. s  58.18  5.23t

e.

s  58.18  5.23t t

f. The slope is 80. a.

s  58.18 5.23 1  0.19; it takes 0.19 year for the mean per-person planned spending to increase by $1.

5.23

T

P

E

9

10

9 10  90

3

15

3 15  45

6

11

6 11  11

5

13

5 13  65

T

P

T P  E

b. TP  E E T P c. T 

420  30; An employee who was paid $420 at $14 an hour worked 30 hours.

14

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174 ISM: A Pathway to Introductory Statistics 80. (continued) d.

480  40 12 270  15 18 40  15  55 hours

82. y  2  3x; Slope: 3; y-intercept: (0, 2)

5 5 84. y   x; Slope: ; y-intercept: (0, 0) 2 2

86. y  5 

7 7 x; Slope: ; y-intercept: (0,5) 4 4

4 4 88. y  3  x; Slope:  ; y-intercept: (0,3) 3 3

3 3 90. y  2  x; Slope: ; y-intercept: (0, 2) 4 4

92. y  3  2 x; Slope: 2; y-intercept: (0, 3)

94. y  3 

5 5 x; Slope:  ; y-intercept: (0,3) 4 4

96. y  5; Slope: 0; y-intercept: (0, 5)

98. y  4  3x; Slope: 3; y-intercept: (0, 4)

100. y  2 

5 5 x; Slope:  ; y-intercept: (0, 2) 4 4

1 1 102. y  1  x; Slope:  ; y-intercept: (0, 1) 3 3

104. a. M  b.

x2  x3 2

x2  x2 2 x2   x2 2 2

c. R  x4  x1 d. If x1  x2  x3  x4 , then x4  x1  x1  x1  0.

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 175 106. a. 3x  5 y  7 5 y  3 x  7 7 3 y  x 5 5

c. ax  by  c by   ax  c c a y  x b b

3 The slope is  .

5 b. 2 x  7 y  3 7 y  2 x  3 3 2 y  x 7 7

a The slope is  . b

2 The slope is  .

7 108. The student is not correct because the constant term is only the y-coordinate of the y-intercept if the equation is in slope-intercept form.

2 y  4  3x 3 y  2 x 2

The y-intercept is (0, 2).

110. a.

b. Answers may vary. x

y

2

7

0

2

2

3

c. For each solution, the sum of 5 times the x-coordinate and 2 times the y-coordinate is 4.

112. Answers may vary.

114. Answers may vary.

Homework 8.5 2. True. When we multiply both sides of an inequality by a negative number, we reverse the inequality symbol.

4. We say that a number is a solution of an inequality in one variable if it satisfies the inequality.

10. x  5  9 x4

6. 4(3)  7  5  1 4(2)  7  1  1

Inequality: x  4

4(0)  7  7  1

Interval Notation: (, 4]

Only 3 satisfies the inequality.

Graph:

8. 3  9  12  12 2  9  7  8 5  9  4  20 All 3 numbers satisfy the inequality.

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176 ISM: A Pathway to Introductory Statistics 12. x  3  1 x2 Inequality: x  2 Interval Notation: [2, ) Graph: 14. 3x  9 x3 Inequality: x  3 Interval Notation: (3, ) Graph: 16. 2 x  10 x  5 Inequality: x  5 Interval Notation: (, 5) Graph: 18. 2w  2 w  1 Inequality: w  1 Interval Notation: [1, ) Graph: 20. 4 x  2 1 x 2 1 Inequality: x  2 1  Interval Notation:  ,   2 

Graph: 22. 3x  0 x0 Inequality: x  0 Interval Notation: (, 0) Graph: 24.  x  1 x 1 Inequality: x  1

5 26.  x  10 2 x  4 Inequality: x  4

Interval Notation: [4, ) Graph: 28. 4 x  7  15 4x  8 x2 Inequality: x  2 Interval Notation: (, 2) Graph: 30. 8  2 x  6 2 x  2 x 1 Inequality: x  1 Interval Notation: [1, ) Graph: 32. 7 w  4  3w 4w  4 w  1 Inequality: w  1 Interval Notation: (1, ) Graph: 34. 4 x  6  2 x 6x  6 x 1 Inequality: x  1 Interval Notation: (,1) Graph: 36. 2.4 x  5.8  8.92 2.4 x  3.12 x  1.3 Inequality: x  1.3 Interval Notation: [1.3, ) Graph:

Interval Notation: [1, ) Graph:

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 177 38. 5k  1  2k  8 3k  9 k 3 Inequality: k  3 Interval Notation: (,3]

48. 2(5 x  3)  2 x  3(2 x  4)  2 12 x  6  6 x  14 6 x  20 10 x 3

Graph:

Inequality: x  

40. 8  x  2  3 x 2 x  6 x  3 Inequality: x  3 Interval Notation: [3, ) Graph: 42. 5( x  2)  15 5 x  10  15 5 x  25 x5

 10  Interval Notation:   ,    3 

Graph: 50. 3.1(2.7  x)  1.55 8.37  3.1x  1.55 3.1x  9.92 x  3.2 Inequality: x  3.2 Interval Notation: (,3.2) Graph:

Inequality: x  5 Interval Notation: [5, )

10 3

52.

Graph: 44. (t  5)  2 t 5  2 t  3

3 1 1 t  4 2 4 3t  2  1 3t  3 t 1 Inequality: t  1

Inequality: t  3

Interval Notation: (,1]

Interval Notation: (3, )

Graph:

Graph: 46. 4(3x  5)  5(2 x  3) 12 x  20  10 x  15 2x  5 x  2.5 Inequality: x  2.5 Interval Notation: (2.5, ) Graph:

54.

1 2 7  x 4 3 12 3  8x  7 8 x  4 1 x 2

Inequality: x  

1 2

1  Interval Notation:  ,    2

Graph:

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178 ISM: A Pathway to Introductory Statistics 62.  1   2  0

3 5 7 5 56.  x    x 4 2 8 2 6 x  20  7  20 x 14 x  13 13 x 14

Inequality: x  

1   2 64. p1  p2  0 p1  p2

13 14

s n

 x  26.9  2.528 

4.9

20 24.1  x  29.7

20

Interval Notation: (24.1, 29.7)

6 p  2 4 p 1  58.

8 6 18 p  6  16 p  4 2 p  10 p  5

Graph: x t

68.

192.3  2.131 

Inequality: p  5

s

n 24.1

 x  x t

Inequality: 179.0  x  205.6

Graph:

Interval Notation: (179.0, 205.6)

4x  7 2x  3 2   15 10 5 8 x  14  6 x  9  12 14 x  5  12 14 x  17 17 x 14

Inequality: x 

Graph:

17 14

17   Interval Notation:  ,   14 

Graph: pˆ  z  0.45  1.96 

pˆ (1  pˆ )  p  pˆ  z  n

pˆ (1  pˆ ) n

0.45(0.55) 0.45(0.55)  p  0.45  1.96  930 930 0.42  p  0.48

Inequality: 0.42  p  0.48 Interval notation: (0.42, 0.48) Graph:

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s n

 x  192.3  2.131 

15 179.0  x  205.6

Interval Notation: [5, )

70.

 x  x t

Inequality: 24.1  x  29.7

Graph:

60.

n 4.9

26.9  2.528 

13   Interval Notation:  ,    14 

s

x t

66.

24.1 15

Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 179 pˆ  z 

72.

0.62  1.96 

pˆ (1  pˆ )  p  pˆ  z  n

pˆ (1  pˆ ) n

0.62(0.38) 0.62(0.38)  p  0.62  1.96  1375 1375 0.59  p  0.65

Inequality: 0.59  p  0.65 Interval notation: (0.59, 0.65) Graph: 74. 2  x  4  3 2 x7 Inequality: 2  x  7

78. 7  5  2 x  13 2  2 x  8 4  x  1

Interval notation: (2, 7)

Inequality: 4  x  1

Graph:

Interval notation: [4, 1)

76. 5  3x  1  13 6  3x  12 2  x  4

Graph: 80.

Inequality: 2  x  4

3 1  1 x  3 4 4 3  4  x  12 1   x  8

Interval notation: (2, 4)

8  x  1

Graph:

Inequality: 8  x  1 Interval notation: (8,1] Graph:

82.

 E  pˆ  p  E p  E  pˆ  p  E

84.

1.3  x    1.3

  1.3  x    1.3

86. a. Since the difference of the heights of student A and student B is positive, student A is taller. b. Since the difference of the heights of student A and student B is negative, student B is taller. c. Since the difference of the heights of student A and student B could be zero, their heights might be equal.

88. a.

b. The model is shown in part (a). It comes close to the data points. c. The slope is 1.34. This means that each year, the number of countries limiting trans fats increases by 1.34. d. n  8.78  1.34(13)  8.6; According to the model, there were about 8.6 countries that limit trans fats in 2013. The residual is 8  8.6  0.6. The observed number of countries that limit trans fats is 0.6 less than the predicted number of countries that limit trans fats.

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180 ISM: A Pathway to Introductory Statistics 88. (continued) e. 8.78  1.34t  10 1.34t  18.78 t  14 Fewer than 10 countries limited trans fats before 2014, although we have little or no faith in this result for years prior to 2010.

90. a.

b. There are no outliers. The association is linear, very strong, and negative; r  0.995. c. Yes, the model comes close to the data points. d. B  64.98  1.67(26)  21.56; The estimated number of births is about  221, 745 births in 2016.

21.56 births 10, 285, 000 1000 women

e. 64.98  1.67t  30 1.67t  34.98 t  21 There were at least 30 births per 1000 women ages 15–19 prior to 2011, although we have little or no faith in this result for years prior to 1995.

92. a.

b. There are no outliers. The association is linear, strong, and positive; r  0.91. c. Yes, the model comes close to the data points. d. The slope is 1.16. For each foot of clarity at Lake Tahoe in the winter, the clarity of Lake Tahoe in the summer of that year will increase by about 1.16 feet. e. 25.47  1.16(80)  67.33; the summer clarity will be less than 67.33 feet.

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 181 94. a. The scatter plot and model are shown below. The model comes close to the data points.

b. The slope is 0.078. For each additional dollar of per-person library funding, the number of per-person library visits will increase by 0.078. c. N  2.36  0.078(34.28)  5; According to the model, there were about 5 per-person library visits in a state with per-person library funding of $34.28. The residual is 4.50  5  0.5. The observed number of per-person library visits is 0.5 less than the predicted number of per-person library visits. d. 2.36  0.078 F  6 0.078 F  3.64 t  46.7 For per-person library funding levels greater than $46.7, the number of per-person library visits will be no fewer than six visits per person. We should have little or no faith in this result for funding levels above $68 per-person. e. The study is observational, which means that we cannot assume causation. Answers may vary.

96. The student incorrectly reversed the inequality when dividing by a positive number.

4 x  24 4 x 24  4 4 x  6 98. a. Answers may vary. b. Answers may vary.

100. a. The statement is true for all values of c. This is the statement of the Addition Property of Inequalities. b. The statement is not true if c is negative. When c is negative,

a b  . c c

102. Answers may vary.

104. a is greater than b. Subtracting a smaller number from a larger number results in a positive value.

