chapter 11€¦ · 7.175 6 1.2 √ — 7.175 — 6 ≈ 1.09 the standard deviation of the old data...

Copyright © Big Ideas Learning, LLC Algebra 1 701All rights reserved. Worked-Out Solutions

Chapter 11

Chapter 11 Maintaining Mathematical Profi ciency (p. 583)

1.

0

2

4

6

8

10

12

0–1 2–3 4–5 6–7

Number of activities

Freq

uen

cy

After School Activities 2.

0

4

8

12

16

20

0–1 2–3 4–5

Number of pets

Freq

uen

cy

Pets

3. Students’ Favorite Subjects

English

Math

HistoryScience

4. Because there are 20 responses, the sum of the frequencies

is a + b = 20. If “maybe” is an option, the sum of the

frequencies is a + b ≤ 20.

Chapter 11 Mathematical Practices (p. 584)

1. Sample answer:

0

5

10

15

20

25

30

35

40

0–14 15–29 30–44 45–59 60–74 75–89

Ages (years)

Freq

uen

cy (

mill

ion

s)

U.S. Female Population

0

5

10

15

20

25

30

35

0–14 15–29 30–44 45–59 60–74 75–89

Male Female

Ages (years)

Pop

ula

tio

n (

mill

ion

s)

U.S. Population

2. Sample answer: Histograms display frequencies in intervals,

and double bar graphs display frequencies for each gender

side-by-side. The populations are roughly equal for the fi rst

four age groups, but decrease signifi cantly for the last two

age groups and the female population is greater than the

male population for the last two age groups.

11.1 Explorations (p. 585)

1. a. The total of the weights of the football players is about

185(5) + 195(4) + 205(6) + 215(6) + 225(2) + 235(4) +

245(3) + 255(3) + 265(4) + 275(2) + 285 + 295(2) +

305(3) + 315(5) + 325(2) + 335 = 13,075 pounds.

The mean of the weights of the football players is

about 13,075 — 53 ≈ 244.7, or about 245 pounds.

335 − 245 = 90 245 − 185 = 60

The weights of the football players vary from the mean of

about 245 pounds by up to 90 pounds.

The total of the weights of the baseball players is about

155 + 175(2) + 185(5) + 195(5) + 205(9) + 215(6) +

225(8) + 235(3) + 245 = 8290 pounds.

The mean of the weights of the football players is

about 8290 — 40 = 207.25, or about 205 pounds.

245 − 205 = 40 205 − 155 = 50

The weights of the baseball players vary from the mean of

about 205 pounds by up to 50 pounds.

b. The weights of the football players vary from the mean

much more than the weights of the baseball players.

c. Each football position seems to be clustered around a

similar weight. So, there does appear to be a correlation

between body weights and the positions of players in

professional football. Each baseball position is more

spread out. So, there does not appear to be a correlation in

professional baseball.

2. Weights of Players on a Basketball Team

175 195 215 235 255 275 295 315 335

Weight(pounds)

Power forwardsSmall forwardsCentersPoint guardsShooting guards

It appears as though guards tend to be lighter and forwards

tend to be heavier. So, there does appear to be a correlation

between the body weights and the positions of players in

professional football.

3. In order to describe the variation of a data set, you can

describe how the data vary from the mean.

702 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 11

11.1 Monitoring Progress (pp. 586–589)

1. a. Mean:

— x = 16.5 + 8.75 + 8.65 + . . . + 8.25 + 9.25 + 8.45 ————

9

= 85.65 —

9

≈ 9.52

So, the value of the mean decreases slightly.

Median:

8.25, 8.25, 8.45, 8.45, 8.65, 8.75, 9.10, 9.25, 16.50

8.65

The middle value, or median, is 8.65. So, the value of the

median decreases slightly.

Mode: 8.25, 8.25, 8.45, 8.45, 8.65, 8.75, 9.10, 9.25, 16.50

8.25 8.45

There are two modes now, 8.25 and 8.45, both of which

are less than the mean and median.

b. The mean is greater than most of the data and the modes

are less than most of the data. So, the median best

represents the data.

