ALL 10 CHAPTERS COVERED
SOLUTIONS MANUAL
, Table of Contents
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472
PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the
prior written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
,SECTION 1.1 1
Chapter 1
Section 1.1
1. (a) The population consists of all the times the process could be run. It is conceptual.
(b) The population consist of all the registered voters in the state. It is tangible.
(c) The population consist of all people with high cholesterol levels. It is tangible.
(d) The population consist of all concrete specimens that could be made from the new formulation. It is
conceptual.
(e) The population consist of all bolts manufactured that day. It is tangible.
2. (iii). It is very unlikely that students whose names happen to fall at the top of a page in the phone
book will differ systematically in height from the population of students as a whole. It is somewhat
more likely that engineering majors will differ, and very likely that students involved with basketball
intramurals will differ.
3. (a) False
(b) True
4. (a) False
(b) True
5. (a) No. What is important is the population proportion of defectives; the sample proportion is only an
approximation. The population proportion for the new process may in fact be greater or less than
that of the old process.
(b) No. The population proportion for the new process may be 0.12 or more, even though the sample
proportion was only 0.11.
(c) Finding 2 defective circuits in the sample.
Page 1
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Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
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,2 CHAPTER 1
6. (a) False
(b) True
(c) True
7. A good knowledge of the process that generated the data.
8. (a) An observational study
(b) It is not well-justified. Because the study is observational, there could be differences between the
groups other than the level of exercise. These other differences (confounders) could cause the difference
in blood pressure.
9. (a) A controlled experiment
(b) It is well-justified, because it is based on a controlled experiment rather than an observational study.
Section 1.2
1. False
2. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all.
3. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all.
4. No. The median of the sample 1, 2, 4, 5 is 3.
5. The sample size can be any odd number.
Page 2
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, SECTION 1.2 3
6. Yes. For example, the list 1, 2, 12 has an average of 5 and a standard deviation of 6.08.
7. Yes. If all the numbers in the list are the same, the standard deviation will equal 0.
8. The mean increases by $50; the standard deviation is unchanged.
9. The mean and standard deviation both increase by 5%.
10. (a) Let X1 , ..., X100 denote the 100 numbers of children.
100
X
Xi = 27(0) + 22(1) + 30(2) + 12(3) + 7(4) + 2(5) = 156
i=1
P100
i=1 Xi 156
X= = = 1.56
100 100
(b) The sample variance is
100
!
2 1 X 2 2
s = X − 100X
99 i=1 i
1
= [(27)02 + (22)12 + (30)22 + (12)32 + (7)42 + (2)52 − 100(1.562 )]
99
= 1.7034
√
The standard deviation is s = s2 = 1.3052.
Alternatively, the sample variance can be computed as
100
1 X
s 2
= (Xi − X)2
99 i=1
1
= [27(0 − 1.56)2 + 22(1 − 1.56)2 + 30(2 − 1.56)2 + 12(3 − 1.56)2 + 7(4 − 1.56)2 + 2(5 − 1.56)2 ]
99
= 1.7034
(c) The sample median is the average of the 50th and 51st value when arranged in order. Both these
values are equal to 2, so the median is 2.
Page 3
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,4 CHAPTER 1
(d) The first quartile is the average of the 25th and 26th value when arranged in order. Both these values
are equal to 0, so the first quartile is 0.
(e) Of the 100 women, 30 + 12 + 7 + 2 = 51 had more than the mean of 1.56 children, so the proportion
is 51/100 = 0.51.
(f) The quantity that is one standard deviation greater than the mean is 1.56 + 1.3052 = 2.8652. Of the
100 women, 12 + 7 + 2 = 21 had more than 2.8652 children, so the proportion is 21/100 = 0.21.
(g) The region within one standard deviation of the mean is 1.56 ± 1.3052 = (0.2548, 2.8652). Of the 100
women, 22 + 30 = 52 are in this range, so the proportion is 52/100 = 0.52.
11. The total height of the 20 men is 20×178 = 3560. The total height of the 30 women is 30×164 = 4920.
The total height of all 50 people is 3560 + 4920 = 8480. There are 20 + 30 = 50 people in total.
Therefore the mean height for both groups put together is 8480/50 = 169.6 cm.
