Essentials of business statistics 5th edition bowerman solutions manual 1 by judy.marcano245

Essentials of Business Statistics 5th Edition Bowerman Solutions Manual

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CHAPTER 6 Continuous Random Variables

§6.1, 6.2 CONCEPTS

6.1 For continuous probability distributions, probabilities are assigned to intervals of values on the real number line

LO06-01

6.2 Probabilities are computed by finding areas under the probability curve.

LO06-01

6.3 (1) f(x)  0 for all values x of the random variable; (2) the area under the curve equals 1.

LO06-01

6.4 The height of the probability curve is not a probability; it is the area under the probability curve that is a probability. Since the area under a single point is always equal to zero, we only find probabilities for intervals of x. The height of the probability curve at any given point represents the relative likelihood that x will be near the given point.

LO06-01

Chapter 06 - Continuous Random Variables 6-1

6.5 The use of the uniform distribution is appropriate when the distribution of the variable x over a certain interval has a rectangular shape That is, when the relative likelihood that x will be near a given point is the same for all points over an interval on the real number line.

LO06-02

§6.1, 6.2 METHODS AND APPLICATIONS

6.6 x is uniform with: c = 2 and d = 8

a.f(x) = 1 �� = 1 8 2 = 1/6 for 2 ≤ x ≤ 8; f(x) = 0 otherwise

Chapter 06 - Continuous Random Variables 6-2

1/6 1 2 3 4 5 6 7 8 x μ

P(3  x  5) = (5 – 3)(1/6) = 2/6 = 0.3333

P(1.5  x  6.5) = P(2  x  6.5) = (6.5 – 2)(1/6) = (9/2)(1/6) = 9/12 = 0.75 e. μ = ��+�� 2 = 2+8 2 = 5 σ2 = (�� ) 12 2 = (8 2) 12 2 = 3 σ =√3

b. f(x)

= 1.7321

6.6 f. [μ ± 2σ] = [5 ± 2(1.7321)] = [1.5359, 8.4641]

P(1.5359 ≤ x ≤ 8.4641) = P(2 ≤ x ≤ 8) = 1

LO06-02

6.7 Because the area under the rectangle must equal 1, we set (175 – 50) h = 1 and solve for h

Therefore, 125h = 1, and so h = 1/125 LO06-02

6.8 a. x is wait time in minutes

x is uniform with: c = 0 and d = 6

= 1 6 0 = 1/6 for 0 ≤ x ≤ 6; f(x) = 0 otherwise

= [1.268, 4.732]

6.10 To be a valid continuous probability distribution, the area must be equal to 1.

The figure is a triangle. The formula for the area is ½bh. The base of the triangle is b = 5 – 0 = 5.

Find the height k so that ½bk = 1

½(5)k = 1 so k = 2/5 LO06-01

Chapter 06 - Continuous Random Variables 6-3

��

f(x) = 1 ��

1/6 0 1 2 3 4 5 6

b. f(x)

x minutes

 x  4) =

– 2)(1/6) = 2

= 0.3333

 x  6) =

–

1/6) = (3/6) = 0.5

3)(

 x 

 x  6}) = (2 – 0)(1/6) + (6 – 5)(1/6) = (2/6) + (1/6) = (3/6) = 0.5 LO06-02 6.9 a. μ = ��+�� 2 = 0+6 2 = 3 σ2 = (6 0) 12 2 = (6) 12 2 = 3 σ =√3 =

[μ ± σ]

P(1.268 ≤

({0

2} or {5

1.732 b.

= [3 ± 1.732]

x ≤ 4.732) = (4.732 - 1.268)(1/6) = 0.5773 LO06-02

6.11 a. x is flight time in minutes

6.13 To be a valid continuous probability distribution, the area must be equal to 1. The formula for the area of the triangle is ½bh.

is b = 17 – 5 =

Chapter 06 - Continuous Random Variables 6-4

f(x) = 1 �� = 1 140 120 = 1/20; f(x) = 0 otherwise b. f(x) 1/20 120 125 130 135 140 x minutes c. P(125  x  135) = (135 – 125)(1/20) = 10(1/20) = 0.5 d. P(x  135) = P(135  x  140) = (140 – 135)(1/20) = 5(1/20) = .25 LO06-02

a. μ = ��+�� 2 = 120+140 2 = 130 σ = �� √12 = 140 120 34641 = 5.7735

[μ ± σ] = [130 ±

P(124.22650

≤

= 1357735–1242265 140 120 = 0.57735 LO06-02

x is uniform with: c = 120 and d = 140

6.12

5.7735]

[124.2265, 135.7735]

≤ x

135.7735)

base of the triangle

12. Find the height h so that ½bh = 1 or ½(12)h = 1 or (12/2)h = 1 so h = 2/12 = 1/6 LO06-01

rainfall in

x is uniform with: c = 3 and d = 6 f(x) = 1 �� = = 1 6 3 = 1/3 for 3 ≤ x ≤ 6; f(x) = 0 otherwise b. P(x ≥ 4) = (6 – 4)(1/3) = 2/3 1/3 3 4 5 6 x inches P(x ≥ 5) = (6 – 5)(1/3) = 1/3 1/3 3 4 5 6 x inches LO06-02 6.15 μ = ��+�� 2 = 3+6 2 = 4.5 inches σ = �� √12 = 6 3 34641 = 0.8660 inches [μ ± 2σ] = [4.5 ±

(2.7680 ≤ x ≤ 6.2320)

P(3 ≤ x ≤ 6) = 1 LO06-02

The

6.14

x is

inches

2(0.8660)] = [2.7680, 6.2320] P

§6.3 CONCEPTS

6.16 There is not just one normal distribution, but rather an entire family of normal distributions, one for each possible value of μ and σ.