2. a. The value $72,000 is much greater than the other wages.

So, it is the outlier.

Mean:

with outlier:

— x = 32 + 41 + 38 + 72 + 42 + 38 + 45 + 35 ————

8

= 343 —

8

= $42.875 thousand, or $42,875

without outlier:

— x = 32 + 41 + 38 + 42 + 38 + 45 + 35

———— 7

≈ $38.714 thousand, or about $38,714

So, the outlier increases the mean by

$42,875 − $38,714 = $4161.

Median:

with outlier: 32, 35, 38, 38, 41, 42, 45, 72

38 + 41

— 2 = $39.5 thousand, or

$39,500

without outlier: 32, 35, 38, 38, 41, 42, 45

$38 thousand, or $38,000

So, the outlier increases the median by

$39,500 − $38,000 = $1500.

Mode:

with outlier: 32, 35, 38, 38, 41, 42, 45, 72


without outlier: 32, 35, 38, 38, 41, 42, 45


So, the outlier does not affect the mode.

b. Sample answer: The outlier could be the salary of

the manager.

3. Show A:

19, 20, 20, 21, 22, 27, 27, 29, 29, 30, 31

So, the new range is still 31–19, or 12 years.

Show B:

19, 20, 21, 22, 22, 24, 25, 25, 27, 27, 32

So, the new range is 32–19, or 13 years.

For Show A the greatest and least values are the same. So,

the range did not change. However, for Show B, the oldest

contestant was voted off. So, the range is smaller, meaning

the ages of the contestants are less spread out.

4. — x = 25 + 20 + 22 + . . . + 22 + 21 + 24 ————

12

= 312 —

12

= 26

x — x x − — x (x − — x )2

25 26 −1 1

20 26 −6 36

22 26 −4 16

27 26 1 1

48 26 22 484

32 26 6 36

19 26 −7 49

27 26 1 1

25 26 −1 1

22 26 −4 16

21 26 −5 25

24 26 −2 4

1 + 36 + 16 + . . . + 16 + 25 + 4

——— 12

= 670 —

12 ≈ 55.83

√— 670

— 12

≈ 7.47

The standard deviation is about 7.47 years. This means that

the typical age of a contestant on Show B differs from the

mean by about 7.47 years.

5. The standard deviation for Show B is greater. So, the ages

are more spread out.


Chapter 11

6. Mean: 1

9 —

10 + 1

2 —

5 + 4

1 —

5 + 1

1 —

2 + 1

1 —

5 +

9 —

10

——— 6 = 11.1

— 6 = 1.85

1.85 + 1.5 = 3.35

So, the mean of the new data set is 3.35 miles.

Median:

9 —

10 , 1

1 —

5 , 1

2 —

5 , 1

1 —

2 , 1

9 —

10 , 4

1 —

5

1 2 —

5 + 1

1 —

2

— 2 = 1.2

— 2 = 1.45

1.45 + 1.5 = 2.95

So, the median of the new data set is 2.95 miles.

Mode:

9 —

10 , 1

1 —

5 , 1

2 —

5 , 1

1 —

2 , 1

9 —

10 , 4

1 —

5

None of the data values are repeated. So, the data set has

no mode.

Range:

9 —

10 , 1

1 —

5 , 1

2 —

5 , 1

1 —

2 , 1

9 —

10 , 4

1 —

5

4 1 — 5 − 9 —

10 = 4.2 − 0.9 = 3.3

So, the range of the old data set is 3.3 miles, and the range of

the new data set is the same.

Standard deviation:

x — x x − — x (x − — x )2

1.9 1.85 0.05 0.0025

1.4 1.85 −0.45 0.2025

4.2 1.85 2.35 5.5225

1.5 1.85 −0.35 0.1225

1.2 1.85 −0.65 0.4225

0.9 1.85 −0.95 0.9025

0.0025 + 0.2025 + 5.5225 + 0.1225 + 0.4225 + 0.9025

————— 6

= 7.175 —

6 ≈1.2

√— 7.175

— 6 ≈ 1.09

The standard deviation of the old data set is about 1.09 miles,

and the standard deviation of the new data set is the same.

