CHAPTER 1
EXERCISE 1 a) Mean = 17.00, SD = 5.53, median = 17.00, Range = 22, IQR = 20 – 13 = 7. b) See boxplot.
c) A typical value for X is 17 which is both the mean and the median. About half the values are between 13 and 20. The boxplot shows that the upper tail is slightly longer, but difference is small and mean and median are equal. This distribution is roughly symmetric.
Chapter 1 – Page 1
EXERCISE 2 Variable
Mean
Median
Variance
Standard Deviation
Shape
WATER
7.125
1.500
452.864
21.28
extremely positively skewed
VEG
1.120
0.00
4.327
2.080
extremely positively skewed
FOWL
75.635
11.5
42197.33
205.420
extremely positively skewed
WATER Stem Leaf
#
Boxplot
14 9 13 12 11 10 9 8 7 6 5 4 3 13 2 1 05667 0 00000000000001111111111112222222222345556779 + + + + + + + + Multiply Stem.Leaf by 10**+1
1
*
2
*
5 44
0 +--+--+
VEG Stem Leaf 9 0 8 8 0 7 7 0 6 6 5 5 2 4 4 0 3 3 0 2 8 2 00002 1 58 1 0002 0 58 0 00000000000000000000000000000002 + + + + + +
# 1
Boxplot *
1
*
1
*
1
0
1
| | | | | +-----+ | + | | | *-----*
1 1 5 2 4 2 32
Chapter 1 – Page 2
FOWL Stem Leaf 14 1 13 12 11 10 9 8 7 6 5 4 3 6 2 124 1 227888 0 00000000000000000000001111112222223335678 + + + + + + + +Multiply Stem.Leaf by 10**+2
# 1
Boxplot *
1 3 6 41
* 0 0 +--+--+
b) Frequency distribution for FOWL Fowl 0 ≤ x < 100 100 ≤ x < 200 200 ≤ x < 300 300 ≤ x < 400 1400 ≤ x < 1500
Frequency 41 6 3 1 1
Midpoint 50.0 150.0 250.0 350.0 1450.0
approximate mean = 5500/52 = 105.8 approximate variance = 50746
c) Scatterplots
fow l 1500 1400 1300 1200 1100 1000 900 800 700 600 500 400 300 200 100 0 0
1
2
3
4
5
6
7
8
9
veg
Aside from the single point with extraordinary number of waterfowl, there is little relationship.
Chapter 1 – Page 3
EXERCISE 3 a) INDEX: mean = -0.149 median = -0.155
variance = 1.771
IQR = 1.59 (show stem and leaf, histogram, or boxplot as part of answer)
FUTURE: mean = -0.208 median = -0.3 variance = 0.601 IQR = 0.81 (show stem and leaf, histogram, or boxplot as part of answer)
b) The Scatter plot is shown below. There is a clear trend for positive changes in the FUTURE contract value to be associated with positive changes in the NYSE INDEX. Yes, FUTURE can be used to help predict changes in the NYSE index.
index 3 2 1 0 -1 -2 -3 -4 -3
-2
-1
0
1
2
future
Chapter 1 – Page 4
EXERCISE 4 Y1
Y2
Y3
Stem Leaf 8 14 7 3 6 1237889 5 338 4 025 3 112 2 478 1 668 +
Stem Leaf 6 5679 6 4 5 789 5 22 4 678 4 0024 3 559 3 12333 + Stem Leaf 9 4 8 0 7 6 777 5 444 4 000 3 2 7777777 1 333333 0 0 +
+
+
+
+
# 2 1 7 3 3 3 3 3
Boxplot | | +-----+ *-----* | + | +-----+ | |
# 4 1 3 2 3 4 3 5
Boxplot | | +-----+ | | *--+--* | | +-----+ |
# 1 1
Boxplot | | | | +-----+ | | | + | *-----* +-----+ |
+
+
3 3 3 7 6 1 +
+
+
Chapter 1 – Page 5
Y4
Stem Leaf 7 4 6 5 0 4 578 3 99 2 123357 1 01468 0 1113699 +
# 1 1 3 2 6 5 7 +
+
Boxplot | | | | +-----+ *--+--* | | +-----+
+
Y1 is slightly negatively (or left) skewed; Y2 is nearly symmetrically distributed, though without a well defined peak; Y3 is positively (or right) skewed; Y4 is extremely positively skewed. comparison to empirical rule, actual percentage of observations in interval mean
Std dev
Shape
±1 SD (expect 68%)
±2 SD (expect 95%)
±3 SD (expect all)
Y1
4.840
2.107
slight left skew
68%
100%
100%
Y2
4.748
1.262
near symm.