The mean μ tells us where a specific normal curve is centered on the real number line.

The mean μ also tells us where the highest point on the normal curve is, as well as the value of the median and the mode.

The standard deviation σ measures the spread of the values described by the normal distribution. Approximately 2/3 of the values are within one standard deviation of the mean; 95% within two standard deviations; and 99%within three deviations.

LO06-03

6.17 To find the z value corresponding to x, we subtract the mean from x and divide the result by the standard deviation. This tells us the number of standard deviations that x is above or below the mean.

LO06-03

§6.3 METHODS AND APPLICATIONS

Chapter 06 - Continuous Random Variables 6-5

6.18 a. 40 30 20 10 0 014 012 010 008 006 004 002 000 X D e n s i t y 3 6 StDev DistributionPlot Normal,Mean=20 b. 40 35 30 25 20 15 10 014 012 010 008 006 004 002 000 X D e n s i t y 20 30 Mean DistributionPlot Normal,StDev=3 c. L06-03 250 200 150 100 50 0.04 0.03 0.02 0.01 0.00 X D e n s i t y 100 10 200 20 Mean StDev DistributionPlot Normal

6.19 a. z = �� σ = 25 30 5 = -1 x is one standard deviation below the mean.

b. z = �� σ = 15 30 5 = -3 x is three standard deviations below the mean.

c. z = �� σ = 30 30 5 = 0 x is equal to the mean.

d. z = �� σ = 40 30 5 = 2 x is two standard deviations above the mean.

e. z = �� σ = 50 30 5 = 4 x is four standard deviations above the mean.

LO06-03

6.20 a. P(0  z  1.5) = 0.9332 – 0.5000 = .4332

b. P(z  2) = 1 – .9772 = .0228

c. P(z  1.5) =.9332

d. P(z  –1) = 1 – .1587 =.8413

e. P(z  –3) = .00135

f. P(–1  z  1) = .8413 – .1587 = .6826

g. P(–2.5  z  .5) = .6915 – .0062 = .6853

Chapter 06 - Continuous Random Variables 6-6

6.20

6.21

e. -z.05 = -1.645

f. -z 10 = -1.28 LO06-03

Then use the table to solve.

z = �� σ = �� 1000 100

Chapter 06 - Continuous Random Variables 6-7

h. P(1.5  z  2) = .9772 – .9332 = .0440 i. P(–2  z  –.5) = .3085 – .0228 = .2857 LO06-03

.01 = 2.33

b. z.05 = 1.645

c. z.02 = 2.054

d. -z.01 = -2.33

a. P(1000  x  1200) = P(1000 1000 100  �� σ  1200 1000 100 ) = P(0 ≤ z ≤ 2) = .9772 – .5 = .4772 b. P(x > 1257) = P(z > 2.57) = 1 – .9949 = .0051 c. P(x < 1035)

P(z < .35) = .6368 d. P(857  x  1183) = P(–1.43  z  1.83) = .9664 – .0764 = .8900 e. P(x  700) = P(z  – 3) = .00135 f. P(812  x  913) = P(–1.88  z  –.87) = .1922 – .0301 = .1621 g. P(x > 891) = P(z > –1.09) = 1 – .1379 = .8621 h. P(1050  x  1250) = P(.5  z  2.5) = .9938 – .6915 = .3023 LO06-04

6.22 Restate each probability in terms of the standard normal random variable

6.23 First find the z-value from the table that makes the statement true. Then calculate k using the formula: k = z + = z(100) + 500

a. P(x  k) = 0.0250

P(z  1.96) = 0.0250

k = 1.96(100) + 500 = 696

P(x  696) = 0.0250

b. P(z  1.645) = 0.0500

k = 1.645(100) + 500 = 664.5

P(x  664.5) = 0.0500

c. P(z < -1.96) = 0.0250

k = -1.96(100) + 500 = 304

P(x < 304) = 0.0250

d. P(z ≤ -2.17) = 0.0150

k = -2.17(100) + 500 = 283

P(x  283) = 0.0150

e. P(z < 2.17) = 0.9850

k = 2.17(100) + 500 = 717

P(x < 717) = 0.9850

f. P(z > -1.645) = 0.9500

k = -1.645(100) + 500 = 335.5

P(x > 335.5) = 0.9500

Chapter 06 - Continuous Random Variables 6-8

0004 0003 0002 0001 0000 X D e n s t y 6960 0025 500 DstributonPlot Norma Mean 500 StDev 100

0004 0003 0002 0001 0000 X D e n s y 005 500 6645 DistributionPlot Norma Mean=500 StDev=100

0004 0003 0002 0001 0000 X D e n s t y 0025 500 3040 DistributionPlot Norma Mean 500 StDev 100

0004 0003 0002 0001 0000 X D e n s y 0015 500 2830 DistributionPlot Norma Mean 500 SDev 100