11.1 Exercises (pp. 590–592)

Vocabulary and Core Concept Check

1. The measure of center represents the center, or typical

value, of the data. The measure of variation represents the

distribution of the data, or how much the data values vary

from the center.

2. When the outlier of a data set is removed, the mean gets

closer to the median.

3. Sample answer: The data set 3, 4, 4, 7, 9, 9 has two modes.

The modes are 4 and 9.

4. An advantage of using the range to describe a data set is

that it is easy to calculate. The standard deviation, however,

is considered a more reliable measure of variation than

the range because all of the values of a data set are used in

calculating the standard deviation.

Monitoring Progress and Modeling with Mathematics

5. a. Mean: — x = 3 + 5 + 1 + 5 + 1 + 1 + 2 + 3 + 15 ————

9

= 36

— 9 = 4

Median: 1, 1, 1, 2, 3, 3, 5, 5, 15

The middle value is 3.

Mode: 1, 1, 1, 2, 3, 3, 5, 5, 15

The data value that occurs most often is 1.

The mean is 4, the median is 3, and the mode is 1.

b. The median best represents the data. The mode is less than

most of the data and the mean is greater than most of

the data.

6. a. Mean: — x = 12 + 9 + 17 + 15 + 10 ——

5 = 63

— 5

= 12.6

Median: 9, 10, 12, 15, 17


Mode: 9, 10, 12, 15, 17

All of the data values occur once.

The mean is 12.6, the median is 12, and the data set has

no mode.

b. The mean best represents the data because there are

no outliers.

7. a. Mean: — x = 13 + 30 + 16 + 19 + 20 + 22 + 25 + 31 ————

8

= 176 —

8 = 22

Median: 13, 16, 19, 20, 22, 25, 30, 31

The mean of the two middle values is

20 + 22 —

2 = 42

— 2 = 21.

Mode: 13, 16, 19, 20, 22, 25, 30, 31


The mean is 22, the median is 21, and the data set has

no mode.

b. The mean best represents the data because there are

no outliers.


Chapter 11

8. a. Mean:

— x = 14 + 15 + 3 + 15 + 14 + 14 + 18 + 15 + 8 + 16 —————

10

= 132 —

10 = 13.2

Median: 3, 8, 14, 14, 14, 15, 15, 15, 16, 18


14 + 15

— 2 = 29

— 2 = 14.5.

Mode: 3, 8, 14, 14, 14, 15, 15, 15, 16, 18

The data values that occur most often are 14 and 15.

The mean is 13.2, the median is 14.5, and the modes are

14 and 15.

b. The median best represents the data. It is the mean of the

two modes, and the mean is less than most of the data.

9. a. Mean: — x =

1 1 —

3 + 3 + 2 + 1

2 —

3 + 2

1 —

3 + 2 + 2 + 1

2 —

3 + 1

2 —

3

———— 9

= 17 2 —

3 ÷ 9 = 53

— 3 ⋅ 1 —

9 = 53

— 27

= 1 26

— 27

Median: 1 1 —

3 , 1

2 —

3 , 1

2 —

3 , 1

2 —

3 , 2, 2, 2, 2

1 —

3 , 3


Mode: 1 1 —

3 , 1

2 — 3 , 1

2 — 3 , 1

2 — 3 , 2, 2, 2, 2

1 —

3 , 3

The data values that occur most often are 1 2 —

3 and 2.

The mean is 1 26

— 27

or about 1.96, the median is 2, and the

modes are 1 2 —

3 and 2.

b. The median best represents the data because the data are

evenly distributed.

10. a. Mean:

— x = 1.05 + 2.64 + 0.66 + . . . + (−2.41) + 1.39 + 2.20 —————

12

= −4.82 —

12 ≈ −0.402

Median: −13.78, −3.01, −2.41, −0.28, 0.66, 0.67, 1.05,

1.39, 2.03, 2.20, 2.64, 4.02


0.67 + 1.05

— 2 = 1.72

— 2 = 0.86.