106. Answers may vary.

108. Answers may vary.

Chapter 8 Review Exercises 1. 9  5w

2. 8  pw 3. (2  k )  y 4. b( xw) 5. 3(8 x  4)  24 x  12

6.

4 (15 y  35)  12 y  28 5

7. (3x  6 y  8)  3x  6 y  8 8. 5a  2  13b  a  4b  9  4a  9b  7 9. 5 y  3(4 x  y )  6 x  5 y  12 x  3 y  6 x  18 x  8 y

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182 ISM: A Pathway to Introductory Statistics 10. 2.6(3.1x  4.5)  8.5  8.06 x  11.7  8.5  8.06 x  20.2 11. (2m  4)  (3m  8)  2m  4  3m  8  5m  4 12. 4(3a  7b)  3(5a  4b)  12a  28b  15a  12b  3a  40b

21. 8m  3  m  2  4m 7 m  3  2  4m 11m  5 5 m 11 22.

13. 4( x  7)  4 x  28 14. 7  2( x  8)  7  2 x  16  2 x  9 15. 5( x  4)  5(4  x)  2( x  10)  3x  5 x  20 16. 2  5(3)   3(4(3)  7) 2  15   3(12  7) 13   3(5) 13  15 No, 3 is not a solution.

17.

a  5  12 a  5  5  12  5 a7

18. 4 x  20 4 x 20  4 4 x  5

23. 6(4 x  1)  3(2 x  5)  2(5 x  3) 24 x  6  6 x  15  10 x  6 18 x  21  10 x  6 8 x  15 15 x 8 24.

25.

19.  p  3  p 3  1 1 p3 20.

8 x  7(2 x  3)  x 8 x  14 x  21  x 8 x  13 x  21 21x  21 x 1

7  a  14 3 3 7  3    a    14 7 3  7 a  6

w 3 5   8 4 6 w 19  8 12 38 w 3 3p  4 5p  2 7   2 4 6  3p  4  5p  2 7 12   12   12    2   4  6 6(3 p  4)  3(5 p  2)  2(7) 18 p  24  15 p  6  14 18 p  24  15 p  20 3 p  44 44 p 3

26. The student should have added 5 to both sides of the equation. x5  2 x55  25 x0  7 x7 27. 2.5(3.8 x  1.9)  83.7 9.5 x  4.75  83.7 9.5 x  78.95 x  8.31 Copyright © 2021 Pearson Education, Inc.

Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 183 28.

5 3 1 7 r   r 6 4 6 2 5  3 1 7  12  r   12    12    12  r  6  4 6 2  10r  9  2  42r 32r  11 11 r 32

29.

5 3 1 7 5 21 9 2 r   r  r r  6 4 6 2 6 6 12 12 26 11  r 6 12 13 11  r 3 12

30. When simplifying an expression, you cannot multiply it by any number other than 1. The expression is already fully simplified.

31. x  1.64 32. x  3

35. 3  2 x 

2 3

7 3 7 x 6

2x  

33. x  1 34. 3  2 x  5 2 x  8 x  4 36. a. n  25.3  2.22t

b. n  25.3  2.22(10)  47.5; According to the model, there were 47.5 thousand applications to Stanford University in 2018.

40  25.3  2.22t c.

14.7  2.22t 14.7 t 7 2.22 According to the model, there were 40 thousand applications to Stanford University seven years after 2008, or 1015.

37. a.

f (t )  1.12  0.04t

f (4)  1.12  0.04(4)  1.28; According to the model, the total spending for March Madness was $1.28 billion in 2017. c. 1.2  1.12  0.04t 0.08  0.04t 0.08 t 2 0.04

b.

According to the model, the total spending for March Madness was $1.2 billion two years after 2013, or 1015.

38. Let n represent the number of Americans that played baseball in 2014. n  0.205n  15.9 1.205n  15.9 15.9 n  13.2 million Americans 1.205

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184 ISM: A Pathway to Introductory Statistics 39. a.

b. The association is linear, very strong, positive, and there are no outliers. The value r  0.997 suggests a very strong linear relationship. c. The model is shown in part (a). It comes close to the data points. p  1.23(30)  23.27  13.63; The percentage of 30-year-old men who have high blood pressure is 13.63%.

50  1.23a  23.27 e.

73.27  1.23a 59.57  a

d.

Half of men have high blood pressure at approximately 59.57 years.

40.

x    z 47.3  52.9  z (2.5)

x1  x2  x3 3 b. The cutoff for a B is 83 points. x x x x 1 2 3 3 72  85  x3 80  3 240  157  x3

41. a. x 

5.6  2.5 z 5.6  2.24 z 2.5

x3  83

44. xi yi  (2  7  1  5)(6  3  4  8)  315

42. xi2  22  7 2  12  52  79 43. xi yi  2  6  7  3  1  4  5  8  77 45.  2   

Ot  Et 2 O1  E1 2 O2  E2 2 O3  E3 2 

Et

8  6.5 6.5

2



E1

15  17.8

2



17.8





E2

12  10.7 

E3

2

10.7

1.52 (2.8) 2 1.32    6.5 17.8 10.7 2.25 7.84 1.69    6.5 17.8 10.7  0.94

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 185 46. a.

E t

s n

E s  t n s

b. s 

E n t

3.623 16  8.3 1.746

47. a. n  5.3  0.16t b.

n  5.3  0.16t 0.16t  5.3  n 5.3  n t 0.16

5.3  4.7  4; The mean number of visits per person was 4.7 visits in 2014.

0.16 d. 4.7  5.3  0.16t 0.6  0.16t 0.6 t 4 0.16

c. t 

The mean number of visits per person was 4.7 visits in 2014. e. The results are the same. The equation from part (b) was easier to use. f. It is easier to evaluate the right-hand side of n  5.3  0.16(7) than to solve the equation 7 

5.3  n .

0.16

In 2017, there were 5.3  0.16(7)  4.2 visits.

3 .

2 The y-intercept of the equation is (0,3).

48. The slope of the equation is

50. x  3  4 x  1 Inequality: x  1 Interval notation:  1,   Graph: 51. 4 x  8 x  2

2 49. The slope of the equation is  .

3 The y-intercept of the equation is (0, 5).

52. 5w  3  3w  9 2 w  6 w  3 Inequality: w  3 Interval notation:  3,   Graph:

Inequality: x  2 Interval notation:  2,   Graph:

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186 ISM: A Pathway to Introductory Statistics 53. 3(2a  5)  5a  2(a  3) 6a  15  5a  2a  6  a  15  2a  6 3a  9 a  3

55.

Inequality: 4  2 x  6 Interval notation:  2,3

Inequality: a  3

Graph:

Interval notation:  , 3 Graph: 54.

2b  4 3b  4  3 4 4(2b  4)  3(3b  4) 8b  16  9b  12 b  4 b  4

Inequality: b  4

1  2 x  5  11 4  2 x  6 4  2 x  6

x t

56.

75.9  1.686 

s

n 12.1

   x t

s n

   75.9  1.686 

38 72.6    79.2

12.1 38

Inequality: 72.6    79.2 Interval notation: 72.6, 79.2 Graph:

Interval notation:  4,   Graph: 57. a.

b. The model is shown in part (a). It comes close to the data points. c. The slope is 8.30. This means that each year, the violent crime rate decreases on average by 8.30 violent crimes per 100,000 people. d. v  507.65  8.30(15)  383.2 violent crimes per 100,000 people; The residual is 374  383.2  9.2 violent crimes per 100,000. The observed violent crime rate is −9.2 violent crimes per 100,000 people less than the predicted violent crime rate. e. The violent crime rates were greater than 450 violent crimes per 100,000 people up until 2007, although we have little or no faith in the results for years before 2004.

Chapter 8 Test 1. 3 p  4

2. (3x) y 2 3.  (6 x  9)  4 x  6 3

4. 5(2 w  7)  3(4w  6)  10 w  35  12 w  18  22 w  53

5. (3a  7b)  (8a  4b  2)  3a  7b  8a  4b  2  11a  3b  2 6. 6 x  3  19 6 x  22 22 11  x 6 3

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 187 7.

12. 8.21x  3.9(4.4 x  2.7) 8.21x  17.16 x  10.53 8.95 x  10.53 x  1.18

3 x6 5 5 3 5  x  6 3 5 3 x  10

13. 9(3x  2)  (4 x  6)  27 x  18  4 x  6  23 x  24

8. 9a  5  8a  2 a7

14. 9(3x  2)  (4 x  6)  x 27 x  18  4 x  6  x 23x  24  x 24  22 x 12  x 11

9. 8  2(3t  1)  7t 8  6t  2  7t 10  6t  7t 10  13t 10 t 13

15. No, the solution of an equation is a number.

10. 3(2 x  5)  2(7 x  9)  49 6 x  15  14 x  18  49 8 x  33  49 8 x  82 41 x 4 11.

16. Answers may vary.

17. x  2 18. x  2 19. 1  7  4 x 6  4x 3 x 2

7 3 1 1 x  x 8 10 4 2 7 3 2 5 x  x 8 10 8 10 5 4 x 8 5 8 5 4 8  x  5 8 5 5 32 x 25

20. a. n  2.02  0.13t b. n  2.02  0.13(7)  2.93; There were 2.93 million U.S. patents in force in 2017. c.

2.5  2.02  0.13t 0.48  0.13t t4

There were 2.5 million U.S. patents in force four years after 2010, or in 2014.

21. Let x represent the sales in 2019. x  0.63 x  17.8 0.937 x  17.8 x  19.0 The sales in 2019 were 19.0 million units.

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188 ISM: A Pathway to Introductory Statistics 22.

0.9  0.4  0.7  P( E and F ) 0.9  1.1  P( E and F ) 0.2   P( E and F ) 0.2  P( E and F )

23.

z

G  G

G z G  G  G z G  G  G

24. a.

b. The model is shown in part (a). It comes close to the data points. c. The slope is 4.69. The revenue is decreasing by $4.69 billion per year. s  54.37  4.69t

d.

54.37  s  4.69t t

54.37  s 54.37  s or t  4.69 4.69

54.37  20  7.3; The annual revenues were $20 billion approximately 7.3 years after 2010, or in 4.69 2017.

e. t 

54.37  25  6.3; The annual revenues were $25 billion approximately 6.3 years after 2010, or in 4.69 2016. t

54.37  30  5.2; The annual revenues were $30 billion approximately 5.2 years after 2010, or in 4.69 2015. t

54.37  35  4.1; The annual revenues were $35 billion approximately 4.1 years after 2010, or in 4.69 2014. t

54.37  40  3.1; The annual revenues were $40 billion approximately 3.1 years after 2010, or in 4.69 2013. t

25. The slope is

2 . The y-intercept is (0, 4).

3

26. 3(2 x  1)  4( x  2)  1 6x  3  4x  8  1 6x  3  4x  7 2x  4 x2 Inequality: x  2 Interval notation:  , 2 Graph:

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Chapter 8: Solving Linear Equations and Inequalities to Make Predictions 189 27.

1 5  3  x  13 2 2 1  6  5 x  26 5  5 x  20 1  x  4 4  x  1

Inequality: 4  x  1 Interval notation:  4,1 Graph:

28. a. The model is shown below. It comes close to the data points.

b. The slope is 2.02. The percentage of childhood brain cancer survivors who have mild cognitive deficits is increasing by 2.02 percentage points per year of age. f (50)  25.00  2.02(50)  76; At age 50 years, 76% of childhood brain cancer survivors have mild cognitive deficits. The residual is 77%  76%  1%. The observed percentage of childhood brain cancer survivors who have mild cognitive deficits is 1 percentage point greater than the predicted percentage. d. 50  25.00  2.02a 75  2.02a a  37.13

c.