0004 0003 0002 0001 0000 X D e n s t y 0985 500 7170 DistributionPlot Norma Mean 500 StDev 100

0004 0003 0002 0001 0000 X D e n s i y 095 500 3355 DistributionPlot Norma Mean 500 SDev 100

g. P(z ≤ -1.96) = 0.9750

k = -1.96(100) + 500 = 304

P(x  696) = 0.9750

h. P(z > 2.00) = 0.0228

k = 2.00(100) + 500 = 700

P(x  700) = 0.0228

i. P(z > -2.00) = 0.9772

k = -2.00(100) + 500 = 300

P(x > 300) = 0.9772

c. (1) P(x > 140) = P(z > 2.5) = 1 – .9938 = .0062

(2) P(x < 88) = P(z < –.75) = .2266

(3) P(72 < x < 128) = P(–1.75 < z < 1.75) = .9599 – .0401= .9198

(4) P(–1.5  z  1.5) = .9332 – .0668 = .8664

d. P(x > 136) = P(z > 2.25) = 1 – .9878 = .0122; 1.22% LO06-03, LO06-04

6-9

Chapter 06 - Continuous Random Variables

0004 0003 0002 0001 0000 X D e n s i y 0975 500 6960 DistributionPot Norma Mean=500 StDev=100

0004 0003 0002 0001 0000 X D e n s t y 6999 00228 500 DistributionPot Norma Mean 500 StDev 100

0004 0003 0002 0001 0000 X D e n s t y 3001 09772 500 DistributionPot Norma Mean 500 StDev 100 LO06-05

6.24 a. b. z = �� = �� 100 16

6.25 a. (1) P(x  959) = P(

(2)

b. P(x > order) = 0.025 and P(z > 1.96) = 0.025

z = ��

= �� 800 75 = 1.96 order = 800 + 1.96(75) = 947 boxes of cereal

L06-04, LO06-05

6.26 Restate each probability in terms of z. Then use the table to solve.

a. P(7  x  9) = P( 7 8 05 ≤

b. P(8.5  x  9.5) = P(1 ≤ z ≤ 3) = .9987 – .8413 = .1574

c. P(6.5  x  7.5) = P(-3 ≤ z ≤ -1) = .1587 – .00135 = .15735

d. P(x  8) = P(z  0) = 1 – .5 = .5

e. P(x  7) = P(z  -2) = .0228

f. P(x  7) = P(z  -2) = 1 – .0228 = .9772

g. P(x  10) = P(z  4) = 1.0 (approximately)

h. P(x > 10) = P(z > 4) = 0 (approximately)

LO06-04

6.27 a. P(x  27) = P(�� σ ≤ 27 30 1 ) = P(z  -3.00) = 0.00135

b. Claim is probably not true, because the probability is very low of randomly purchasing a car getting no more than 27 mpg if the mean is actually 30 mpg.

LO06-04

Chapter 06 - Continuous Random Variables 6-10

�� σ ≤ 959

P(z  2.12)

800

) =

= .9830

P(�� σ > 1004 800 75 ) = P(z > 2.72) = 1 – .9967 = .0033

z < –2)

P(x > 1004) =

(3) P(x < 650) + P(x > 950) = P(

+ P(z > 2) = .0228 + (1 – .9772) = .0456

�� σ ≤

) =P(-2.0 ≤ z ≤ 2.0) =

.9772 – .0228 = .9544

6.28

Chapter 06 - Continuous Random Variables 6-11

x = Common stocks: x = 12.4, x = 20.6 y = Tax free municipal bond: y = 5.2, y = 8.6 a. P(x > 0) = P(�� σ > 0 124 20.6 ) = P(z > -0.60) = 1 – 0.2743 = 0.7257 b. P(y > 0) = P(�� σ > 0 5.2 86 ) = P(z > -0.60) = 1 – 0.2743 = 0.7257 c. P(x > 10) = P(�� σ > 10 12.4 206 ) = P(z > -0.12) = 1 – 0.4522 = 0.5478 d. P(y > 10) = P(�� σ > 10 52 86 ) = P(z > 0.56) = 1 – 0.7123 = 0.2877 e. P(x  -10) = P(�� σ ≤ 10 12.4 206 ) = P(z  -1.09) = 0.1379 f. P(y  -10) = P(�� σ ≤ 10 52 8.6 ) = P(z  -1.77) = 0.0384 LO06-04 6.29 P(in specification) = P(15.95 ≤ x  16.05) = P( 1595 160024 0.02454 ≤ �� σ ≤ 1605 160024 0.02454 ) = P(-2.14 ≤ z  1.94) = 0.9738 - 0.0162 = .9576 P(not in specification) = 1 – 0.9576 = 0.0424 LO06-04

the mileage

k: P(x < k) = 0.02 P(z < -2.05)

z = �� σ = �� 40,000 4,000 =

k = 40,000

LO06-05

6.30 Find

k so that only

of tires have mileage less than

= 0.02

-2.05

– 2.05(4,000) = 31,800 miles

6.31 a. By definition, the bottom 10% of the yearly returns are less than the tenth percentile and the bottom 90% of the yearly returns are less than the ninetieth percentile.