Mode: −13.78, −3.01, −2.41, −0.28, 0.66, 0.67, 1.05,

1.39, 2.03, 2.20, 2.64, 4.02


The mean is about −0.402, the median is 0.86, and the

data set has no mode.

b. The median best represents the data because the mean is

less than most of the data and there is no mode.

c. Mean:

— x = 1.05 + 2.64 + 0.66 + . . . + 1.39 + 2.20 + 4.28 ————

13

= −0.54

— 13

≈ −0.042

Median: −13.78, −3.01, −2.41, −0.28, 0.66, 0.67, 1.05,

1.39, 2.03, 2.20, 2.64, 4.02, 4.28

The middle value is 1.05.

Mode: −13.78, −3.01, −2.41, −0.28, 0.66, 0.67, 1.05,

1.39, 2.03, 2.20, 2.64, 4.02, 4.28


The mean is about −0.042, the median is 1.05, and the

data set has no mode. Because $4.28 is greater than the

mean and median, they both increase.

11. 2 + 8 + 9 + 7 + 6 + x

—— 6 = 6

32 + x

— 6 = 6

6 ⋅ (32 + x) —

6 = 6 ⋅ 6

32 + x = 36

−32 −32

x = 4

The value of x is 4.

12. 12.5 + (−10) + (−7.5) + x

——— 4 = 11.5

−5 + x

— 4 = 11.5

4 ⋅ (−5 + x)

— 4 = 4 ⋅ 11.5

−5 + x = 46

+5 +5

x = 51


13. 9, 10, 12, x, 20, 25

12 + x —

2 = 14

2 ⋅ (12 + x)

— 2

= 2 ⋅ 14

12 + x = 28

−12 −12

x = 16



Chapter 11

14. 30, 45, x, 100

45 + x —

2 = 51

2 ⋅ (45 + x)

— 2 = 2 ⋅ 51

45 + x = 102

−45 −45

x = 57


15. a. The value 62 is much less than the other masses. So, it is

the outlier.

Mean:

With outlier:

— x = 455 + 364 + 262 + 553 + 471 + 62 + 358 + 351 —————

8

= 2876 —

8 = 359.5

Without outlier:

— x = 455 + 364 + 262 + 553 + 471 + 358 + 351 ————

7

= 2814 —

7 = 402

So, the outlier decreases the mean by 402 − 359.5 = 42.5.

Median:

With outlier: 62, 262, 351, 358, 364, 455, 471, 553


358 + 364

— 2 = 722

— 2 = 361.

Without outlier: 262, 351, 358, 364, 455, 471, 553


So, the outlier decreases the median by 364 − 361 = 3.

Mode:

With outlier: 62, 262, 351, 358, 364, 455, 471, 553

Without outlier: 262, 351, 358, 364, 455, 471, 553

All of the data values occur once in each set. So, the

outlier does not affect the mode.

b. Sample answer: The outlier could be the mass of a baby

polar bear.

16. a. The value 46 is much greater than the other sizes. So, it is

the outlier.

Mean:

With outlier:

— x = 2 + 3 + 5 + 2 + 1 + 46 + 3 + 7 + 2 + 1 ————

10

= 72 —

10 = 7.2

Without outlier:

— x = 2 + 3 + 5 + 2 + 1 + 3 + 7 + 2 + 1 ————

9

= 26 —

9 ≈ 2.9

So, the outlier increases the mean by about 7.2 − 2.9 = 4.3.

Median:

With outlier: 1, 1, 2, 2, 2, 3, 3, 5, 7, 46


2 + 3

— 2 = 5 —

2 = 2.5.

Without outlier: 1, 1, 2, 2, 2, 3, 3, 5, 7


So, the outlier increases the median by 2.5 − 2 = 0.5.