12. (a) The mean for A is
(18.0+18.0+18.0+20.0+22.0+22.0+22.5+23.0+24.0+24.0+25.0+25.0+25.0+25.0+26.0+26.4)/16 = 22.744
The mean for B is
(18.8+18.9+18.9+19.6+20.1+20.4+20.4+20.4+20.4+20.5+21.2+22.0+22.0+22.0+22.0+23.6)/16 = 20.700
The mean for C is
(20.2+20.5+20.5+20.7+20.8+20.9+21.0+21.0+21.0+21.0+21.0+21.5+21.5+21.5+21.5+21.6)/16 = 20.013
The mean for D is
(20.0+20.0+20.0+20.0+20.2+20.5+20.5+20.7+20.7+20.7+21.0+21.1+21.5+21.6+22.1+22.3)/16 = 20.806
(b) The median for A is (23.0 + 24.0)/2 = 23.5. The median for B is (20.4 + 20.4)/2 = 20.4. The median
for C is (21.0 + 21.0)/2 = 21.0. The median for D is (20.7 + 20.7)/2 = 20.7.
Page 4
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,SECTION 1.2 5
(c) 0.20(16) = 3.2 ≈ 3. Trim the 3 highest and 3 lowest observations.
The 20% trimmed mean for A is
(20.0 + 22.0 + 22.0 + 22.5 + 23.0 + 24.0 + 24.0 + 25.0 + 25.0 + 25.0)/10 = 23.25
The 20% trimmed mean for B is
(19.6 + 20.1 + 20.4 + 20.4 + 20.4 + 20.4 + 20.5 + 21.2 + 22.0 + 22.0)/10 = 20.70
The 20% trimmed mean for C is
(20.7 + 20.8 + 20.9 + 21.0 + 21.0 + 21.0 + 21.0 + 21.0 + 21.5 + 21.5)/10 = 21.04
The 20% trimmed mean for D is
(20.0 + 20.2 + 20.5 + 20.5 + 20.7 + 20.7 + 20.7 + 21.0 + 21.1 + 21.5)/10 = 20.69
(d) 0.25(17) = 4.25. Therefore the first quartile is the average of the numbers in positions 4 and 5.
0.75(17) = 12.75. Therefore the third quartile is the average of the numbers in positions 12 and 13.
A: Q1 = 21.0, Q3 = 25.0; B: Q1 = 19.85, Q3 = 22.0; C: Q1 = 20.75, Q3 = 21.5; D: Q1 = 20.1,
Q3 = 21.3
(e) The variance for A is
1
s2 = [18.02 + 18.02 + 18.02 + 20.02 + 22.02 + 22.02 + 22.52 + 23.02 + 24.02
15
+ 24.02 + 25.02 + 25.02 + 25.02 + 25.02 + 26.02 + 26.42 − 16(22.7442 )] = 8.2506
√
The standard deviation for A is s = 8.2506 = 2.8724.
The variance for B is
1
s2 = [18.82 + 18.92 + 18.92 + 19.62 + 20.12 + 20.42 + 20.42 + 20.42 + 20.42
15
+ 20.52 + 21.22 + 22.02 + 22.02 + 22.02 + 22.02 + 23.62 − 16(20.7002 )] = 1.8320
√
The standard deviation for B is s = 1.8320 = 1.3535.
The variance for C is
1
s2 = [20.22 + 20.52 + 20.52 + 20.72 + 20.82 + 20.92 + 21.02 + 21.02 + 21.02
15
+ 21.02 + 21.02 + 21.52 + 21.52 + 21.52 + 21.52 + 21.62 − 16(20.0132 )] = 0.17583
√
The standard deviation for C is s = 0.17583 = 0.4193.
The variance for D is
1
s2 = [20.02 + 20.02 + 20.02 + 20.02 + 20.22 + 20.52 + 20.52 + 20.72 + 20.72
15
+ 20.72 + 21.02 + 21.12 + 21.52 + 21.62 + 22.12 + 22.32 − 16(20.8062 )] = 0.55529
√
The standard deviation for D is s = 0.55529 = 0.7542.
Page 5
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Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
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, 6 CHAPTER 1
(f) Method A has the largest standard deviation. This could be expected, because of the four methods,
this is the crudest. Therefore we could expect to see more variation in the way in which this method
is carried out, resulting in more spread in the results.
(g) Other things being equal, a smaller standard deviation is better. With any measurement method, the
result is somewhat different each time a measurement is made. When the standard deviation is small,
a single measurement is more valuable, since we know that subsequent measurements would probably
not be much different.
13. (a) All would be divided by 2.54.
(b) Not exactly the same, because the measurements would be a little different the second time.
14. (a) We will work in units of $1000. Let S0 be the sum of the original 10 numbers and let S1 be the sum
after the change. Then S0 /10 = 70, so S0 = 700. Now S1 = S0 − 100 + 1000 = 1600, so the new mean
is S1 /10 = 160.
(b) The median is unchanged at 55.