Find the yearly return on common stocks such that the bottom 10% of the yearly returns are below this number:

P(x < k) = 0.10

P(z < -1.28) = 0.10

z = �� σ = �� 124 206 = -1.28

k = -13.968

b. Find Q1 the yearly return on common stocks such that the bottom 25% of the yearly returns are below this number:

P(x < k) = 0.25

P(z < -0.67) = 0.25

z = �� σ = �� 12.4 206 = -0.67

k = -1.402

Find Q3 the yearly return on common stocks such that the bottom 75% of the yearly returns are below this number:

P(x < k) = 0.75

P(z < 0.67) = 0.75

z = �� σ = �� 124 206 = 0.67

k = 26.202 LO06-05

The test average is 75 with standard deviation 12. LO06-03

Chapter 06 - Continuous Random Variables 6-12

6.32 Student-1’s test score x1 = 63 and z score: z1 = -1 so z1 = �� σ = 63 μ σ = -1.00 Student-2’s test score x2 = 93 and z score: z2 = 1.5 so z2 = �� σ = 93 μ σ = 1.50

63 �� σ = -1.00 93 �� σ = 1.50 -μ = -1.00σ – 63 -μ = 1.50σ – 93 μ = 1.00σ + 63 μ = -1.50σ + 93 therefore 1.00σ + 63 = -1.5σ + 93 2.50σ = 30 σ = 12 μ

1.00σ

Solve the two equations for the two unknowns:

+ 63 = 12 + 63 = 75

6.33 a. Process A in control: μA = $0

Process B is investigated more often.

Process A is investigated more often.

B will be investigated more often.

= $10,000. Thus the probability of investigating an out of control Process B is:

we use

Chapter 06 - Continuous Random Variables 6-13

Process

P(xA > 2,500) = P(�� σ > 2500 0 5000 ) = P(zA > 0.50) = 1 – 0.6915 = 0.3085 P(xB > 2,500) = P(�� σ > 2500 0 10000 ) = P(zB > 0.25) = 1 – 0.5987 = 0.4013

σA = $5,000

B in control: μB = $0 σB = $10,000

σA =

Process B

of control: μB =

σB = $10,000 P(xA > 2,500) = P(�� σ > 2500 7500 5000 ) = P(zA > -1.00) = 1 – 0.1587 = 0.8413 P(xB > 2,500) = P(�� σ > 2500 7500 10000 ) = P(zB > -0.50) = 1 – 0.3085 = 0.6915

Process A out of control: μA = $7,500

$5,000

out

$7,500

Process

P(xB > k)

control so use μB = $0 and σB = $10,000

(z > 0.50) =

z = �� σ = �� 0 10000 = 0.50 Thus k = 5000,

will

μB = $7,500

σB

P(xB > 5,000) = P(�� σ > 5000 7500 10000 ) = P(zB > -0.25) = 1 – 0.4013 = 0.5987 LO06-04, LO06-05 6.34 P(x  k) = 0.33 P(z < -0.44) = 0.33 z = �� σ = �� 3000 500 = -0.44 k = $2780 LO06-05

Find

so that

= .3085 when process B is in

.3085 implies that

and we

investigate Process B if the cost variance exceeds $5000. If Process B is out of control

and

6.35 P(x  656) = 0.33

-0.44) = 0.33

§6.4

= -0.44σ – 656

μ = 1.96σ – 896 μ = 0.44σ + 656 μ = -1.96σ + 896

therefore 0.44σ + 656 = -1.96σ + 896 2.40σ = 240

= 100

μ = 0.44σ + 656 = 0.44(100) + 656 = 700

LO06-05

CONCEPTS

6.36 Binomial tables are often unavailable for large values of n, and it can be very time consuming to compute exact binomial probabilities when n is large.

LO06-06

6.37 We may use the normal distribution to approximate the binomial distribution when both np and n(1 – p) are at least 5. That is, np ≥ 5 and n(1 – p) ≥ 5.

LO06-06

6.38 If we attempt to find P(x = i) for some integer i under any continuous probability distribution, the answer will always be zero. We estimate P(x = i) by finding area under the continuous curve for the interval i – 0.5 and i + 0.5

When estimating probabilities for more general binomial events such as P(a ≤ z  b), we subtract 0.5 from the smallest integer and add 0.5 to the largest integer and then find the area under the normal curve of the new interval.

The correction is necessary because the binomial distribution, which is discrete, is being approximated by a normal distribution, which is continuous.