At about age 37, 50% of childhood brain cancer survivors have mild cognitive deficits. e. At ages less than 32 years, although we have little faith for our result for ages less than 20. Also, if we assume an adult is at least 18 years of age, then there is model breakdown for ages less than 18 years.

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Chapter 9: Finding Equations of Linear Models

Chapter 9: Finding Equations of Linear Models Homework 9.1 2. It is true that if a line has slope 4 and contains the point (0, 7), then an equation of the line is y  7  4 x.

4. False. The graph of the equation y  2  6( x  4) is a line that has slope 6 and contains the point (4, 2).

6. y  4  3( x  2) y  4  3x  6 y  2  3 x 8. y  (8)  5( x  3) y  8  5 x  15 y  7  5x 10. y  (4)  1( x  (7)) y  4  x  7 y  11  x 1 ( x  5) 2 1 5 y3 x 2 2 1 1 y  x 2 2

26. y  (93.3)  25.3( x  (42.8)) y  93.3  25.3 x  1082.84 y  1176.14  25.3 28.

30.

12. y  3 

7 14. y  (4)   ( x  (1)) 3 7 7 y4  x 3 3 19 7 y  x 3 3 5 16. y  (2)   ( x  (4)) 3 5 20 y2  x 3 3 26 5 y  x 3 3

18. Horizontal line: y  3 20. Vertical line: x  3 22. y  3.8  1.3( x  6.6) y  3.8  1.3 x  8.58 y  4.87  1.3 x 24. y  9.60  2.07( x  (4.73)) y  9.60  2.07 x  9.79 y  0.19  2.07 x

1 4  3 2 1 y  1  3( x  2) y  1  3 x  6 y  7  3x b

5  (10) 3 3  (2) y  5  3( x  3) b

y  5  3x  9 y  4  3 x

32.

8  (2)  3 1  (3) y  (2)  3( x  (3)) b

y  2  3 x  9 y  11  3 x

34.

1  (5)  2 3  0 y  (5)  2( x  0) b

y  5  2 x y  5  2 x

36. b 

3  (3)  0; Horizontal line: y  3 1  (5)

38. b 

6 1 undefined; Vertical line: x  7 77

40.

1 3 1  62 2 1 y  3   ( x  2) 2 1 y  3   x 1 2 1 y  4 x 2 b

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191

192 ISM: A Pathway to Introductory Statistics 42.

44.

46.

48.

5  (2) 3  62 4 3 y  (2)   ( x  2) 4 3 3 y2  x 4 2 1 3 y  x 2 4 b

25 3  4 2  (6) 3 y  2   ( x  (2)) 4 3 3 y2  x 4 2 1 3 y  x 2 4

50.

y  2.7  0.54167 x  1.78751 y  2.7  0.54167 x  1.78751 y  0.91  0.54 x

52.

b

6  (2) 4  5  (1) 3 4 y  6  ( x  5) 3 4 20 y6  x 3 3 2 4 y  x 3 3

5.3  2.7  0.54167 8.1  3.3 y  2.7  0.54167( x  3.3) b

4.82  3.92  0.58895 8.99  ( 5.85) y  3.92  0.58895( x  (5.85)) b

3.92  0.58895 x  3.44536 y  0.48  0.59 x

54.

883.7  (650.2)  1.19011 287.6  (483.8) y  (883.7)  1.19011( x  ( 287.6)) y  883.7  1.19011x  342.27564 b

y  1225.98  1.19 x

b

56.

50 5  1  (3) 2 5 y  0  ( x  (3)) 2 5 15 y  x 2 2 b

4  (9) 5  2  (6) 4 5 y  (4)  ( x  (2)) 4 5 5 y4 x 4 2 3 5 y  x 2 4 b

58. The student forgot to check whether both points satisfy the equation. The correct equation is y  3  2 x.

60. a.

2 2  4  6  (3) 3 2 y  4   ( x  (3)) 3 2 y4  x2 3 2 y  2 x 3 b

b.

c. Answers may vary.

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Chapter 9: Finding Equations of Linear Models

193

62. a. It is possible for a line to have no y-intercepts. Answers may vary. b. It is possible for a line to have exactly one y-intercept. Answers may vary. c. It is not possible for a line to have exactly two y-intercepts. Answers may vary. d. It is possible for a line to have an infinite number of y-intercepts. The equation of this line is x  0.

64. a.

b. The association is linear, very strong, positive, and there are no outliers. c.

32  5 3 10  1 s  17  3(t  5) b

s  17  3t  15 s  2  3t

d. The equation is shown graphed on the scatterplot in part (a).

66. b 

5  3 4  2  (4) 3

a. i.

 4 3  a     (4)  3 3 a

4 ii. y  3   ( x  (4)) 3 4 16 y3   x 3 3 7 4 y  x 3 3

16 3

7 3 7 4 y  x 3 3 b. The equations are the same. a

68. It is a vertical line because only vertical lines have an undefined slope.

70. The y-intercept will be 0.5 unit higher: y  3.5  4 x.

72. Answers may vary.

74. a.

b. Answers may vary. The equation created by taking any two points on the line will be y  3  2 x.

76. Answers may vary.

Homework 9.2 2. It is true that viewing a scatterplot of the data can help determine which two points to use to find an equation of a reasonable linear model.

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194 ISM: A Pathway to Introductory Statistics 4. It is true that if the variables w and h are linearly associated and w is the explanatory variable, then the situation can be modeled well by an equation of the form h  a  bw.

6. Answers may vary. An equation of a line that comes close to the points is y  5.04  1.34 x.

8. Answers may vary. An equation of a line that comes close to the points is y  17  1.5 x.

10. a.

b. The association is linear, strong, positive, and there are no outliers. The value r  0.992 supports the conclusion of a strong positive association. c. An equation of a linear model is E  7.68  1.96t. Answers may vary.

d.

12. a.

b. The association is linear, strong, positive, and there are no outliers. The value r  0.999 supports the conclusion of a strong positive association. c. An equation of a linear model is p  4.7  4.7t. Answers may vary. d.

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Chapter 9: Finding Equations of Linear Models 195 12. (continued) The p-intercept of this model is much lower than the p-intercept found in Exercise 11. This makes sense because the slope has not changed in the model, so extending the distance from the data points to the p-intercept will make the p-intercept significantly lower.

14. a.

b. The association is linear, strong, negative, and there are no outliers. The value r  0.973 supports the conclusion of a strong negative association. c. An equation of a linear model is y  47.90  0.05t. Answers may vary. d. The graph is shown on the scatterplot in part (a). The model comes close to all the data points.

16. a. N (t )  15.47  5.40t and D(t )  104.70  1.91t b.

c. 15.47  5.40t  104.70  1.91t 7.31t  89.23 t  12.2 Yes, the models estimate that Netflix and the Disney Channel will both have about 83.4 million subscribers in 2013. d. Yes, the number of Netflix subscribers will be greater than the number of Disney Channel subscribers after 2013.

18. a.

b. The association is linear, fairly strong, negative, and there are no outliers. The value r  0.976 supports the conclusion of a fairly strong negative association. c. An equation of a linear model is p  51.7  1.2t. Answers may vary. d. slope of the model is –1.2. The percentage of unpaid internships/co-ops is decreasing by 1.2 percentage point per year. e.

p  51.7  1.2(7)  51.7  8.4  43.3; According the model, about 43.3% of internships/co-ops will be unpaid. The residual is 0. The predicted and observed percentages of unpaid internships/co-ops are the same.

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196 ISM: A Pathway to Introductory Statistics 20. a.

b. The association is linear, very strong, negative, and there are no outliers. The value r  0.99 supports the conclusion that there is a very strong negative association. c.

f (t )  41.10  0.32t

f (32)  41.10  0.32(32)  41.10  10.24  30.86; This means that, according to the model, about 30.9% of households consisted of married couples with children in the year 2021. e. 41.10  0.32t  32 0.32t  9.1 t  28.4

d.

This means that, according to the model, 32% of households consisted of married couples with children in the year 2009.

22. a.

b. The association is linear, very strong, negative, and there are no outliers. The value r  0.999 supports the conclusion that there is a very strong negative association. c.

f (a)  9.20  0.46a

f (50)  9.20  0.46(50)  9.20  23  13.8; This means that, according to the model, about 13.8% of Americans age 50 have been diagnosed with diabetes. e. 9.20  0.46a  12

d.

0.46a  21.2 a  46.1

This means that, according to the model, 12% of Americans between ages 46 and 47 have been diagnosed with diabetes.

24. a.

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Chapter 9: Finding Equations of Linear Models 197 24. (continued) b. The treatment group’s mean well-being score increased more than that of the control group. c. A reasonable linear model is s  37.32n  153.1. Answers may vary. d. The slope of the treatment group model is 37.32. This means that, according to the model, the mean wellbeing score of the treatment group increases by about 37.32 after each visit. e. s  37.32(10)  153.1  373.2  153.1  526.3; This means that, according to the model, the treatment group’s well-being score would be 526.3 after 10 visits. Model breakdown has occurred because the maximum possible well-being score is 500.

26. a. The association is linear: P  3907.68  91.83t. b. The association is not linear, the scatterplot shows a curved pattern. c. The association is linear: C  4396.60  99.40t. d. Answers will vary. The association of the annual rate of all crimes would probably be the most convincing. The association of the murder rate would probably be the least convincing since there was an increase after 2014. e. The property crime data points for the time intervals are (8, 3215) and (12, 2868) for 2008 to 2012 and (12, 2868) and (17, 2362) for 2012 to 2017.

2868  3215  86.75 12  8 2362  2868  64.8 17  12 The slopes in both intervals are negative and are relatively close in steepness. The association from part (a) would appear to support Levitt’s theory.

The murder rate data points for the time intervals are (8, 5.4) and (12, 4.7) for 2008 to 2012 and (12, 4.7) and (17, 5.3) for 2012 to 2017.

4.7  5.4  0.175 12  8 5.3  4.7  0.12 17  12 The slope is negative in the first interval and positive in the second interval. The association from part (b) would appear to not support Levitt’s theory.

The crime rate data points for the time intervals are (8, 3673) and (12, 3256) for 2008 to 2012 and (12, 3256) and (17, 2745) for 2012 to 2017.

3265  3673  102 12  8 2745  3256  102.2 17  12 The slopes are negative and almost equal in both intervals. The association from part (c) would appear to support Levitt’s theory.

28. a. The association is linear, very strong, positive, and there are no outliers. b. Yes, the line appears to be a reasonable model because it follows the trend and is close to all of the data points. c. Answers may vary. The equation should be about H  0.320  0.509 L. d. Answers may vary. The estimate should be about 22.9 meters. This is about 12 times the mean height of American men.

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198 ISM: A Pathway to Introductory Statistics 30. a. The association is negative. This makes sense because, as a golfer hits the ball a longer distance, it is generally harder to control the accuracy of the shot. b. The association is linear, fairly strong, and there are no outliers. c. Answers may vary.