60%
100%
100%
Y3
3.696
2.434
right skew
76%
96%
100%
Y4
2.328
1.888
extreme right skew
64%
96%
100%
The empirical rule works reasonably well even for the skewed distributions.
Chapter 1 – Page 6
EXERCISE 5 a) DAYS: Mean= 15.854 median= 17, variance= 24.324
TEMP: Mean= 39.348 median= 40 variance= 11.152
b) Scatterplot:
days 30
20
10
0 30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
temp
Chapter 1 – Page 7
From the scatterplot, there appears to be no definitive relationship between the average temperature and the number of rainy January days.
EXERCISE 6 a) The distribution of EXPEND is extremely right skewed with a mean ($1816M) that is much greater than the median ($1169M), so the best measure of location is the median and the best measure of dispersion is the IQR ($1938M). There are three outlier states with California being an extreme outlier at $15348M. b) Per Capita expenditures are also right skewed, but slightly less so than expenditures. The median is $0.269k ($2,690) and the IQR is $0.139k. c) For Per Capita expenditures, there is one state that is an extreme outlier (Alaska) and another that is a moderate outlier (Delaware). Both these are states with small populations, which may mean that there are a lot of fixed costs in the systems.
EXERCISE 7 Scatter of DFOOT vs. HCRN:
dfoot 6
5
4
3
2 1
2
3
4
5
hcrn
6
7
8
9
Scatter of
Chapter 1 – Page 8
dfoot vs. ht:
dfoot 6
5
4
3
2 20
21
22
23
24
25
26
27
28
29
30
27
28
29
30
31
32
ht
Scatter of HCRN vs. HT:
hcrn 9 8 7 6 5 4 3 2 1 20
21
22
23
24
25
26
ht
31
32
The
strongest relationship exists between DFOOT, the diameter of the tree at one foot above ground level and HT, the total height of the tree. (notice from the scatter plot that it is closest to a linear relationship) One would expect that as the base of the tree increases in diameter the tree would increase in height as well.
Chapter 1 – Page 9
EXERCISE 8 a and b) The histogram for the ages of the cases slightly left-tailed (it would be less so except for the artificial truncation of the uppermost class). But the histogram for the ages of those who died is extremely left skewed – almost all the deaths are among older cases.
c) The death rate increases dramatically after age 50.
Chapter 1 – Page 10
EXERCISE 9 a) Median was 249 and mean 292.4. The mean is larger than (to the right of) the median, indicating a distribution skewed to the right. Yes, both the stem and leaf plot and the box plot reveal the skewness of the distribution. b) The outliers 955 and 1160 may have resulted from younger patients or from patients diagnosed earlier or in whom the disease was less severe. c) 38 out of 51 patients were in remission for less than one year, so approximately 75%.
EXERCISE 10 a) For the combined group, scores are almost evenly distributed over the interval from 0 to 100.
1 7 .5
1 5 .0
1 2 .5 P e 0 .0 r 1 c e .5 n 7 t 5 .0
2 .5
0 0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 0
1 0 0
sco re
Chapter 1 – Page 11
b)
30 25 P e 20 r 0 c 1 5 e n t 10 s t a t u s
5 0 30 25 P e 20 r 1 c 1 5 e n t 10 5 0 0
10
20
30
40
50
60
70
80
90
100
6 0
7 0
8 0
9 0
1 0 0
sco re
30 25 P e 20 r 2 c 1 5 e n t 10 s t a t u s
5 0 0
1 0
2 0
3 0
4 0
5 0 sco re
As expected, the placement scores among those who passed the class (Status=0) tend to be higher than the scores of those who failed the class (Status=1). Surprisingly, the scores for those who withdrew (Status=2) are as high as those for those who passed. Perhaps this is because those who withdrew do so for a variety of reasons frequently unrelated to their math ability.
NOTE TO INSTRUCTOR: The differences between the groups might best be displayed with boxplots, though boxplots have a tendency to hide multiple modes.
Chapter 1 – Page 12
100
75
s c o r e
5 0
25
0 0
1
2
sta tu s
Chapter 1 – Page 13
EXERCISE 11 a) Since the data is sequentialby year, use a time series plot.
b) Prices have been increasing, with an especially sharp jump in the period from 2007 to 2007.
EXERCISE 12 a) Reliability (36.7%) and Enthusiasm (25%) are the traits most frequently thought of as most important.