LO06-06

Chapter 06 - Continuous Random Variables 6-14

(x 

P(z ≤

0.975 z = �� σ = 656 �� σ = -0.44 z = �� σ = 896 �� σ = 1.96

896) = 0.975 P

≤

1.96) =

§6.4

LO06-06

Chapter 06 - Continuous Random Variables 6-15

ETHODS

PPLICATIONS 6.39 a. np = (200)(.4) = 80 and n(1 – p) = (200)(.6) = 120 so both np and n(1 – p) are  5 b. μ = np = (200)(.4) = 80 and σ = √�� = √(200)(.4)(.6) = √48 = 6.9282 (1) P(x = 80) = P(79.5  x  80.5) = P( 79.5 80 69282 ≤ �� σ ≤ 80.5 80 69282 ) = P(–.07  z  .07) = 0.5279 – 0.4721 = 0.0558 (2) P(x  95) = P(x  95.5) = P(z  2.24) = 0.9875 (3) P(x < 65) = P(x  64.5) = P(z  –2.24) = 0.0125 (4) P(x  100) = P(x  99.5) = P(z  2.81) = 1 – 0.9975 = 0.0025 (5) P(x > 100) = P(x  100.5) = P(z  2.96) = 1 – 0.9985 = 0.0015 LO06-06 6.40 a. np = (200)(.5) = 100 and n(1 – p) = (200)(.5) = 100 so both np and n(1 – p) are  5 b. μ = np = (200)(.5) = 100 and σ = √�� = √(200)( 5)(5) = √50 = 7.0711 P(x = 80) = P(79.5  x  80.5) = P( 795 100 7.0711 ≤ �� σ ≤ 805 100 7.0711 ) = P(-2.90  z  -2.76) = 0.0029 – 0.0019 = 0.0010 P(x  95) = P(x  95.5) = P(z  -0.64) = 0.2611 P(x < 65) = P(x ≤ 64.5) = P(z ≤ -5.02) = 0.0000 (approximately) P(x  100) = P(x  99.5) = P(z  -0.07) = 1 – 0.4721 = 0.5279 P(x > 100) = P(x ≥ 100.5) = P(z ≥ 0.07) = 1 – 0.5279 = 0.4721

AND A

LO06-06

(1 – p

(1000)(.8)

np and n(1 – p) are  5 (2) μ = np = (1000)(.2) = 200 and σ = √�� = √(1000)(.2)(.8) = √160 = 12.6491

P(x 

P(x  150.5) = P(z  -3.91) = 1 – 0.99995 = 0.00005

6.41 a. (1) np = (1000)(.2) = 200 and n

) =

= 800

both

(3)

150) =

b. No. If the claim were true, the probability of observing this survey result is only 0.00005

6.42 a. μ = np = (205)(.65) = 133.25 and

65)(35) = √46375 = 6.8292

b. No. If the claim were true, the probability of observing this result is very small. LO06-06

6.43 a. μ = np = (250)(.05) = 12.5 and

b. No LO06-06

6.44 μ = np = (2000)(.20) = 400 and

P(x  stk) = 0.01

P(z  2.33) = 0.01

z = ��

400 178885 = 2.33

20)(80) = √320 = 17.8885

= 3.4460

stk = 2.33(17.8885) + 400 = 441.7 Stock 442 items to be 99% sure of not running out.

LO06-06

§6.5 CONCEPTS

6.45 Explanations will vary.

LO06-07

6.46 x is exponential with parameter λ

x is the number of units of time or space between successive events

e is the base of Napierian logarithms

The mean of the exponential random variable is μ = 1/λ

The the standard deviation of the exponential random variable is σ = 1/λ so σ2 = (1/λ)2

f(x) = λe-λx for x ≥ 0; f(x) = 0 otherwise.

LO06-07

6.47 If the number of events occurring per unit of time or space has a Poisson distribution with mean λ, then the number of units of time or space between successive events has an exponential distribution with mean 1/λ.

LO06-07

Chapter 06 - Continuous Random Variables 6-16

P(x  117) = P(x  117.5) = P(�� σ ≤ 1175 13325 68292 ) = P(z  -2.31) = 0.0104

√��

√(205)(

√��

√(250)(.05)(.95)

≥

x ≥ 39.5) = P(�� σ ≥ 39.5 12.5 34460 ) = P(z ≥ 7.84) = 0 (approximately)

= √11.875

P(x

40) = P(

√�� =

√(2000)(

��

§6.5 METHODS AND APPLICATIONS

6.48

Chapter 06 - Continuous Random Variables 6-17

a. x

exponential with λ = 2 f(x) = λe-λx = 2e-2x for x ≥ 0

Graph

included in this manual. c. P(x ≤ 1) = P(0 ≤ x ≤ 1) = e -2•0 – e -2•1 = e0 – e-2 = 1 – 2.7183-2 = 1 – 0.1353 = 0.8647 d. P(0.25 ≤ x ≤ 1) = e-2(.25) – e -2•1 = e -.5 – e-2 = 0.6065 – 0.1353 = 0.4712 6.48 e. P(x ≥ 2) = e -λa = e -2•2 = 2.7183-4 = 0.0183 f. μ = 1/λ = ½ = 0.5 σ2 = (1/λ)2 = (½)2 = ¼ = 0.25 σ = 1/λ = ½ = 0.5 g. [μ ± 2σ] = [0.5 ± 2(0.5)] = [-0.5, 1.5] P(-0.5 ≤ x ≤ 1.5) = P(0 ≤ x ≤ 1.5) = e-2(0) – e-2(1.5) = e0 – e-3 = 1 – 0.0498 = 0.9502 LO06-07 6.49 a. x is exponentially distributed with λ = 3 f(x) = λe-λx = 3e-3x for x ≥ 0 b. Graph not included in this manual. c. P(x  1) = .9502 d. P(.25  x  1) = .4226 e. P(x  2) = .0025 f. μ = 1/λ = 1/3 σ2 = (1/λ)2 = (1/3)2 = 1/9 σ = 1/λ = 1/3 g. [μ ± 2σ] = [1/3 ± 2(1/3)] = [-1/3, 1] P(-1/3 ≤ x ≤ 1) = P(0 ≤ x ≤ 1) = e-3(0) – e-3(1) = e0 – e-3 = 1 – 0.0498 = 0.9502 LO06-07

not

6.50 a. y is the number of bank customers arriving per 15 minutes

y is Poisson with μ = 7 minutes

x = time between bank customers arriving in minutes

x is exponentially distributed with λ = 7/15 minutes

f(x) = λe-λx = (7/15) e-7x/15 for x ≥ 0

b. Graph not included in this manual.