P10 2
(c) Let X1 , ..., X10 be the original 10 numbers. Let T0 = i=1 Xi . Then the variance is (1/9)[T0 −
2 2
10(70 )] = 20 = 400, so T0 = 52, 600. Let T1 be the sum of the p squares after the change. Then
T1 = T0 − 1002 + 10002 = 1, 042, 600. The new standard deviation is (1/9)[T1 − 10(1602 )] = 295.63.
15. (a) The sample size is n = 16. The tertiles have cutpoints (1/3)(17) = 5.67 and (2/3)(17) = 11.33. The
first tertile is therefore the average of the sample values in positions 5 and 6, which is (44+46)/2 = 45.
The second tertile is the average of the sample values in positions 11 and 12, which is (76+79)/2 = 77.5.
(b) The sample size is n = 16. The quintiles have cutpoints (i/5)(17) for i = 1, 2, 3, 4. The quintiles are
therefore the averages of the sample values in positions 3 and 4, in positions 6 and 7, in positions 10
and 11, and in positions 13 and 14. The quintiles are therefore (23 + 41)/2 = 32, (46 + 49)/2 = 47.5,
(74 + 76)/2 = 75, and (82 + 89)/2 = 85.5.
Page 6
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,SECTION 1.3 7
16. (a) Seems certain to be an error.
(b) Could be correct.
Section 1.3
Stem Leaf
0 011112235677
1. (a) 1 235579
2 468
3 11257
4 14699
5 5
6 16
7 9
8 0099
9
10
11 0
12 7
13 7
0.45
0.4
0.35
Relative Frequency
0.3
0.25
(b) Here is one histogram. Other 0.2
choices for the endpoints are pos- 0.15
sible.
0.1
0.05
0
0 2 4 6 8 10 12 14
Rainfall (inches)
(c) 0 2 4 6 8 10 12 14
Rainfall (inches)
Page 7
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permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
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, 8 CHAPTER 1
(d) 15
Rainfall (inches)
10
The boxplot shows one outlier.
5
0
2. (a)
Stem Leaf
14 12333444467788889
15 0023344567799
16 124456
17 2222338
18
19
20
21 9
22 4
(b) Here is one histogram. Other choices for the endpoints are possible.
0.35
0.3
Relative Frequency
0.25
0.2
0.15
0.1
0.05
0
14 15 16 17 18 19 20 21 22 23
Sulfur Trioxide (percent)
Page 8
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SOLUTIONS MANUAL
, Table of Contents
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472
PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the
prior written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
,SECTION 1.1 1
Chapter 1
Section 1.1
1. (a) The population consists of all the times the process could be run. It is conceptual.
(b) The population consist of all the registered voters in the state. It is tangible.
(c) The population consist of all people with high cholesterol levels. It is tangible.
(d) The population consist of all concrete specimens that could be made from the new formulation. It is
conceptual.
(e) The population consist of all bolts manufactured that day. It is tangible.
2. (iii). It is very unlikely that students whose names happen to fall at the top of a page in the phone
book will differ systematically in height from the population of students as a whole. It is somewhat
more likely that engineering majors will differ, and very likely that students involved with basketball
intramurals will differ.
3. (a) False
(b) True
4. (a) False
(b) True
5. (a) No. What is important is the population proportion of defectives; the sample proportion is only an
approximation. The population proportion for the new process may in fact be greater or less than
that of the old process.
(b) No. The population proportion for the new process may be 0.12 or more, even though the sample
proportion was only 0.11.
(c) Finding 2 defective circuits in the sample.
Page 1
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Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
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,2 CHAPTER 1
6. (a) False
(b) True
(c) True
7. A good knowledge of the process that generated the data.
8. (a) An observational study
(b) It is not well-justified. Because the study is observational, there could be differences between the
groups other than the level of exercise. These other differences (confounders) could cause the difference
in blood pressure.
9. (a) A controlled experiment
(b) It is well-justified, because it is based on a controlled experiment rather than an observational study.
Section 1.2
1. False
2. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all.
3. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all.
4. No. The median of the sample 1, 2, 4, 5 is 3.
5. The sample size can be any odd number.
Page 2
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, SECTION 1.2 3
6. Yes. For example, the list 1, 2, 12 has an average of 5 and a standard deviation of 6.08.
7. Yes. If all the numbers in the list are the same, the standard deviation will equal 0.
8. The mean increases by $50; the standard deviation is unchanged.
9. The mean and standard deviation both increase by 5%.
10. (a) Let X1 , ..., X100 denote the 100 numbers of children.
100
X
Xi = 27(0) + 22(1) + 30(2) + 12(3) + 7(4) + 2(5) = 156
i=1
P100
i=1 Xi 156
X= = = 1.56
100 100
(b) The sample variance is
100
!