4 0
Relative Frequency
3 0
2 0
1 0
0 R E LB LE N T H US LF S TG G R M HE LO Q TP U S H Y
TRAIT
Chapter 1 – Page 14
EXERCISE 13a) 4
Half-Life of drug
3
2
1
0 1
2
3
4
5
6
7
8
9
10
11
Dosage in mg per kg
The initial dosage seems to be higher for drug A but the half-lives are not much different. b) There is not any strong relationship between half-life and dosage, either overall or within the individual drugs. c) Drug A: M = 9.21, SD = 1.14;
Drug B: M = 2.67, SD = 0.44;
Typical dosage is much higher for
Drug A. This supports the conclusion in part (a).
Chapter 1 – Page 15
EXERCISE 14 a) IQR = 119-18 = 101, upper fence = 119+1.5*101 = 270.5, upper whisker will stop at 253 and there is one outlier at 300. See boxplot below. This distribution is very right (or positively) skewed. A small number of homes linger on the market for a long time. b) The mean will be greater than the median.
EXERCISE 15. The boxplots show that typical wind speeds are higher at 4PM than at 4AM. There is not much difference in the dispersions (the IQR is slightly higher at 4AM but the range slightly higher at 4PM). Both distributions are very right skewed, with numerous outliers on the right side.
Chapter 1 – Page 16
CHAPTER 2 EXERCISE 1 a) Y
50,000
10,000
1,000
10
0
P(y)
1/150,000
5/150,000
25/150,000
1,000/150,000
148,969/150,000
b) mean= (1/150000) * 50000 + (5/150000) * 10000 + (25/150000) * 1000 + (1000/150000) * 10 + (148,969/150,000)*0 = $0.90 c) no, expected value is only $0.90 so there would be an expected (average over many tickets) loss of $0.10 per ticket. d) σ²= (50000-.9)²*(1/150000) + (10000-.9)²*(5/150000) + (1000-.9)²*(25/150000) + (10-.9)²*(1000/150000) + (0-.9)²*(148969/150000) = 20166.5233 σ = $142.01 This distribution is very skewed, so the standard deviation does not reflect the very large (but low probability) prizes, even if you use an interval of mean +3 standard deviations.
EXERCISE 2 a) 0.5, using area of rectangle with base from 0 to 1. b) using formulas for uniform distribution in section 2.4, µ = 1, Variance = 4/12 = 1/3.
EXERCISE 3 a) mean=(0*.3277)+ (1*.4096) + (2*.2048) + (3*.0512) + (4*.0064) + (5*.0003) = 0.9999 Variance= 0.7997 b) µ= np = (5)(.2) = 1
σ² = np(1-p) = (5)(.2)(.8)= .8
Yes, they agree except for small rounding error.
EXERCISE 4 The system will fail if at least one component fails (since they are in series). Hence, they will fail if they do not all work. Probability of failing = 1 – probability all work = 1 – (0.999)10 = 0.00996
Chapter 2 – Page 1
EXERCISE 5 Arrangement 1: probability of failure = prob [(A1 fails and A2 fails) OR (B1 fails and B2 fails)] = (0.01)2 + (0.01)2 – (0.01)4 = 0.0001999 Arrangement 2: probability of failure = Prob[(A1 fails or B1 fails) AND (A2 fails or B2 fails)] = (0.01 + 0.01 – 0.012)2 = 0.000396 Arrangement 1 is most reliable.
EXERCISE 6 If you use a scientific calculator rather than table A.1, your values may differ slightly. a) 0.1587 b) 0.8413 c) 0.5 – 0.1587 = 0.3413 d) 1 – 0.9332 – 0.0668 e) 0.9808 – 0.1635 = 0.8173 f) look for 0.05 in the column for probability, A is between 1.64 and 1.65 (or use Table A.1A to find it is 1.645, or an inverse normal function on a scientific calculator or software). g) look for 0.025 in the column for probability, C is 1.96
EXERCISE 7
a) P(Y >15) = P(Z > 1)= 0.1587 on TI-84, normcdf(15,1E99,10,5)
b) P(8 < Y < 12) = P( -2/5 < Z < 2/5) = .6554 - .3446 = 0.3108 on TI-84, normcdf(8,12,10,5) = 0.3108 c) P(Z < c) = .90 P( Z > c) = .1 c = 1.282 = (Y – 10)/5 Y = 16.41 On TI-84 calculator, invnorm(0.90,10,5) = 16.41
EXERCISE 8 Treating grades as a continuous random variable. a) Prob (Y < 60) = P(Z < (60-76)/14) = P(Z < -1.14) = 0.1265 b) (Y – 76)/14 = 1.28, so lowest score for an A is 93.92, or 94. c) Adding points changes the mean, not the standard deviation. If only 5% are to fall at or below 60, then the new mean must be such that
Chapter 2 – Page 2