c. P(a ≤ x ≤ b) = e -λa – e -λb = e-7a/15 – e-7b/15

(1) P(1  x  2) = .2338

(2) P(x < 1) = .3729

(3) P(x > 3) = .2466

(4) P(.5  x  3.5) = .5966

d. μ = 1/λ = 15/7 = 2.1429

σ2 = (1/λ)2 = (15/7)2 = 225/49 = 4.5918

σ = 1/λ = 15/7= 2.1429

e. [μ ± σ] = [15/7 ± (15/7)] = [0, 30/7] P(0 ≤ x ≤ 30/

6.51 x = phone call length in minutes x is exponential with μ = 1.5 minutes

min

= (2/3) e-2x/3 for x ≥ 0

b. Graph not included in this manual.

c. P(a ≤ x ≤ b) = e -λa – e -λb = e-2a/3 – e-2b/3

(1) P(x  3) = .8647

(2) P(1  x  2) = .2498

(3) P(x > 4) = .0695

(4) P(x < .5) = .2835

LO06-07

Chapter 06 - Continuous Random Variables

6-18

7) = e0 – e-(7/15)(30/7) = 1 – e-2 = 1 – 0.1353 = .8647 [μ ± 2σ] = [15/7 ± 2(15/7)] = [-15/7, 45/7] P(-15/7 ≤ x ≤ 45/7) = P(0 ≤ x ≤ 45/7) =e-(7/15)(0) – e-(7/15)(45/7) = 1 – e-3 = 1 – 0.0498 = .9502 LO06-07

λe

λx

λ = 1/μ = 1/1.5 = 2/3

(

) =

6.52 a. b = # breakdowns per hour

b is Poisson with μb = λ = 2/500 = 0.004 breakdowns per hour

x = time between breakdown in hours

x is Exponential with λ = 0.004 μ = 1/λ = 250 hours between breakdowns

b. f(x) = λe-λx = 0.004e-.004x for x ≥ 0

c. Graph not included in this manual.

d. P(x ≤ 5) = 1 – e-.004(5) = 1 – e-.02 = 1 – 0.9802 = 0.0198

e. P(100 ≤ x ≤ 300) = e-(.004)(100) – e-(.004)(300) = e-.4 – e-1.2 = 0.6703 – 0.3012 = 0 3691

f. Probably not; the probability of this result is quite small (.0198) if the claim is true.

LO06-07

6.53 a. a = # accidents per month

a is Poisson with μa = λ = 1 accident per month

x = time between accidents in months

x is Exponential with μx = 1/λ = 1 month between accidents

λ = 1 f(x) = λe-λx = e-x for x ≥ 0

(1) P(x> 2) = e-2 = 2.7183-2 = 0.1353

(2) P(1 ≤ x ≤ 2) = e-1 – e-2 = 0.3679 – 0.1353 = 0.2326

(3) P(x ≤ 0.25) = 1 – e-.25 = 1 – 0.7788 = 0.2212

b. Probably not; the probability of this happening is .2212 (which is not terribly small).

LO06-07

§6.6 CONCEPTS

6.54 (1) Arrange the data in order from smallest to largest

(2) For each observation number i, compute the quantity i/(n+1)

(3) For each observation number i, compute the Standardized Normal Quantile Value

(4) Plot the data on the y-axis vs the Standardized Normal Quantile Value on the x-axis

LO06-08

6.55 That the data approximately follows a normal distribution

LO06-08

Chapter 06 - Continuous Random Variables

6-19

§6.6 METHODS AND APPLICATIONS

6.56 a. & b.

c. Data are skewed because the normal probability plot is not approximately linear.

LO06-08

6.57 Since the normal curve plot below is not approximately linear, the data do not follow a normal distribution.

LO06-08

Chapter 06 - Continuous Random Variables

6-20

Income (1000's) i/(n+1) z-value 7.524 0.0769 -1.43 11.07 0.1538 -1.02 18.211 0.2308 -0.74 26.817 0.3077 -0.50 36.551 0.3846 -0.29 41.286 0.4615 -0.10 49.312 0.5385 0.10 57.283 0.6154 0.29 72.814 0.6923 0.50 90.416 0.7692 0.74 135.54 0.8462 1.02 190.25 0.9231 1.43

NormalProb.Plot 0.000 50.000 100.000 150.000 200.000 -1.43 -1.02 -0.74 -0.50 -0.29 -0.10 0.10 0.29 0.50 0.74 1.02 1.43 z-score Income Normal Curve Plot 0 2 4 6 8 10 12 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 Normal Score Ratings