2 1 X 2 2
s = X − 100X
99 i=1 i
1
= [(27)02 + (22)12 + (30)22 + (12)32 + (7)42 + (2)52 − 100(1.562 )]
99
= 1.7034
√
The standard deviation is s = s2 = 1.3052.
Alternatively, the sample variance can be computed as
100
1 X
s 2
= (Xi − X)2
99 i=1
1
= [27(0 − 1.56)2 + 22(1 − 1.56)2 + 30(2 − 1.56)2 + 12(3 − 1.56)2 + 7(4 − 1.56)2 + 2(5 − 1.56)2 ]
99
= 1.7034
(c) The sample median is the average of the 50th and 51st value when arranged in order. Both these
values are equal to 2, so the median is 2.
Page 3
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,4 CHAPTER 1
(d) The first quartile is the average of the 25th and 26th value when arranged in order. Both these values
are equal to 0, so the first quartile is 0.
(e) Of the 100 women, 30 + 12 + 7 + 2 = 51 had more than the mean of 1.56 children, so the proportion
is 51/100 = 0.51.
(f) The quantity that is one standard deviation greater than the mean is 1.56 + 1.3052 = 2.8652. Of the
100 women, 12 + 7 + 2 = 21 had more than 2.8652 children, so the proportion is 21/100 = 0.21.
(g) The region within one standard deviation of the mean is 1.56 ± 1.3052 = (0.2548, 2.8652). Of the 100
women, 22 + 30 = 52 are in this range, so the proportion is 52/100 = 0.52.
11. The total height of the 20 men is 20×178 = 3560. The total height of the 30 women is 30×164 = 4920.
The total height of all 50 people is 3560 + 4920 = 8480. There are 20 + 30 = 50 people in total.
Therefore the mean height for both groups put together is 8480/50 = 169.6 cm.
12. (a) The mean for A is
(18.0+18.0+18.0+20.0+22.0+22.0+22.5+23.0+24.0+24.0+25.0+25.0+25.0+25.0+26.0+26.4)/16 = 22.744
The mean for B is
(18.8+18.9+18.9+19.6+20.1+20.4+20.4+20.4+20.4+20.5+21.2+22.0+22.0+22.0+22.0+23.6)/16 = 20.700
The mean for C is
(20.2+20.5+20.5+20.7+20.8+20.9+21.0+21.0+21.0+21.0+21.0+21.5+21.5+21.5+21.5+21.6)/16 = 20.013
The mean for D is
(20.0+20.0+20.0+20.0+20.2+20.5+20.5+20.7+20.7+20.7+21.0+21.1+21.5+21.6+22.1+22.3)/16 = 20.806
(b) The median for A is (23.0 + 24.0)/2 = 23.5. The median for B is (20.4 + 20.4)/2 = 20.4. The median
for C is (21.0 + 21.0)/2 = 21.0. The median for D is (20.7 + 20.7)/2 = 20.7.
Page 4
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,SECTION 1.2 5
(c) 0.20(16) = 3.2 ≈ 3. Trim the 3 highest and 3 lowest observations.
The 20% trimmed mean for A is
(20.0 + 22.0 + 22.0 + 22.5 + 23.0 + 24.0 + 24.0 + 25.0 + 25.0 + 25.0)/10 = 23.25
The 20% trimmed mean for B is
(19.6 + 20.1 + 20.4 + 20.4 + 20.4 + 20.4 + 20.5 + 21.2 + 22.0 + 22.0)/10 = 20.70
The 20% trimmed mean for C is
(20.7 + 20.8 + 20.9 + 21.0 + 21.0 + 21.0 + 21.0 + 21.0 + 21.5 + 21.5)/10 = 21.04
The 20% trimmed mean for D is
(20.0 + 20.2 + 20.5 + 20.5 + 20.7 + 20.7 + 20.7 + 21.0 + 21.1 + 21.5)/10 = 20.69
(d) 0.25(17) = 4.25. Therefore the first quartile is the average of the numbers in positions 4 and 5.
0.75(17) = 12.75. Therefore the third quartile is the average of the numbers in positions 12 and 13.
A: Q1 = 21.0, Q3 = 25.0; B: Q1 = 19.85, Q3 = 22.0; C: Q1 = 20.75, Q3 = 21.5; D: Q1 = 20.1,
Q3 = 21.3
(e) The variance for A is
1
s2 = [18.02 + 18.02 + 18.02 + 20.02 + 22.02 + 22.02 + 22.52 + 23.02 + 24.02
15
+ 24.02 + 25.02 + 25.02 + 25.02 + 25.02 + 26.02 + 26.42 − 16(22.7442 )] = 8.2506
√
The standard deviation for A is s = 8.2506 = 2.8724.