692  0.751; Ryan Moore had a fairway accuracy of about 75.5%, so his mean driving distance would be 922 about 270 yards. e. The association is stronger for men. Answers may vary.

d.

32. Student C has chosen the points that create the best linear model for the data. Answers may vary.

34.

The value of b in the original model can be increased so that the line describes the data better.

36. Answers may vary.

Homework 9.3 2. For a group of data points, the linear regression model is the linear model with the least sum of squared residuals.

4. The coefficient of determination is the proportion of the variation in the response variable that is explained by the regression line.

6. (d) because the pattern of points on the scatterplot going over and back under the regression line matches the pattern on the residual plot.

8. (b) because the points in the scatterplot all lie on the regression line, which matches the points in the residual plot all lying on the zero residual line.

10. There is a large outlier.

12. All of the conditions are met.

14. The red dot is not an influential point; the regression line does not change significantly when the red dot is removed.

16. The red dot is an influential point; the regression line’s slope and intercept change significantly when the red dot is removed.

18. (b) because an r 2 value of 0.36 indicates a weak association.

20. (d) because an r 2 value of 0 indicates no association.

22. yˆ  29.47  1.41x; The regression line comes close to all of the points on the scatterplot.

24. a.

b. yˆ  0.181  0.821x

c. Both lines are shown on the scatterplot in part (a). The regression line comes close to all of the data points, while the line yˆ  0.12  092 x only comes close to the two points used to fit the model. d. 1.86

e. 0.9

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Chapter 9: Finding Equations of Linear Models 199 24. (continued) f. The sum of squared residuals for the regression line is lower. This is not surprising since for a group of data points, the linear regression model is the linear model with the least sum of squared residuals.

26. a.

b. yˆ  9.22  0.46 x c. Both lines are shown on the scatterplot in part (a). The regression line comes close to all of the data points, while the line yˆ  28.55  0.31x only comes close to the two points used to fit the model. d. 6.0 e. 0.3 f. The sum of squared residuals for the regression line is lower. This is not surprising since for a group of data points, the linear regression model is the linear model with the least sum of squared residuals.

28. a.

b. yˆ  10, 284.34  532.96 x; The graph of the line is shown on the scatterplot in part (a). The regression line comes close to the data points. c. yˆ  10, 284.34  532.96(35)  28,938; $28,938 in 2015 is comparable to $10,000 in 1980. d. The residual is 28,858  28, 938  80; The model overestimated the comparable cost by $80. e. 10, 284.34  532.96 x  25, 000 532.96 x  14, 715.66 x  27.6 A cost of $10,000 in 1980 was equivalent to a cost of $25,000 in 2008.

30. a.

b. yˆ  3.25  0.468 x; The graph of the line is shown on the scatterplot in part (a). The regression line comes close to the data points. c. yˆ  3.25  0.468(23.5)  7.7; The Green Bay Packers would have 8 wins. d. The graph of yˆ  6.49  0.66 x is shown on the scatterplot in part (a). The line does not come close to the data points. The regression line from part (b) fits the data better. e. The association changed from 2017 to 2018. Answers will vary.

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200 ISM: A Pathway to Introductory Statistics b. yˆ  42.72  1.088 x; The graph of the regression line comes close to the data points.

32. a.

c. 42.72  1.088(70)  28 chirps

d. xˆ  39.87  0.897 y

e. xˆ  y  40; This rule of thumb is similar to the model found in part (d).

34. a.

b. yˆ  6.94  0.58 x; This model is shown on the scatterplot in part (a). c.

d. The lowest dot in the residual plot represents an age of 54.5 years and a residual of 4.7%. The corresponding data point is 4.7% below the regression line. e. The highest dot represents an age of 70 years and a residual of 3.3%. The corresponding data point is 3.3% above the regression line.

36. a. 6.94  0.58(30)  10.46; According to the model, about 10.46% of 30-year-old Americans gamble online. b. The slope of the model is 0.58. This means that, according to the model, the percentage of Americans of a given age that gamble online is about 0.58 greater than the percentage of Americans one year younger than the given age that gamble online. c. 6.94  0.58 x  16 0.58 x  22.94 x  39.55 According to the model, about 16% of Americans between the ages of 39 and 40 gamble online. d. r 2  0.918; 91.8% of the variation in the response variable is explained by the regression line.

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Chapter 9: Finding Equations of Linear Models 201 38. a.

b. There are no outliers. The association is very strong, linear, and negative; r  0.995. c.

yˆ  1.665  0.028 x

d.

The residual plot confirms that the association is very strong and linear, and that there are no outliers. e. r 2  0.989; 98.9% of the variation in the response variable is explained by the regression line.

40. a.

b. There are no outliers. The association is strong, linear, and positive; r  0.956. c.

yˆ  392.1  19.3 x

d.

The residual plot confirms that the association is moderately strong and linear. There may be two outliers.

The vertical spread of the residual plot does increase somewhat as the values of the explanatory variable increase. e. r 2  0.914; 91.4% of the variation in the response variable is explained by the regression line. f. No. While there appears to be an association between the two variables, it is unlikely that there is a causal relationship between these two variables. There may be a lurking variable instead.

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202 ISM: A Pathway to Introductory Statistics 42. a.

s

Stopping distance

20

44  25  69

30

66  57  123

40

88  101  189

50

110  158  268

60

132  227  359

70

154  310  464

80

176  404  580

b.

c. Dˆ  132.75  8.52 s d. 2.2 s  (132.75  6.32s )  132.75  8.52 s; The expressions are equal. This makes sense because D  R  B. 44. a.

b. There are no outliers. The association is strong, linear, and positive; r  0.98. c.

The residual plot supports the conclusion that there are no outliers. The pattern in the residual plot contradicts the conclusion that the association is linear. d. The lowest dot represents an SAT score of 1345 and a residual of 4.45. The corresponding data point is 4.45% below the regression line. e. Since the residual plot shows a pattern that is not near the zero residual line, a linear model may not be the most appropriate model for these data.

46. a. The percent receiving free or reduced-fee lunch is about 85%, and the percent wearing bicycle helmet is about 45%. b. Los Arboles is a low-income neighborhood, but has a high percentage of bicycle helmet wearers despite that. Since it is an outlier, it may be useful to determine why it is an outlier. c. The slope of the regression line changed significantly, so the point is an influential point. d. No, the vertical spread is not about the same for each value of the explanatory variable. This violates one of the conditions for regression, so this model should not be used to make predictions.

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Chapter 9: Finding Equations of Linear Models 203 48. a.

b. The outlier is for the year 2001; the terrorist attacks in 2001 resulted in all flights being grounded for several days, and resulted in people flying less for some time afterward. c. No; removing the outlier does not change the regression line significantly. d. yˆ  23.9 x  424.5  23.9(11)  424.5  687.4 622  687.4  65.4 The terrorist attacks reduced commercial airline boardings by about 65.4 million. e.

65, 400, 000  340  $5.559 billion 4

50. Answers may vary.

52. If a dot in a residual plot lies below the zero residual line, then the residual is negative, which means the dot corresponds to a data point that is below the regression line.

54. Answers may vary.

56. Answers may vary.

58. The coefficient of determination only explains what percent of the variation is explained by the model. It does not identify how close the points are to the line.

60. a. Answers will vary. b. Answers will vary. c. Answers will vary. d. Answers will vary. e. Answers will vary.

f. Answers will vary. g. Answers will vary. h. Answers will vary. i. Answers will vary.

62. a.

b. Men tend to have longer driving distances, since the values of the explanatory variable on the scatterplot for males are larger than for females. c. For female golfers, the golfer with a driving distance of 225.8 yards and a fairway accuracy of 49.3% may be an outlier. The association is weak, linear, and negative; r  0.42.

For male golfers, there are no outliers. The association is weak, linear, and negative; r  0.61.

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204 ISM: A Pathway to Introductory Statistics 62. (continued) d. For female golfers, r 2  0.18; about 18% of the variation in the response variable is explained by the regression line.

For male golfers, r 2  0.37; about 37% of the variation in the response variable is explained by the regression line. e. For females, yˆ  0.256 x  130.9. For males, yˆ  0.388 x  174.9. For all golfers, yˆ  0.139 x  102.8. f. This makes sense because the combined data set has the same range in the response variable, but much greater range in the explanatory variable. g.

796  75.0% driving accuracy 1061 0.388 x  174.9  75.0

0.388 x  99.9 x  257 yards

Chapter 9 Review Exercises

1. y  (1)  4( x  2) y  1  4 x  8 y  7  4x 2 2. y  (4)   ( x  (6)) 3 2 y4  x4 3 2 y  8  x 3

8.

y  5  5 x  20 y  15  5 x

9.

3. Vertical line: x  3 4. Horizontal line: y  4 5. y  8.82  5.29( x  (4.93)) y  8.82  5.29 x  26.0797 y  17.26  5.29 x 6. y  (7.13)  a  1.45( x  ( 2.79)) y  7.13  1.45 x  4.0455 y  3.08  1.45 x 7.

2  (7) 3 1  (2) y  2  3( x  1) y  2  3x  3 b

y  1  3 x

5  (5) 5 42 y  5  5( x  4) b

10.

5 6  9  6  (3) 3 5 y  (6)   ( x  6) 3 5 y  6   x  10 3 5 y  4 x 3 b

7  (10) 3  2  (4) 2 3 y  (7)  ( x  (2)) 2 3 y7  x3 2 3 y  4  x 2 b

11. b 

3  2 ; undefined; Vertical line: x  5 55

12. b 

3  (3)  0; Horizontal line: y  3 1  (4)

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Chapter 9: Finding Equations of Linear Models 13.

4.8  9.2  0.84615 8.7  3.5 y  9.2  0.84615( x  3.5) b

15.

y  9.2  0.8461x  2.96153 y  12.16  0.85 x

14.

3.99  2.49  0.91915 1.83  (5.22) y  2.49  0.91915( x  (5.22)) b

205

4  (3) 1  3  (2) 5 1 y  (4)   ( x  3) 5 1 3 y4  x 5 5 17 1 y  x 5 5 b

y  2.49  0.91915 x  4.79796 y  2.31  0.92 x

16. Answers may vary. The regression equation is yˆ  30.38  2.13x.

17. a.

b. Answers may vary. The regression equation is yˆ  61.08  0.22 x. c. The slope is 0.22. The percentage of American adults who are in the middle class decreased on average by 0.22 percentage points per year. d. 61.08  0.22(45)  51.2% e.

61.08  0.22 x  55 0.22 x  6.08 x  27.6

The percentage of American adults who are in the middle class was 55% in 1998.

18. a.

b. There are no outliers. The association is linear, positive, and strong; r  0.99, which confirms that the association is strong. c. Answers may vary. The regression equation is yˆ  12.31  0.49 x. the observed mean number of students who received free school lunches in 2017 is 0.64 million students less than the predicted mean number of students who received free school lunches in 2017. c. yˆ  12.31  0.49(17)  20.64 million students; The residual is 20.0  20.64  0.64 million students. e. The y-intercept is (0, 25.41). In 2000, the mean number of students who received free school lunches is 12.31 million students.

19. There is an outlier when the explanatory variable is 26.

20. The point is an influential point, because the slope changes significantly when it is removed.

21. 0.9 is a reasonable estimate, since the association is strong and linear.