6.58 Data are approximately normal because the normal curve plot below is close to a straight line.

LO06-08

SUPPLEMENTARY EXERCISES

6.59 x = bottle fill in ounces

normally distributed with

LO06-04

6.60 x = men’s height in inches

x is normally distributed with = 73.5” and  = 1”. a. P(x < 71 or x >

Chapter 06 - Continuous Random Variables

6-21

�� σ < 1595 16 002 ) = P(z < 2.5) = 0 0062 or 0.62%

x is

= 16 oz. and  = 0.02 oz. P(x < 15.95) =

(

1 – P(71 ≤ x ≤ 76) = P( 71 735 1 ≤ �� σ ≤ 76 735 1 ) = 1 – P(-2.5 ≤ z ≤ 2.5) = 1 – (.9938 – 0.0062) = 1 – 0.9876 = 0.0124

P(x  74) = P(z  .5) = 1 – .6915 = .3085

.00135

Normal Curve Plot 29.5 30.0 30.5 31.0 31.5 32.0 32.5 33.0 33.5 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 Normal Score MPG

76) =

c. P(x < 70.5) = P(z < -3) =

LO06-04

6.61 x = IQ

x is normally distributed with = 100 and  = 16.

a. P(x  80) = P(z  -1.25) = 1 – 0.1056 = 0.8944

b. find IQ so that P(x  k) = 0.95

P(z < -1.645) = 0.05 therefore P(z > -1.645) = 0.95

z = �� σ = �� 100 16 = -1.645 k = 100 – 1.645(16) = 73.68

The minimum IQ required to attend public school would be 73.68 or 73 since IQs are integers. LO06-04, LO06-05

6.62 x = round-off error in cents

x is uniformly distributed with c = -0.5 and d = 0.5 cents a.

Chapter 06 - Continuous Random Variables 6-22

��

.5+.5 =

-0.5 ≤ x ≤ 0.5, f(x) = 0 otherwise

f(x) 1.0 -0.5 -0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 x cents c. P({x > 0.3} or {x < -0.3}) = .2 + .2 = .4 d. P({x > 0.1} or {x < -0.1}) = .4 + .4 = .8 6.62 e. μ = ��+�� 2 = 5+5 2 = 0 cents σ2 = (�� ) 12 2 = (�� ) 12 2 = σ = 5 ( 5) √12 = 1 34641 = 0.2887 f. [μ ± σ] = [0 ± 0.2887] = [-0.2887, 0.2887] P(-0.2887 ≤ x ≤ 0.2887) = (0.2887 – (-0.2887))(1) = 0.5774 LO06-02

f(x) = 1

�� = 1 .5 ( .5) = 1

1 for

6.63 x = interest rate forecast in percent

x is normally distributed with x = 5 percent and x = 1.2 percent

a. P(x  3.5) = P(z  –1.25) = 1-.1056 = .8944

b. P(x  6) = P(z  0.83) = .7967

c. P(3.5  x  6) = .7967 – .1056 = .6911

LO06-04

6.64 x = interest rate forecast in percent

x is normally distributed with = 5 percent and  = 1.2 percent

a. By definition, 10% of the individual forecasts fall below the 10th percentile, x10th, of the distribution of forecasts. This means that 90% of the individual forecasts fall at or above the 10th percentile.

If the top 10% of the z-scores are above z.10 = 1.282, then, by symmetry, the bottom 10% of the z-scores will be below -z.10 = -1.282.

6.65 x = test scores

x is normally distributed with = 200 points and  = 50 points

Find the test score k so that only 2.5% of test takers pass: P(x

= 0.025 P

k = 200 + 1.96(50) = 298 Set the lowest passing score to 298 points

Chapter 06 - Continuous Random Variables 6-23

P(x  x10th) = P(�� σ  ��10th 5 12 ) = P(z  -1.282) = 0.10 ��10th 5 12 = -1.282 x10th = 5 + (-1.28)(1.2) =

percent

3.464

Q1 is the 25th percentile, P(x  Q1) = P(�� σ  ��1 5 12 ) = P(z ≤ -0.67) = 0.25 ��1 5 12 = -0.67 Q1 = 5 + (-0.67)(1.2) = 4.196 percent Since Q3 is the 75th percentile, P (x  Q3) = P(�� σ  ��3 5 12 ) = P(z ≤ 0.67) = 0.75 ��3 5 1.2 = 0.67 Q3 = 5 + (0.67)(1.2) = 5.804 percent LO06-05

b. Since

(z >

z 025

�� σ = �� 200 50

≥ k)

1.96) = 0.025

= 1.96

LO06-05

6.66 x = lengths

x is normally distributed with = μ and  = 0.02”

Find the mean μ so that only 0.4% of the parts will be shorter than 15”

P(x < 15) = P(z < -z 004) = P(z < -2.65) = 0.004

-z 004 = �� σ = 15 �� 002 = -2.65

μ = 15 + 0.053 = 15.053 Set the machine to average 15.053”

LO06-04

6.67 x = number of people with reservations who show up

x is binomially distributed with n = 325 p = 0.9 q = 0.1

x is approximately normally distributed

since np = 325(.9) = 292.5 is more than 5 and nq = 325(.1) = 32.5 is more than 5 with = np = 325(.9) = 292.5 people and  = √�� = √325( 9)(1) = √2925 = 5.4083 people