Mode:

With outlier: 1, 1, 2, 2, 2, 3, 3, 5, 7, 46

Without outlier: 1, 1, 2, 2, 2, 3, 3, 5, 7

The mode is 2 in each set. So, the outlier does not affect

the mode.

b. Sample answer: The email could have contained a picture.

17. Golfer A: 83, 84, 87, 88, 89, 90, 91, 95, 95, 98

So, the range is 98 − 83, or 15 strokes.

Golfer B: 87, 88, 89, 89, 91, 92, 92, 93, 94, 95

So, the range is 95 − 87, or 8 strokes.

The range of the scores for Golfer A is 15 strokes, and the

range of scores for Golfer B is 8 strokes. So, the scores for

Golfer A are more spread out.

18. Rookie season: 0, 0, 1, 2, 3, 6

So, the range is 6 − 0, or 6 home runs.

This season: 4, 4, 6, 7, 8, 13

So, the range is 13 − 4, or 9 home runs.

The range of the monthly home run totals for the player’s

rookie season is 6 home runs, and the range this season is

9 home runs. So, the monthly home run totals for the player

are more spread out this season than in the player’s rookie

season.

19. a. 35, 40, 45, 55, 60

So, the range is 60 − 35, or 25.

b. — x = 40 + 35 + 45 + 55 + 60 ———

5 = 235

— 5 = 47

x — x x − — x (x − — x )2

40 47 −7 49

35 47 −12 144

45 47 −2 4

55 47 8 64

60 47 13 169

49 + 144 + 4 + 64 + 169

——— 5 = 430

— 5

= 86

√—

86 ≈ 9.27

The standard deviation is about 9.27.


Chapter 11

20. a. 116, 117, 121, 126, 135, 141


b. — x = 141 + 116 + 117 + 135 + 126 + 121 ————

6 = 756

— 6 = 126

x — x x − — x (x − — x )2

141 126 15 225

116 126 −10 100

117 126 −9 81

135 126 9 81

126 126 0 0

121 126 −5 25

225 + 100 + 81 + 81 + 0 + 25

——— 6 = 512

— 6 ≈ 85.33

√—

512

— 6 ≈ 9.24


21. a. 0.5, 1.0, 1.5, 1.5, 2, 2.5

So, the range is 2.5 − 0.5, or 2.

b. — x = 0.5 + 2.0 + 2.5 + 1.5 + 1.0 + 1.5 ———

6 = 9 —

6 = 1.5

x — x x − — x (x − — x )2

0.5 1.5 −1 1

2.0 1.5 0.5 0.25

2.5 1.5 1 1

1.5 1.5 0 0

1.0 1.5 −0.5 0.25

1.5 1.5 0 0

1 + 0.25 + 1 + 0 + 0.25 + 0

——— 6 = 2.5

— 6 ≈ 0.417

√—

2.5

— 6 ≈ 0.646


22. a. 2.4, 2.6, 3.3, 4.8, 5.6, 7.0, 8.2, 10.1

So, the range is 10.1 − 2.4, or 7.7.

b. — x = 8.2 + 10.1 + 2.6 + 4.8 + 2.4 + 5.6 + 7.0 + 3.3 ————

8

= 44 —

8 = 5.5

x — x x − — x (x − — x )2

8.2 5.5 2.7 7.29

10.1 5.5 4.6 21.16

2.6 5.5 −2.9 8.41

4.8 5.5 −0.7 0.49

2.4 5.5 −3.1 9.61

5.6 5.5 0.1 0.01

7.0 5.5 1.5 2.25

3.3 5.5 −2.2 4.84

7.29 + 21.16 + 8.41 + 0.49 + 9.61 + 0.01 + 2.25 + 4.84 —————

8

= 54.06 —

8 = 6.7575

√— 54.06

— 8 ≈ 2.60


23. a. Golfer A:

— x = 83 + 84 + 91 + 90 + 98 + 88 + 95 + 89 + 87 + 95 —————

10

= 900 —

10 = 90

x — x x − — x (x − — x )2

83 90 −7 49

84 90 −6 36

91 90 1 1

90 90 0 0

98 90 8 64

88 90 −2 4

95 90 5 25

89 90 −1 1

87 90 −3 9

95 90 5 25

49 + 36 + 1 + 0 + 64 + 4 + 25 + 1 + 9 +25

———— 10

= 214

— 10

= 21.4

√—

21.4 ≈ 4.6

So, Golfer A’s typical score differs from the mean by

about 4.6 strokes.