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206 ISM: A Pathway to Introductory Statistics 22. a.

b. yˆ  17.64  1.01x c. The two lines are shown on the same graph of the scatterplot in part (a). The line yˆ  23.75  2.25 x fits the data poorly, while the regression line fits the data very well. d. 197.56 e. 22.21 f. The regression line has a lower sum of squared residuals than the line yˆ  2.25 x  23.75; this makes sense because the regression line has the least sum of squared residuals of all lines.

23. a.

yˆ  6.04  0.61x

b.

c. The highest point is for the year 2010, with a residual of 26.8  (6.045  0.607(30))  26.8  24.26  2.54; For the year 2010, the corresponding data point is 2.54 million former Catholics above the regression line. d. yˆ  50.27  0.51x

e. The slope of the former-Catholic model (0.61) is greater than the slope of the Catholic model (0.51); the number of former Catholics is increasing at a greater rate than the number of Catholics.

24. a.

b. The association is negative; the older a child is when he or she moves, the less time the wealthy neighborhood or lurking variables (if any) will have to affect the child. c. yˆ  93.31  3.85 x d. The coefficient of determination is 0.97; 97% of the variation in future-income percentiles is explained by the regression line. e. 3.85(15)  93.31  35.56; 35th percentile. The residual is 7 percentiles; the actual future income is 7 percentiles greater than the income predicted by the regression line.

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Chapter 9: Finding Equations of Linear Models

207

Chapter 9 Test

1. y  (4)  7( x  (2)) y  4  7c  14 y  10  7 x 2 2. y  (1)   ( x  6) 3 2 y 1   x  4 3 2 y  3 3

3.

36 1  2  (4) 2 1 y  3   ( x  2) 2 1 y  3   x 1 2 1 y  4 x 2

4.

7.1  2.9  1.92308 1.8  (3.4) y  2.9  1.92308( x  ( 3.4)) b

y  2.9  1.92308 x  6.53847 y  3.64  1.92 x

5.

b

1 4  (3)  5 2 3 1 y  (3)   ( x  2) 3 1 2 y3   x 3 3 7 1 y  x 3 3 b

6.

Increase a and decrease b.

7. a.

b. An equation of a linear model is y  184.35  6.39 x. Answers may vary. c. The slope is 6.39. the number of worldwide commercial airline accidents decreased by 6.39 accidents per year. d. (0,184.35); In 2000, there were 184 worldwide commercial airline accidents. We should have little or no faith in a result when we perform extrapolation. e.

y  184.35  6.39(17)  184.35  108.63  76 accidents; The residual is 88  76  12 accidents. The observed number of accidents in 2017 is 12 accidents more than the predicted number of accidents.

8. For the residual plot, the vertical spread decreases from left to right.

9. yˆ  44.87  3.33 x; Yes, the graph comes close to the scatterplot.

10. a.

b. yˆ  40.33  1.79 x; The regression equation is shown on the scatterplot in part (a). c.

yˆ  40.33  1.79(17)  40.33  30.43  9.9 million viewers

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208 ISM: A Pathway to Introductory Statistics 10. (continued) d.

e. 2014; –3.87; The observed mean number of viewers in 2014 is 3.87 million viewers less than the predicted mean number of viewers. f. 2008; 1.79; The observed mean number of viewers in 2008 is 1.79 million viewers more than the predicted mean number of viewers.

11. a.

b. yˆ  2.73  1.04 x; The regression equation is shown on the scatterplot in part (a). c.

d. r 2  0.81; 81% of the variation in the percentage of nutritionists who say a certain food is healthy can be explained by the regression line. e. For all the data points below the regression line except (94, 95), the percentage of survey respondents who think a certain food is healthy is greater than the percentage of nutritionists who think the food is healthy.

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Chapter 10: Using Exponential Models to Make Predictions 209

Chapter 10: Using Exponential Models to Make Predictions Homework 10.1 2. If m and n are integers and b  0, then (b m ) n  b mn .

4. An exponential function is a function whose equation can be put into the form f ( x)  ab x , where a  0, b  0, and b  1.

6. b 2b7  b 2 7  b9

40.

8. (10b7 c 2 )(3b5 c 2 )  10  3  b7 5 c 2 2  30b12 c 4 3

3 3 3

42.

3 3

10. (3bc)  3 b c  27b c 12.

14.

b8

 b8  6  b 2

b6

14b5 c9



3 6

21b c

b2 c

7

 b 2 c7

9b 1

3c5   6 c 5 d 7 2bd 7

44. (b5 ) 8  b5( 8)  b 40 

2 53 9 6 2b 2 c3  b c 3 3

1 b 40

46. (3b 2 c 6 )(b9 c 1 )  3b 29 c 6  ( 1)  3b7 c 7 

4

3b7 c7

b4 b 16.    4 c c

48. 42003 42000  42003 2000  43  64

18. (b3 )8  b38  b 24

50. (1301 ) 1  1301( 1)  1301  130

20. (2b3b 4 )5  (2b7 )5  25 b75  32b35

52.

22. (5b3c9 )0  1 24. (bc 6 )3 b 2 c 4  (b3c18 )b 2 c 4  b3 2 c18 4  b5 c 22 2

5 3

2 2

3 15

26. 4b (2b )  2 b (2 b )  2 28.

15b9 b 2



8

20b

3b11 4b

 8

2  3 2 15

b

17

 32b

56.

3 118 3b3 b  4 4

58.

6 2

 4bc 42 b 2 c12 16b 2 c12 30.  3   2 6  7 d 49d 6  7d 

60.

2 0

 5c 32.  4   1  9b  4 3 3

34.

(2b c )

5 4 2

(6b c )

36. b 2  38.

1 b 5

1 b

2

b



54.

b 4 b

7

b6 b

1

 b 4  7  b 11 

 5

7 5 7 6

4 2  ( 5) 4b3 b  9 9

 7 5 ( 6)  71  7

25 b 7 2 2

2 b

 25 2 b 7  ( 2)  23 b 5 

62. (2b 4 ) 4  24 b16  3 12 9

2 b c

2 10 8

6 b c



b11

 b6 ( 1)  b7

4b 2 9b

1

b16 2

4



b5

b16 16

2

8 1210 98 2b c  b c 36 9

8

64. 6(bc 4 ) 3  6(b 3c 12 )  

6 3 12

b c

66. (7b 4 c 1 ) 2 (2b3c 2 )5  (7 2 b8 c 2 )(25 b15 c 10 ) 

5



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25 72

b815 c 2 ( 10)

32b 23 49c8

210 ISM: A Pathway to Introductory Statistics

68.

70.

72.

18b5 c3 d 7

3 5 ( 6) 33 7  ( 2) 3 11 0 5 3b11 b c d b c d       4 4 4d 5 24b 6 c3 d 2 (16b 2 c)(25b 4 c 5 )



(15b5 c 1 )(8b 7 c 2 ) (3bc 2 )2 3

(3b c)

1



16  25  b 2 c 4

10 2 ( 2) 4 ( 3) 10 4 1 10b 4 b c   b c  3 3 3c 15  8  b 2 c 3

32 b 2 c 4

 32 ( 1) b 2 3c 4  ( 1)  31 b 5 c5 

1 3 1

3 b c

3

 2 2

 8b c  76.    12b 5 c 3 

3

1 b

1

b c

53



2    b3 c 5  3  33



78.

6 15

8b6 125c15

3



3 9 15

2 b c

3b5

90. a.

3  2bc 7  2  74.  1 2    b 2 c 5  5   5b c 

23

c5



2

3

33

27 8b9 c15

b 9 c 15

x

f ( x)

x

f ( x)

4

16

0

1

3

8

1

2

4

2

1

2

3

b.

1  1  bc c

80. b 1  c 1 

1/2

1 1 bc   b c bc

1 c.   2

82. f (2)  42  16 84. f (3)  43 

 0.7

92. t   X  1  1 3

4



1 64

X 

94.  02  (n  1) s 2 02 

0

86. g (0)  2(3)  2 1  2 2

2 88. g (3)  2(3) 3  3  27 3

100. a. b.

102. a. b.

1 2 1 4 1 8

f ( L)  720 L1 

s 2 (n  1)

 02



s2n  s2

 02

96. V  8000(1.07)6  $12, 005.84 98. P  7000(1.04) 6 

7000 (1.04)6

 $5532.20

720 L

720  60; You must exert 60 pounds of force on a wrench with a handle length of 12 inches to 12 loosen a bolt. f (12) 

f ( L)  8910d 2  f (80) 

8910 d2

8910

 1.39; A 50-watt bulb will produce about 1.39 milliwatts per square centimeter of light at a 802 distance of 80 from the bulb.

Copyright © 2021 Pearson Education, Inc.

Chapter 10: Using Exponential Models to Make Predictions 211 104. a.

b. There is a nonlinear exponential association. c. The model is shown on the scatterplot in part (a). The model comes close to the data points. d. yˆ  2.49(1.29)11  $40.99 billion; The residual is 48.08  40.99  $7.09 billion. e.

yˆ  2.49(1.29)18  $243.68 billion; The residual is 232.89  243.68  $10.79 billion.

106. The student added the exponents instead of multiplying: (b 4 )6  b 46  b 24 .

108. The student multiplied 2 by 4 instead of raising 2 to the 4th power: (2b 2 ) 4  (24 )b 24  16b8 .

110. Student 1 simplified correctly. Student 2 incorrectly added the exponents of the numerator and denominator, instead of subtracting. b 112. a.   c

2

b b.   c

n

114. a. b 1 

 

b 2

c2

c

 2

b2

bn

cn

c

n



5

c5 b c.    5 c b

bn

1 b

c. ((b 1 ) 1 ) 1  b 1 

b. (b 1 ) 1  b1  b

1 b

d. (((b 1 ) 1 ) 1 )1  b1  b

e. ((((b 1 )1 )1 )1...) 1  b if n is even, or

1 if n is odd. b

116. Answers may vary.

Homework 10.2

2. False. For the counting numbers m and n, where n  1 and b is any real number for which b1/n is a real number, b m / n  (b1/ n ) m .

4. If m and n are rational numbers and b is any real number for which b m and b n are real numbers, then (b m ) m  b mn .

6. 271/3  3

18. 27 4/3  (271/3 ) 4  34  81

8. 321/5  2

20. 813/4  (811/4 )3  33  27

10. 811/4  3

22. 161/4 

12. 641/6  2 14. 163/4  (161/4 )3  23  8 16. 64

2/3

1/3 2

 (64

1 1/4

16

24. 321/5  

2

)  4  16

Copyright © 2021 Pearson Education, Inc.



1 2

1 1/5

32



1 2

212 ISM: A Pathway to Introductory Statistics 26. (9) 3/2 

1 3/2

9



1 1/2 3

(9

1

28. (32) 3/5 

)

1





3

3

52. (8b 6 c12 ) 2/3  82/3 b 6(2/3) c12(2/3)

1 27

 4b 4 c8 

1

 3/5

((32)1/5 )3 1   3 8 (2)

7/5 3/5

30. 3

3

(32) 1

7/5 3/5

3

10/5

3

34.