The variance for B is
1
s2 = [18.82 + 18.92 + 18.92 + 19.62 + 20.12 + 20.42 + 20.42 + 20.42 + 20.42
15
+ 20.52 + 21.22 + 22.02 + 22.02 + 22.02 + 22.02 + 23.62 − 16(20.7002 )] = 1.8320
√
The standard deviation for B is s = 1.8320 = 1.3535.
The variance for C is
1
s2 = [20.22 + 20.52 + 20.52 + 20.72 + 20.82 + 20.92 + 21.02 + 21.02 + 21.02
15
+ 21.02 + 21.02 + 21.52 + 21.52 + 21.52 + 21.52 + 21.62 − 16(20.0132 )] = 0.17583
√
The standard deviation for C is s = 0.17583 = 0.4193.
The variance for D is
1
s2 = [20.02 + 20.02 + 20.02 + 20.02 + 20.22 + 20.52 + 20.52 + 20.72 + 20.72
15
+ 20.72 + 21.02 + 21.12 + 21.52 + 21.62 + 22.12 + 22.32 − 16(20.8062 )] = 0.55529
√
The standard deviation for D is s = 0.55529 = 0.7542.
Page 5
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Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
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, 6 CHAPTER 1
(f) Method A has the largest standard deviation. This could be expected, because of the four methods,
this is the crudest. Therefore we could expect to see more variation in the way in which this method
is carried out, resulting in more spread in the results.
(g) Other things being equal, a smaller standard deviation is better. With any measurement method, the
result is somewhat different each time a measurement is made. When the standard deviation is small,
a single measurement is more valuable, since we know that subsequent measurements would probably
not be much different.
13. (a) All would be divided by 2.54.
(b) Not exactly the same, because the measurements would be a little different the second time.
14. (a) We will work in units of $1000. Let S0 be the sum of the original 10 numbers and let S1 be the sum
after the change. Then S0 /10 = 70, so S0 = 700. Now S1 = S0 − 100 + 1000 = 1600, so the new mean
is S1 /10 = 160.
(b) The median is unchanged at 55.
P10 2
(c) Let X1 , ..., X10 be the original 10 numbers. Let T0 = i=1 Xi . Then the variance is (1/9)[T0 −
2 2
10(70 )] = 20 = 400, so T0 = 52, 600. Let T1 be the sum of the p squares after the change. Then
T1 = T0 − 1002 + 10002 = 1, 042, 600. The new standard deviation is (1/9)[T1 − 10(1602 )] = 295.63.
15. (a) The sample size is n = 16. The tertiles have cutpoints (1/3)(17) = 5.67 and (2/3)(17) = 11.33. The
first tertile is therefore the average of the sample values in positions 5 and 6, which is (44+46)/2 = 45.
The second tertile is the average of the sample values in positions 11 and 12, which is (76+79)/2 = 77.5.
(b) The sample size is n = 16. The quintiles have cutpoints (i/5)(17) for i = 1, 2, 3, 4. The quintiles are
therefore the averages of the sample values in positions 3 and 4, in positions 6 and 7, in positions 10
and 11, and in positions 13 and 14. The quintiles are therefore (23 + 41)/2 = 32, (46 + 49)/2 = 47.5,
(74 + 76)/2 = 75, and (82 + 89)/2 = 85.5.
Page 6
PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
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,SECTION 1.3 7
16. (a) Seems certain to be an error.
(b) Could be correct.
Section 1.3
Stem Leaf
0 011112235677
1. (a) 1 235579
2 468
3 11257
4 14699
5 5
6 16
7 9
8 0099
9
10
11 0
12 7
13 7
0.45
0.4
0.35
Relative Frequency
0.3
0.25
(b) Here is one histogram. Other 0.2
choices for the endpoints are pos- 0.15
sible.
0.1
0.05
0
0 2 4 6 8 10 12 14
Rainfall (inches)
(c) 0 2 4 6 8 10 12 14
Rainfall (inches)
Page 7
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Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
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, 8 CHAPTER 1
(d) 15
Rainfall (inches)
10
The boxplot shows one outlier.
5
0
2. (a)
Stem Leaf
14 12333444467788889
15 0023344567799
16 124456
17 2222338
18
19
20
21 9
22 4
(b) Here is one histogram. Other choices for the endpoints are possible.
0.35
0.3
Relative Frequency
0.25
0.2
0.15
0.1
0.05
0
14 15 16 17 18 19 20 21 22 23
Sulfur Trioxide (percent)
Page 8
PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
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