Chapter 11

b. Golfer B:

— x = 89 + 93 + 92 + 88 + 89 + 87 + 95 + 94 + 91 + 92 —————

10

= 910

— 10

= 91

x — x x − — x (x − — x )2

89 91 −2 4

93 91 2 4

92 91 1 1

88 91 −3 9

89 91 −2 4

87 91 −4 16

95 91 4 16

94 91 3 9

91 91 0 0

92 91 1 1

4 + 4 + 1 + 9 + 4 + 16 + 16 + 9 + 0 + 1

———— 10

= 64 —

10 = 6.4

√—

6.4 ≈ 2.5

So, Golfer B’s typical score differs from the mean by

about 2.5 strokes.

c. The standard deviation for Golfer A is greater. So, Golfer A’s

scores are more spread out, and Golfer B is more consistent.

24. a. Rookie season: — x = 1 + 0 + 6 + 2 + 0 + 3 ——

6 = 12

— 6 = 2

x — x x − — x (x − — x )2

1 2 −1 1

0 2 −2 4

6 2 4 16

2 2 0 0

0 2 −2 4

3 2 1 1

1 + 4 + 16 + 0 + 4 + 1

——— 6 =

26 —

6 ≈ 4.33

√—

26

— 6 ≈ 2.08

So, the typical number of home runs differs from the

mean by about 2.08 home runs.

b. This season: — x = 4 + 6 + 4 + 8 + 7 + 13 ———

6 = 42

— 6 = 7

x — x x − — x (x − — x )2

4 7 −3 9

6 7 −1 1

4 7 −3 9

8 7 1 1

7 7 0 0

13 7 6 36

9 + 1 + 9 + 1 + 0 + 36

——— 6 =

56 —

6 ≈ 9.33

√—

56

— 6 ≈ 3.06

So, the typical number of home runs differs from the

mean by about 3.06 home runs.

c. The standard deviation for this season is greater. So, the

monthly home run totals are more spread out this season.

25. Mean: — x = 4 + 4 = 8

Median: The middle value is 3 + 4 = 7.

Mode: The data value that occurs most often is 1 + 4 = 5.

The new mean is 8, the new median is 7, and the new

mode is 5.

26. Mean: — x = 12.6 ⋅ (1 + 0.20) = 12.6 ⋅ 1.2 = 15.12

Median: The middle value is 12 ⋅ (1 + 0.20)

= 12 ⋅ 1.2 = 14.4.

Mode: All of the data values occur once.

The new mean is 15.12, the new median is 14.4, and the data

set still has no mode.

27. Mean: — x = 62 + 14 = 76

Median: 55 + 14 = 69

Mode: 49 + 14 = 63

The range and standard deviation stay the same.

So, the new mean is 76, the new median is 69, the new mode

is 63, the range is still 46, and the standard deviation is still

15.5.

28. Mean: — x = 320(0.5) = 160

Median: 300(0.5) = 150

Mode: none

Range: 210(0.5) = 105

Standard deviation: 70.6(0.5) = 35.3

So, the new mean is 160, the new median is 150, the data set

still has no mode, the new range is 105, and the new standard

deviation is 35.3.


Chapter 11

29. The numbers should be put in numerical order before fi nding

the median.

2, 3, 4, 4, 6, 6, 7, 8, 8


So, the median is 6.

30. When a number is added to each value in a data set, the

range stays the same.

−13, −12, −7, 2, 10, 13

So, the range is 13 − (−13), or 26.

31. a. Mean:

Team A: — x = 172 + 130 + 173 + 212 ———

4 = 687

— 4 = 171.75

Team B: — x = 136 + 184 + 168 + 192 ———

4 = 680

— 4 = 170

Team A has the greater mean score. So, Team A wins.