51/3

 b 6/3c 2/2  b 2 c 1 

2

3 9

 54/31/3  53/3  51  5

1/4

1/4

4

4  2  9 3

60.

b3/4 c1/2 b

5 6 2  3 1  2 1  3 1  6 0

1 32 1 16 1 8 1 4 1 2 1

1 6 1 3 1 2 2 3 5 6 1



c

 b1c1  bc  16b12 c 2  62.  3 4   2b c 

1/3



 8b15 c 6  

f ( x)



1/3

 81/3 b 5 c 2

1 1/3 5 2

8 b c 1 2b5 c 2

64. 161/4 b1/4b1/3  2b3/12b 4/12

2

 2b3/12 4/12

4

 2b7/12

8

66.

16

b 2/3 b1/7



b 14/21 b3/21

 b 14/213/21

 b 17/21 

32

68.

64

(32b3 )3/5 3 3/2

(16b )

46. b1/5b3/5  b1/53/5  b 4/5 48. b 2/7 b 6/7  b 2/7  ( 6/7)  b 4/7 

 2b 2 c3

 b3/4 ( 1/4) c1/2  ( 1/2)  b 4/4 c 2/2

1/4 1/2

5 42. h    2(4)5/2  2(41/2 )5  2(2)5  64 2 x

1/4

  (16b6 )(b 2 c12 )   16b8 c12 

4 4  2 40. g     4(27) 2/3  2/3  1/3  3 27 (27 ) 2

f ( x)

b2 c

56. (6bc 2 )5/7 (6bc 2 ) 2/7  (6bc 2 )5/7  2/7  (6bc 2 )7/7

58.  (4b3 ) 2 (b 2 c12 ) 

2 38. g    4(27) 2/3  4(271/3 ) 2  4(3) 2  36 3

x

1

 (6bc 2 )1  6bc 2

1 36. f    811/4  3 4

44.

b4

54. (b 4/3c1/2 )(b 2/3c 3/2 )  b 4/3 ( 2/3) c1/2 ( 3/2)

32. (22/351/3 )3  2(2/3)35(1/3)3  2251  20 54/3

4c8



1 b

50. (27b 27 )1/3  271/3 b 27(1/3)  3b9

Copyright © 2021 Pearson Education, Inc.

b

323/5 b3(3/5) 3/2 3(3/2)

16

b

8b18/10



8b9/5 64b9/2

1  b18/10 45/10 8 64b 1 1  b 27/10  27/10 8 8b 

4/7

1 17/21

45/10

Chapter 10: Using Exponential Models to Make Predictions 213  27b1/3c3/4  70.  2/3 1/2  c   8b

4/3

 33 b1/3c3/4    3 2/3 2/4  c  2 b  33 b1c1/4     238    

27

4/3

72.

(1000b 7 c8 ) 2/3



(32b15 c 4 )3/5

4/3

 

3(4/3) 4/3 (1/4)(4/3)

b

c

8

4 4/3 1/3

c

2

4



81b 4/3c1/3 16

74. c1/3 (c8/3  c5/3 )  c8/31/3  c5/31/3  c9/3  c 6/3  c3  c 2

323/5 b15(3/5) c 4(3/5) 100b 14/3c16/3 8b9 c12/5 25b 14/3c80/15

2b 27/3c36/15 25 14/3 27/3 80/1536/15  b c 2

3(4/3)

3 b

10002/3 b 7(2/3) c8(2/3)

76. E  tsn 1/2 

25 41/3 44/15 25c 44/15  b c 2 2b 41/3

ts n1/2

78. a.

b. There is a nonlinear, exponential association. c. The model is shown on the scatterplot in part (a). The model comes close to the data points. d. 0.2(1.095)30  3.0 million tests. The residual is 3.5  30.  0.5 million, or about 500 thousand tests. The predicted value is 0.5 million tests less than the observed number of tests. e. 0.2(1.095)35  4.8 million tests. The residual is 4.7  4.8  0.1 million, or about 100 thousand tests.

The predicted value is 0.1 million tests more than the observed number of tests.

80. The student has mistaken an exponent for multiplication.

82. The student incorrectly evaluated the rational exponent. The correct expression is 642/3  (641/3 ) 2  42  16.

84. The student interpreted the negative exponent as making the base negative. The correct expression is 1 251/2  (251/2 ) 1  .

5 1/3

1 86.   2

1/2

1  0.794;   3

1/3

1  0.577. This makes sense because   2



1/2

1 ,   1/3 3 2 1



1 1/2

3

, and 21/3  31/2.

88. Answers may vary.

Homework 10.3

2. Let f ( x)  ab x , where a  0. If 0  b  1, then the function f is decreasing.

4. An exponential model is an exponential function, or its graph, that describes an authentic association.

6.

8.

Copyright © 2021 Pearson Education, Inc.

214 ISM: A Pathway to Introductory Statistics 26. Domain: all real numbers or (, ) Range: y  0 or (0, )

10.

12.

28. a. Answers may vary. b.

14.

c. For each input-output pair, the output is 16 times one-half raised to the power of the input.

16.

30.

18.

32.

20.

22.

x

f ( x)

g ( x)

h( x )

k ( x)

0

3

64

2

100

1

6

32

6

10

2

12

16

18

1

3

24

8

54

1 10

4

48

4

162

1 100

x

f ( x)

g ( x)

h( x )

k ( x)

0

625

37

3

400

1

125

3

30

200

2

25

21

300

100

3

5

147

3000

50

4

1

1029

30, 000

25

34. f (1)  2 36. f (1) 

24. Domain: all real numbers or (, ) Range: y  0 or (, 0)

1 2

38. f (1)  2 40. No solution; 2 is not in the range of f .

42. f (6)  192 44. f (0)  3 46. f (1)  3 48. f (5)  3

50. a.

Copyright © 2021 Pearson Education, Inc.

Chapter 10: Using Exponential Models to Make Predictions 215 50. (continued) b. The graph is shown on the scatterplot in part (a). c. There are no outliers; the association is strong, exponential, and negative. d. The coefficient is 22.9. In 2010, the total Wii and Wii U console annual sales were 22.9 million units. e. The base of the model is 0.67; The total Wii and Wii U console annual sales decrease by 33% per year.

52. a.

b. The graph is shown on the scatterplot in part (a). The association is strong, exponential, and negative. c. The coefficient of the model is 217; in 1960, there were 217 men’s colleges. Since this is an extrapolation, we should place little or no faith in this estimate. d. The base of the model is 0.93; the number of men’s colleges has decayed by about 7% per year. e. 217(0.93) 25  35.4 men’s colleges. The residual is 27  35.4  8.4. The predicted number of men’s colleges in 1985 was 8.4 more than the observed number of men’s colleges.

54. a.

b. The graph is shown on the scatterplot in part (a). c. There are no outliers. The association is strong, exponential, and positive. d. The base is 1.119; for seniors at a certain age, the percentage with severe memory impairment is 11.9% greater than for seniors who are one year younger. e. Answers may vary.

56. There is no x-intercept. The y-intercept is (0, 2).

58. There is no x-intercept. The y-intercept is (0, 9).

60. f (0)  20  30  1  1  2 62. f (1)  21  31 

68. f ( x) 

32 x 3

x

64. 3x  9 3x  32 x2

66. 3x 

1 1 5   2 3 6

1 3

3x  31 x  1

 32 x  x  3x  g ( x); The values of f and g are the same. x

1 70. f ( x)  2 x  (21 ) x     g ( x); The values of f and g are the same. 2

Copyright © 2021 Pearson Education, Inc.

216 ISM: A Pathway to Introductory Statistics 72. f ( x) 

6x

(2  3) x

3

x

 x

3



2 x 3x 3

x

 2 x  g ( x); The values of f and g are the same.

74. f ( x)  2 x 3x  (2  3) x  6 x  g ( x); The values of f and g are the same.

76. f ( x)  8 x /3  (81/3 ) x  2 x  g ( x); The values of f and g are the same.

78. f ( x)  25 x /2  5 x  (251/2 ) x  5 x  5 x  5 x  (5  5) x  25 x  g ( x); The values of f and g are the same.

80. a. c is greater because the y-intercept of g is higher than the y-intercept of f . b. b is greater because the graph of f grows at a faster rate than the graph of g.

82. The equations are all of the family y  2b x , where b is any positive number.

84. f ( x)  2(2.1) x 4(3) 2  4(9)  36

86. a. i.

ii. 42  32  16  9  144

122  144

122  144

b. i. Answers may vary. g ( x) can be rewritten as 12 x  (4  3) x  4 x 3x , which increases faster than f ( x). ii. Answers may vary. g ( x) can be rewritten as 12 x  (4  3) x  4 x 3x , which is identical to f ( x).

88. There is no x-intercept. The y-intercept is (0, a).

90. a.

b.

g (2  5)  325  37

g (2  4)  32 4  36

c.

g (2)  g (5)  3235  37

g (2)  g (4)  3234  36

The statement is true.

The statement is true.

46

g (4  6)  3

10

3

g ( x  y )  3x  y

d.

g (4)  g (6)  34 36  310

g ( x)  g ( y )  3x 3 y  3x  y

The statement is true.

The statement is true.

92. a.

1 f   is undefined because in the expression (4)1/2 , the base is negative and n is even. 2

b.

1 f   is undefined because in the expression (4)1/4 , the base is negative and n is even. 4

c. Answers may vary.

94. Answers may vary.

Homework 10.4

2. If n is odd, then the real-number solution of an equation of the form b n  k is k1/ n .

4. True. In words, the quotient of the left sides of two equations is equal to the quotient of the right sides.

6. b 4  81 b  811/4 b  3

8. b5  100, 000 b  100, 0001/5 b  10

Copyright © 2021 Pearson Education, Inc.

Chapter 10: Using Exponential Models to Make Predictions 217 10. 5b 2  45

24.

2

b 9 1/2

b  9 b  3

26.

12. 44b3  12 3 b3  11

28.

14. 1.7b 4  86.4 86.4 b4  1.7

30.

1/4

 86.4  b    1.7  b  2.67

16. 2.1b5  8.2  237.5

2b 2

15b

7

 2

10b 7

3b



4 7 15b 2 4  5 7 3b 7 b5  6 2



1/5

2.1b  245.7 b5  117

b  1.03

32. a  5 y  5b x 15  5b1 b3

4

b  10  33 4

b  23 b  231/4 b  2.19

b

10b 7

7 b  6

1 4 5 11 b   6 3 2

3



5

b  1171/5 b  2.59

b

65 3 b 65 b4  3 4

b  2.16

b  0.65

20.

b8

 b8  4  b 4

1/4

1/3

10

b

4

 65  b   3

3 b   11 

18.

b8

 2187

b7  2187

y  5(3) x

34.

a8 y  8b x 79  8b 4 b4 

79 8 1/4

b  21871/3

 79  b    8 

b3

b  1.77 y  8(1.77) x

b9

2 22. 6  9 b 2 b3  9 1/3

2 b  9

b  0.61

Copyright © 2021 Pearson Education, Inc.

2 3b5

218 ISM: A Pathway to Introductory Statistics 36.

a  256 y  256b

44.

x

23  256b7 23 b7  256

46.