P(x ≤ 300.5) = P(�� σ ≤ 3005 2925 54083 ) = P(z ≤ 1.48) = 0.9306

LO06-06

6.68 y = # errors

y is Poisson with 4 errors per 1000 lines of code: μy = 4/1000 = 1/250

x = #lines between errors

x is exponentially distributed with λ = 1/250

f(x) = λe-λx = (1/250)e-x/250 for x ≥ 0

a. P(x ≥ 400) = e -λa = e-400/250 = 0.2019

b. P(x ≤ 100) = 1 - e -λb = e-100/250 = 1 – 0.6703 = 0.3297

LO06-07

Chapter 06 - Continuous Random Variables 6-24

6.69 a. b. The probabilities below were computer generated Answers obtained using the normal table may be slightly different.

c. The probability of a return greater than 50% is:

essentially zero for Fixed Annuities, Cash Equivalents, US Treasury Bonds, US Investment Grade Corporate Bonds, Non-US Government Bonds, and Domestic Large Cap Issues. greater than 1% for International Equities, Domestic MidCap Stocks, and Domestic Small Cap Stocks.

greater than 5% for Domestic Small Cap Stocks.

d. The probability of a loss is:

essentially zero for Fixed Annuities and Cash Equivalents. greater than 1% for all investment-classes except Fixed Annuities and Cash Equivalents greater than 10% for all investment classes except Fixed Annuities, Cash Equivalents, and US Treasury Bonds.

greater than 20% for Domestic Large Cap Stocks, International Equities, Domestic MidCap Stocks, and Domestic Small Cap Stocks. LO06-04 6.70

Chapter 06 - Continuous Random Variables

6-25

Fixed Annuities Cash Equivalents US Treasury Bonds US Corporate Bonds Non-US Government Bonds Domestic Large Cap Stocks International Equities Domestic Midcap Stocks Domestic Small Cap Stocks μ 8.31% 7.73% 8.80% 9.33% 10.95% 11.71% 14.02% 13.64% 14.93% σ 0.54% 0.81% 5.98% 7.92% 10.47% 15.30% 17.16% 18.19% 21.82% P(x < 0) 0 0 0.0708 0.1190 0.1469 0.2206 0.2061 0.2266 0.2483 P(x > 5%) 1.0000 0.9996 0.7374 0.7077 0.7151 0.6695 0.7004 0.6826 0.6755 P(x > 10%) 0.0009 0.0025 0.4205 0.4663 0.5361 0.5445 0.5926 0.5793 0.5894 P(x > 20%) 0 0 0.0305 0.0890 0.1937 0.2940 0.3637 0.3633 0.4081 P(x > 50%) 0 0 0 0 0.0001 0.0062 0.0180 0.0228 0.0540

x = water consumption x is normally distributed with = 800,000 gallons and  = 80,000 gallons P(x > 984,000) = P(�� σ > 984,000 800,000 80,000 ) = P(z > 2.30) = 1 – 0 9893 = 0.0107 LO06-04

x = time of repair x is uniformly distributed with c = 10 minutes and d = 25 minutes P(x ≥ 15) =P(15 ≤ x ≤ 25) = �� = 25 15 25 10 = 10/15 = 2/3 LO06-02

6.71

6.72 This problem can be solved by defining the random variable, its parameters and the probability to be computed in time units of either minutes or hours.

x = waiting time in seconds

x is exponentially distributed with μ = 60 seconds and λ = 1/μ = 1/60 seconds

a. P(x ≥ 90) = e -λa = e-(1/60)(90) = 0.2231

b. P(x ≥ 120) = e -λa = e-(1/60)(120) = 0.1353

x = waiting time in minutes

x is exponentially distributed with μ = 1 minute and λ = 1/μ = 1/1 = 1 minute

a. P(x ≥ 1.5) = e -λa = e-(1)(1.5) = 0.2231

b. P(x ≥ 2) = e -λa = e-(1)(2) = 0.1353

LO06-07

6.73 x = net interest margin

x is normally distributed with = 4.15 percent and  = 0.5 percent

a. P(x > 5.40) = P(��

> 540 415 05 ) = P(z > 2.50) = 1 – 0.9983 = 0.0062

b. P(x < 4.40) = P(

c. Find the margin k so that k is smaller than 95% of all margins:

(x > k) = 0.95 which is the same as

z = �� σ = �� 415 05 = -1.645

k = 4.15 – 1.645(0.5) = 3.3275%

LO06-04, LO06-05

6.74 x = # people who purchased blue jeans

x is binomially distributed with n = 400 p = 0.5 q = 0.5

x is approximately normally distributed since np = 400(.5) = 200 > 5 and nq = 400(.5) = 200 > 5 with = np = 400(.5) = 200 and  = √�� = √400(.5)(.5) = √100 = 10

a. P(x ≤ 180.5) = P(�� σ ≤ 180.5 200 10 ) = P(z ≤ -1.95) = 0.0256

b. Yes since the probability in part a. is small.

LO06-06

Chapter 06 - Continuous Random Variables 6-26

��

�� σ <

) = P(z <

440 415 05

0.50) = 0.6915

P(x < k) = 0.05

(

< -1.645) = 0.05

6.75 x = marriage age

x is normally distributed with = 26 years and  = 4 years

Chapter 06 - Continuous Random

6-27

Variables

P(20 ≤ x < 30) = P(20 26 4 ≤ �� σ < 30 26 4 ) = P(-1.50 ≤ z < 1.00) = 0.8413 – 0.0668 = 0.7745

LO06-04