Median

Team A: 130, 172, 173, 212


172 + 173

— 2 = 345

— 2 = 172.5.

Team B: 136, 168, 184, 192


168 + 184

— 2 = 352

— 2 = 176.

Team B has a greater median score. So, the result would

be different if the rule were changed to say the team with

the greater median score wins.

b. Team A:

130, 172, 173, 212


— x = 171.75

x — x x − — x (x − — x )2

172 171.75 0.25 0.0625

130 171.75 −41.75 1743.0625

173 171.75 1.25 1.5625

212 171.75 40.25 1620.0625

0.0625 + 1743.0625 + 1.5625 + 1620.0625

———— 4

= 3364.75 —

4 ≈ 841.1875

√—

841.1875 ≈ 29.00

Team A’s scores differ from the mean by about 29 pins.

Team B:

136, 168, 184, 192


— x = 170

x — x x − — x (x − — x )2

136 170 −34 1156

184 170 14 196

168 170 −2 4

192 170 22 484

1156 + 196 + 4 + 484

—— 2 = 1840

— 4 = 460

√—

460 ≈ 21.45

Team B’s scores differ from the mean by about 21.5 pins.

So, Team B is more consistent because their range and

standard deviation are less than those of Team A.

c. Mean:

Team A: — x = 171.75 + 15 = 186.75

Team B: — x = 170(1 + 0.125) = 170(1.125) = 191.25

Team B’s new mean score is greater than that of Team A.

So, Team B wins.

32. Even though two data sets have the same range, the numbers

in between may be distributed differently. So, your friend is

incorrect.

33. The data is not numeric. So, the only measure of central

tendency that can be used to describe the data is the mode.

34. Dot plot B has the greatest range, 18 − 12 = 6. So, the

data values are spread out the most on this dot plot, and it

will have the greatest standard deviation. Dot plot C has

the smallest range, 16 − 14 = 2. So, the data values are

spread out the least on this dot plot, and it will have the least

standard deviation.

35. Mean: — x = 27(3) + 8 = 81 + 8 = 89

Median: 32(3) + 8 = 96 + 8 = 104

Mode: 18(3) + 8 = 54 + 8 = 62

Range: 41(3) = 123

Standard deviation: 9(3) = 27

So, the new mean is 89, the new median is 104, the new

mode is 62, the new range is 123, and the new standard

deviation is 27.

36. All the data values could be the same. Then, the standard

deviation would be 0. The standard deviation cannot be

negative, because it is a positive square root.


Chapter 11

37. a. Answers will vary.

b. Sample answer: You would expect the mean, median,

range, and standard deviation to increase, but because it

is highly unlikely that another student is 7 feet tall, the

mode will not be affected.

38. Sample answer: Find the geometric mean of the data from

Exercises 5 and 6, and compare them to their respective

arithmetic means.

Exercise 5

— x = 3 + 5 + 1 + 5 + 1 + 1 + 2 + 3 + 15 ————

9 = 36

— 9 = 4

9 √———

3(5)(1)(5)(1)(1)(2)(3)(15) = 9 √—

6750 ≈ 2.66

Exercise 6

— x = 12 + 9 + 17 + 15 + 10 ——

5 = 63

— 5 = 12.6

5 √——

12(9)(17)(15)(10) 5 √—

275,400 ≈ 12.25

The geometric mean is always less than or equal to the

arithmetic mean.

39. a. Mean:

18(0.35 ⋅ 200) + . . . + 37(0.01 ⋅ 200)

———— 200

= 18(70) + 19(60) + 20(28) + 21(40) + 37(2) ————

200

= 1260 + 1140 + 560 + 840 + 74 ———

200

= 3874 —

200

= 19.37

Median: The median is the mean of the 100th and 101st

ages, both of which are 19. So, the median age is

19 + 19

— 2 = 38

— 2 = 19.

Mode: There are 70 students who are 18 years old. This

is the most common age in the class. So, the mode is 18.