1/7

 23  b  256 

a  2.1

48.

x

9.7 b5  2.1

a(1.91)2  1 a  0.27 y  0.27(1.91) x

ab17

8 492 ab 2 b6  123 11

492  a (0.50)11



a  1, 007, 616 y  1, 007, 616(0.50) x

b  0.50

b  1.36

50.

y  2.1(1.36) x a  97.2 y  97.2b x 17.1  97.2b

ab5 1.3  ab 3.5 1.3 b4  3.5

a(0.78)  3.5 a  4.49 y  4.49(0.78) x 1/4

 1.3  b    3.5 

4

b  0.78

17.1 97.2 1/4

 17.1  b    97.2 

52.

b  0.65 y  97.2(0.65) x

ab10

250.8 6.3 ab 250.8 b6  6.3 4

a(1.85) 4  6.3



a  0.54 y  0.54(1.85) x 1/6

 250.8  b    6.3 

a  12.94

b  1.85

y  12.94b x 2.53  12.94b 20 b 20 

7 1

y  1.25(2) x

1/6

1/5

42.



a  1.25

 2  b    123 

 9.7  b  2.1 

b4 

ab

2

a(2)2  5

b  1.91

9.7  2.1b5

40.

ab5



b  71/3

y  256(0.71) x y  2.1b

10 5 ab b2 2

b3  7

b  0.71

38.

ab3

54.

2.53 12.94 1/20

 2.53  b    12.94  b  0.92

y  12.94(0.92) x

ab12

6.52 39.43 ab 6.52 b4  39.43 8

a(0.64)8  39.43



a  1442.07 y  1442.07(0.64) x 1/4

 6.52  b    39.43  b  0.64

Copyright © 2021 Pearson Education, Inc.

Chapter 10: Using Exponential Models to Make Predictions 219 56. a.

b. The regression equation is yˆ  2.11(1.769) x . Answers may vary. c. The model is shown on the scatterplot in part (a). The model comes close to all of the data points. d. The coefficient is 2.11; in 2010, the revenue from body cameras was about $2.11 million. e.

192  278.26%; Answers will vary.

69

58. a.

b. The regression equation is yˆ  0.0324(1.482) x . Answers may vary. c. The model is shown on the scatterplot from in (a). The model comes close to all of the data points. d. The base is 1.482. The U.S. investment in AI startups has increased by 48.2% per year since 2005. e.

yˆ  0.0324(1.482)11  $2.5 billion.

60. a.

b. The regression equation is yˆ  148.38(0.97) x . The association is strong, exponential, and negative.

Answers may vary. c. The base is 0.97. Infant mortality rates in the United States have decreased by 3% per year. d. 148.38(0.97)105  6.06 deaths per 1000 infants e. The mortality rate in the United States in 2015 was 148.38(0.97)115  4.47 deaths per 1000 infants. The 4.47  2.1; the infant mortality rate in the United States is about 2.1 times the infant mortality rate 2.1 in Singapore.

ratio is

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220 ISM: A Pathway to Introductory Statistics 62. a.

b. Answers may vary. The regression equation is yˆ  22.63(1.00088) x . The association is strong, exponential, and positive. c. The coefficient a is 22.63; a gift card from an expenditure of $0 will be worth $22.63. Model breakdown has occurred. d. The base b is 1.00088. For each additional dollar spent, the value of the gift card increases by 0.088%. e. Customer A is predicted to receive a gift card worth $131.43. They will actually receive a gift card worth $200; the residual is $68.57.

Customer B is predicted to receive a gift card worth $204.04. They will actually receive a gift card worth $200; the residual is $4.04.

Customer C is predicted to receive a gift card worth $316.47. They will actually receive a $200 gift card; the residual is $116.47.

66. The points (0, 7) and (2,3) are on the curve.

64. The points (0, 2) and (1, 6) are on the curve. a2

a7 x

y  7b x

6  2b1

3  7b 2

y  2b b3

b2 

y  2(3) x

3 7 1/2

3 b    7 b  0.65

y  7(0.65) x

68. a. i. No. The value of b in the exponential equation would be 1, but this is not a valid value for the base. ii. No. The value of b in the exponential equation would be 1, but this is not a valid value for the base. b. No. The value of b in the exponential equation would be 1, but this is not a valid value for the base.

70. L( x)  12  2 x, E ( x)  20.16(0.63) x 28  2 52 y  8  2( x  2) b

y  8  2 x  4 y  12  2 x

ab5

2 8 ab 1 b3  4 2

a(0.63) 2  8



a  20.16 y  20.16(0.63) x 1/3

1 b  4

b  0.63

72. The function can be linear but not exponential, since the two points have the same y-value.

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Chapter 10: Using Exponential Models to Make Predictions 221 74. a. No. Answers may vary. b. Yes. Answers may vary.

c. Yes. Answers may vary. d. Yes. Answers may vary.

76. Answers may vary.

Homework 10.5 2. False. The residual is the predicted value subtracted from the observed value, so if the point lies above the curve, the predicted value is less than the observed value, and the residual will be positive.

4. The exponential coefficient of determination is the proportion of the variation in the response variable that is explained by the exponential regression curve.

6. (c) because when r  1, the points will follow an exponential curve exactly.

8. (d) because the association is strong and negative.

10. (a) because the pattern of data points above and below the curve matches the pattern of residuals above and below the zero residual line.

12. (c) because the points all lie exactly on the curve, which is consistent with the residuals all being zero.

14. There is an outlier.

16. The vertical spread of the residual plot is not about the same for all values of the explanatory variable.

18. The outlier is not an influential point, because the exponential regression curve does not appear to change when the outlier is removed.

20. (a) because an r 2 value of 0.8 indicates a strong association.

22. (c) because an r 2 value of 0 indicates no association.

24. yˆ  317.78(0.714) x ; Yes, the model comes close to the data points.

26. a.

b. yˆ  10.37(1.55) x c. There are no outliers. The association is very strong, exponential, and positive; r  0.978. d. yˆ  10.37(1.55)6  144 programs; The residual is 134  144  10 programs. The observed number of U.S. bike-sharing programs in 2016 is 10 less than the predicted number of bike-sharing programs. e.

28. a.

yˆ  10.37(1.55)9  536; The mean number of U.S. bike-sharing programs per state in 2019 is 536 / 50  10.7. Even if a model fits the data points exactly, we still have little or no faith in our results when we extrapolate. yˆ  0.851(1.23) x

b. The coefficient a is 0.851; a person with a Framingham point score of 0 has an 0.85% risk of a heart attack in the next 10 years. c. 0.851(1.23)11  8.3% d. 0.851(1.23)16  23.4%

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222 ISM: A Pathway to Introductory Statistics 30. a.

b. yˆ  0.81(1.29) x c. The coefficient is 0.81; in 2005, 0.81 million, or 810,000 vinyl albums were sold in the U.S. d. The base is 1.29; the number of vinyl albums sold in the U.S increased by 29% per year. e. 0.81(1.29)15  36.9 million albums; This is extrapolation, so there is not much faith in this estimate.

32. a.

yˆ  0.66(1.096) x

b.

Both curves fit the model well. c. 66.39; Values may differ due to rounding. d. 711.21; Values may differ due to rounding. e. The sum of squared residuals for the regression model is over ten times that of the given model. This is very surprising, because we have come to expect the regression equation to have the smallest possible sum of squared residuals.

34. a.

b. yˆ  15.37(0.618) x c. There are no outliers. The association is strong, exponential, and negative; r  0.99. d. 15.37(0.618) 4  2.24 centimeters

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Chapter 10: Using Exponential Models to Make Predictions 223 34. (continued) e.

The point (3, 4.4) has the largest residual. The data point that is farthest above the model is (3, 4.4).

36. a. E  16.2(0.825) x , A  15.4(0.618) x , B  15.9(0.651) x b. The vertical intercept of E is (0,16.2). The vertical intercept of A is (0,15.4). The vertical intercept of B is (0,15.9). E has the highest intercept, so Erdinger Weissbier had the highest amount of froth immediately after being poured. This is consistent with the table. c. The percentage of decay for E is 17.5%. The percentage of decay for A is 38.2%. The percentage of decay for B is 34.9%. Augustinerbräu München has the highest rate of decay, which is consistent with the table.

It had the largest difference in froth height between 0 minutes and 6 minutes. d. The rate of decay of the froth depends on the area of the top of the froth. If the width of the container is variable, this area can change. e. h  15.9(0.651)5.5  1.5; V   (3.6) 2 (1.5)  61.07 cm3 38. a.

yˆ  336(1.00416) x

b. 336(1.00416) 454  2212.4 Newtons; The residual is 2659  2212.4  446.6 Newtons. This data point is above the graph of the model. c. 336(1.00416)842  11, 076.3 Newtons; The residual is 10, 754  11, 076.3  322.3 Newtons. This data point is below the graph of the model. d. r 2  0.967; about 96.7% of the variation in the force of the fired gunpowder is explained by the regression curve. e.

For large values of the explanatory variable, there is a pattern on the residual plot where the residuals become very large.

40. a.

yˆ  2.43(1.194) x

b. 2.43(1.194)16.6  46.12 tons per day c. 45  46.12  1.12 tons per day; The amount of fuel used was 1.12 tons less than predicted. d. r 2  0.986; 98.6% of the variation in the fuel consumption is explained by the regression curve.

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224 ISM: A Pathway to Introductory Statistics 40. (continued) e.

There appears to be a linear pattern on the residual plot where the dots are not all close to the zero residual line.

42. a.

yˆ  0.41(1.487) x

b. The coefficient is 0.41; in 2010, the annual revenue (in billions of dollars) from streaming music was 0.41 billion dollars. c. 0.41(1.487)5  3.0 billion dollars; The residual is 2.8  3  0.2 billion dollars. This data point is below the graph of the model. In 2005, the observed revenue was 0.2 billion dollars less than the predicted revenue. d.

The largest residual is 1  0.91  0.09 billion dollars. In 2012, the observed annual revenue from streaming music is 0.09 billion dollars more than the predicted centimeters greater than the predicted annual revenue from streaming music. e. r 2  0.997; 99.7% of the variation in annual revenue is explained by the regression curve.

44. a.

b. The linear regression equation is yˆ  8.28  0.68 x; the exponential regression equation is yˆ  3.68(1.03) x . Based on the scatterplot, the exponential regression equation seems to be a better match.

c. 3.68(1.03) 40  12.00 million; The residual is 9.82  12.00  2.18 million. Due to the war, it is not surprising that attendance was down, since both ballplayers and fans would be in the military at the time. d. 3.68(1.03)110  95.05 million; The residual is 73.06  95.05  21.99 million. Due to economic conditions, people may have skipped out on going to games.

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Chapter 10: Using Exponential Models to Make Predictions 225 44. (continued) e. 3.68(1.03)117  116.90 million; The residual is 72.68  116.90  44.22 million, which is larger than in 2010 despite the weaker economy. This does not suggest that MLB attendance is strongly affected by the economy. f. No, because causation cannot be inferred from an observational study such as this.

46. a.

b. The association is negative; the higher the percentage of the population involved in agriculture, the lower the GNP per person. This makes sense because fewer people involved in agriculture means more people in more advanced jobs. c.

yˆ  47,382.60(0.92) x ; the association is strong, exponential, and negative.

d. The base is 0.92; for each percentage point increase in a country’s agricultural labor, that country’s GNP per person decreases by 8%. e. The largest residual is Canada, with a residual of $19,334. This country has an advanced economy despite a relatively large percentage of its population working in agriculture.

48. Decrease a, leaving it as a positive number, and leave b the same.

50. This makes sense because if the model fits some data points more closely, this reduces the squared residuals for those points.

52. Answers may vary.

54. For each unit increase in x, y is multiplied by b. This is the same as increasing by b  1 percent.

56. If the residual has a pattern where the dots do not lie close to the zero residual line, if a dot represents an outlier, or if the vertical spread of the residual plot is not about the same for all values of the explanatory variable, then there may be problems with the exponential regression.

58. Association does not imply causation.

60. a.

b. Private non-profit: yˆ  8978(1.03) x ; public: yˆ  1924(1.035) x

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226 ISM: A Pathway to Introductory Statistics 60. (continued) c. For the private non-profit colleges, there are no outliers, and the association is strong, exponential, and positive; r  0.99.

For the public colleges, there are no outliers, and the association is strong, exponential, and positive; r  0.97. d. The base of the model for private non-profit colleges is 1.03; tuition and fees at private non-profit colleges have increased by 3% per year.

The base of the model for public colleges is 1.035; tuition and fees at public colleges have increased by 3.5% per year. e. The average one-year percentage change for the private non-profit colleges was 2.6% per year, which is less than b for this model.

The average one-year percentage change for the public colleges was 3.2% per year, which is less than b for this model. f. Private non-profit colleges had a higher mean tuition and fees than public colleges in 2014–2015, but since the growth rate for public colleges is greater than for private non-profit colleges, this may change in the future. We have little or no faith in this prediction since this assumes that the rates of change will remain the same in the future.

Chapter 10 Review Exercises

1. (2b5b 4 )3  (2b9 )3  23 b 27  8b 27

9. 324/5  (321/5 ) 4  24  16

2. (b 2 c) 4 (bc5 ) 2  (b8 c 4 )(b 2 c10 )

10. 163/4 

8  2 4 10

b

3.

c

10 14

b c

 2400 ( 405)  25  32 405 3 5

9 2

4. (8b c )(6b c )  48b

6.

12.

3 ( 9) 5 ( 2)

c

 48b 12 c3 

5.

48c

3

13.

12

b

b 1/3 b

4/3



(2b 5 c 2 )3

8b 15 c 6 8 158 6  ( 12) c   b 9 (3b 4 c 6 ) 2 9b8 c 12

 (37b 3c 4 )0  1 3 1

1 4

(18b c )(30b c )

 

20  27  b 18  30  b 3 6

b c

b 2 c 5



1 8

2  5 9 3

c

1/4

 32b 2 c5  14.  6 1   2b c 

b

b  b1c 1  c

3 2 6  ( 5)

b







  2b2c

 16b 2 ( 6) c51

1/4

1/4

15. (82/3 b 1/3c3/4 )(641/3 b1/2 c 5/2 )  (4b 2/6 c3/4 )(41 b3/6 c 10/4 )  40 b1/6 c 7/4 

3 ( 1) 1 ( 4)

c

1 5/3

2 11 7 2b11 b c  125 125c 7

 16b8 c 4

7. (37b 3c 4 ) 97 (37b 3c 4 )97  (37b 3c 4 ) 97 97

(20b c )(27b c )

1 23

2b 2 c 1 2 2 ( 9) 1 6   b c 6 4 3/2 9 6 125 (25b c ) 125b c

8 8c 6  b 23c6  23 9 9b

8.



(16b8 c 4 )1/4

1 3 ( 4) 123 bc9 b c   4 16b 4 c3 4

5 3

(161/4 )3

 b 1/3 4/3  b 5/3 

4b 3c12

2 9

1



11. b 4/5b 2/3  b 12/15b10/15  b 2/15  b 2/15

2400 2

1 163/4

b1/6 c 7/4

16. f (3)  43  64

c

17. g (2)  3(5) 2 

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3 5

2



3 25

Chapter 10: Using Exponential Models to Make Predictions 227 1 18. f    491/2  7 2

22. Domain: all real numbers or (, ) Range: y  0 or (, 0)

2 2  3 19. g     2(81) 3/4  3/4   4 27 81

20.

23. Domain: all real numbers or (, ) Range: y  0 or (0, ) 21.

24. x f ( x) g ( x) h( x) k ( x) 2 3 162 2 96 3 15 54 8 48 4 75 18 32 24 5 375 6 128 12 6 1875 2 512 6 25. a.

b. The graph of the equation is shown on the scatterplot in part (a). The association is exponential and strong. c. The y-intercept is (0,1.23); in 1950, the mean ticket price was $1.23. d. 2016: 1.23(1.05)66  $30.79; 2017: 1.23(1.05)67  $32.33 e.

$32.33  1.05; This ratio is equal to the base 1.05. This makes sense because on the basis of the base $30.79 multiplier property, if the year is increased by 1, the mean ticket price is multiplied by the base 1.05.

26. b3  8 b  81/3 b2

27. 2b5  60 b5  30

28. 3.9b7  283.5 283.5 b7  3.9 1/7

 283.5  b  3.9  b  1.84

b  301/5 b  1.97

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228 ISM: A Pathway to Introductory Statistics a  3.8

33.

29. 5b 4  13  67

y  3.8b x

5b 4  80

113.2  3.8b 4 113.2 b4  3.8

b 4  16 b  161/4 b  2

1/4

 113.2  b    3.8 

1 1 2 30. b 2   3 5 3

b  2.34

5b 2  3  10

y  3.8(2.34) x

2

5b  13 13 b2  5

34.

1/2

 13  b    5 b7

83 6 b 83 b5  6 2

a(0.7846)3  30 a  62.11 y  62.11(0.7846) x 1/6

b  0.7846



35.

1/5

 83  b  6 b  1.69

32.

7 30 ab 7 b6  30 

3

7 b     30 

b  1.61

31.

ab9

a2

ab 20

78.6 6.9 ab 78.6 b15  6.9 5

a  3.07 y  3.07(1.18) x 1/15

 78.6  b    6.9  b  1.176

y  2b x

a(1.176)5  6.9



3  2b5 3 b5  2 1/5

3 b  2

b  1.08 y  2(1.08) x

36. a.

b. Answers may vary. The regression equation is yˆ  16.16(1.0125) x . c. There are no outliers. The association is exponential, strong, and positive.

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Chapter 10: Using Exponential Models to Make Predictions 229 36. (continued) d. The coefficient is 16.16; 16.16% of newborns are inactive; because we have performed extrapolation, we have little or no faith in our result. Also, depending on how inactivity is interpreted for newborns, the result might be model breakdown. e. The base is 1.0125; the percentage of people who are inactive grows exponentially by 1.25% per year of age.

37. The plot has a pattern where the points do not lie close to the zero residual line.

38. a.

b. yˆ  32.66(0.849) x c. There are no outliers. The association is strong, exponential, and negative; r  0.97. d. The coefficient of determination is 0.95. 95% of the variation in the unemployment rate is explained by the exponential regression curve. e. The predicted unemployment rate is 32.66(0.849)16  2.38%; The residual is 2.5  2.38  0.12 percentage point. The unemployment rate for people who have 16 full-time-equivalent years of education is 0.12 percentage point greater than the model’s prediction.

39. a. Wˆ ( x)  1.79(1.077) x

b. Mˆ ( x)  1.11(1.093) x

c. The base of Wˆ , 1.077, is less than the base of Mˆ , 1.093. As ages increase, the quarterly rates for women grow exponentially at a rate less than the quarterly rate for men. d. Wˆ (52)  1.79(1.077)52  $84.74 Mˆ (52)  1.11(1.093)52  $113.13 e. Wˆ (62)  1.79(1.077)62  $177.92 Mˆ (62)  1.11(1.093)62  $275.28 Mˆ (62)  Wˆ (62)  $97.36

40. a.

c.

b. yˆ  3.23(1.0112) x d. The point (57, 6.27) has the largest residual; it is the data point that is farthest above the model. e. The residual plot has a pattern where the dots do not lie close to the zero residual line.

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230 ISM: A Pathway to Introductory Statistics 41.

Increase a and decrease b.

Chapter 10 Test

1. 322/5  (321/5 ) 2  22  4

9. f (2)  42 

1 1 1 1 2. 84/3   4/3   1/3 4    4   16 8 (8 ) 2 3. (2b3c8 )3  23 b33c83  8b3c 24  4b 3c  4.   1  25b5 c 9 

6.



1 16

1 1  3 10. f     43/2  3/2   2 8 4

b1/2

b3/6   b3/6 2/6  b1/6 b1/3 b 2/6 12. Domain: all real numbers or (, ) Range: y  0 or (0, )

25b 9 c 8

5  b 9  ( 10) c 8 ( 3) 10 3 7 35b c 5 5b  b1c 5  5 7 7c 2

2

2

 6b(b3c 2 )   6b 4 c 2   2b 2  4b 4 7.          c14  3b 2 c5   3b 2 c5   c7 

8.

4

2

11. Domain: all real numbers or (, ) Range: y  0 or (, 0)

0

5.

1

(25b8 c 6 )3/2

125b12 c 9  (7b 2 )(2c3 ) 1 7b 2 21 c 3 125  2 12 ( 2) 9 ( 3)  b c 7 250 14 6  b c 7 

250b14 7c 6

13. Answers may vary.

14. f (0)  6 15. f (1)  3 16.

a6 f ( x)  6b x 3  6b1 1 b 2 1 f ( x)  6   2

x

17. a.

b. The graph of the model is shown on the scatterplot in part (a). The model comes close to the data points.

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Chapter 10: Using Exponential Models to Make Predictions 231 17. (continued) c. The coefficient is 24.24. 24.24% of newborn drivers have been in speeding-related car accidents in the past five years. Model breakdown has occurred. d. The base is 0.956; the percentage of American drivers who have been in speeding-related car accidents decays exponentially by 4.4% per year of age. e. 24.24(0.956) 25  7.87%; According to the model, 7.87% of American drivers at age 25 years have been in a speeding-related car accident in the past five years 19.

18. 3b6  5  84

a  70 y  70b x

3b6  79 79 b6  3

20  70b6 2 b6  7

1/6

 79  b     3

1/6

2 b    7

b  1.72

b  0.81 y  70(0.81) x

20.

ab7

50 9 ab 50 b3  9 4

a (01.7711) 4  9 a  0.91



y  0.91(1.77) x 1/3

 50  b  9

b  1.7711

21. a.

b. Answers may vary. The exponential regression equation is yˆ  139.04(0.598) x . c. The exponential regression curve is shown on the scatterplot in part (a). The curve comes close to all of the data points. d. The coefficient is 139.04; in 2010, there were 139 bank failures. e. 139.04(0.598)7  4 bank failures; The residual is 6  4  2 bank failures. In 2017, the number of bank failures was 2 more than the model’s estimate.

22. Because the exponential curve changes significantly when the point is removed, this is an influential point.

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232 ISM: A Pathway to Introductory Statistics 23. a.

b. yˆ  9537.47(0.928) x c. The coefficient of determination is 0.97; 97% of the variation in the mean numbers of texts sent and received can be explained by the regression line. d. 9537.47(0.928)18  2485 texts e.

The point (15,3705) has the largest residual; it is the data point that is farthest above the model.

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