The mean age is 19.37, the median age is 19, and the

mode is 18 years.

b. Two students are 37 years old, and they are signifi cantly

older than the other students in the class. So, the outliers

are 37 and 37.

Mean:

= 18(0.35 ⋅ 200) + . . . + 21(0.20 ⋅ 200) ————

198

= 18(70) + 19(60) + 20(28) + 21(40)

——— 198

= 1260 + 1140 + 560 + 840

——— 198

= 3800

— 198

≈ 19.19

Median: The median is the mean of the 99th and 100th

ages, both of which are still 19. So, the median age is

still 19 + 19 —

2 = 38

— 2 = 19.

Mode: There are 70 students who are 18 years old. This is

still the most common age in the class. So, the mode is 18.

The outliers increase the mean but do not affect the

median or mode.

c.

22 yr:20%

38 yr:1%

21 yr:14%

20 yr:30%

19 yr:35%

College Student Ages

Mean: 19.37 + 1 = 20.37

Median: 19 + 1 = 20

Mode: 18 + 1 = 19

The new mean is 20.37, the new median is 20, and the new

mode is 19.

Maintaining Mathematical Profi ciency

40. 6x + 1 ≤ 4x − 9

−4x −4x

2x + 1 ≤ −9

−1 −1

2x ≤ −10

2x

— 2 ≤ −10

— 2

x ≤ −5

The solution is x ≤ −5.

41. −3(3y − 2) < 1 − 9y

−3(3y) − 3(−2) < 1 − 9y

−9y + 6 < 1 − 9y

+9y +9y

6 < 1

The statement 6 < 1 is never true. So, the inequality has no

solution.

42. 2(5c − 4) ≥ 5(2c + 8)

2(5c) + 2(−4) ≥ 5(2c) + 5(8)

10c − 8 ≥ 10c + 40

−10c −10c

−8 ≥ 40

The statement −8 ≥ 40 is never true. So, the inequality has

no solution.


Chapter 11

43. 4(3 − w) > 3(4w − 4)

4(3) − 4(w) > 3(4w) − 3(4)

12 − 4w > 12w − 12

+4w +4w

12 > 16w − 12

+12 +12

24 > 16w

24

— 16

> 16w

— 16

3 —

2 > w

The solution is w < 3

— 2

.

44. f (x) = 4x

f (3) = 43

= (4)(4)(4)

= 64

So, f(x) = 64 when x = 3.

45. f(x) = 7x

f(−2) = 7−2

= 1 —

72

= 1 —

49

So, f(x) = 1 —

49 when x = −2.

46. f (x) = 5(2)x

f (6) = 5(2)6

= 5(64)

= 320

So, f(x) = 320 when x = 6.

47. f (x) = −2(3)x

f (4) = −2(3)4

= −2(81)

= −162

So, f(x) = −162 when x = 4.


1. a.

0 1 3 3 3 3 5 6 8 8 9 10 10

12 13 13 14 16 18 19 23 24 32 45

The median is 10.

b.

0 1 3 3 3 3 5 6 8 8 9 10 10

12 13 13 14 16 18 19 23 24 32 45

The least value is 0, the fi rst quartile is 3 + 5 —

2 = 8 —

2 = 4,

the third quartile is 16 + 18 —

2 = 34

— 2 = 17, and the greatest

value is 45.

c. The box represents the middle half of the data, and the

whiskers represent the bottom and the top quarters of the

data.

2. A box-and-whisker plot can be used to show the median, the

fi rst and third quartiles, and the least and greatest values of a

data set.

3. a. The least value is 17, the fi rst quartile is 19, the median is

21, the third quartile is 22, and the greatest value is 28. So,

a quarter of the students have a BMI between 17 and 19,

half of the ninth-grade students have a BMI between 19

and 22, and the last quarter have a BMI between 22 and 28.

b. The least value is 120, the fi rst quartile is 140, the median

is 180, the third quartile is 220, and the greatest value

is 240. So, a quarter of the roller coasters have a height

between 120 and 140 feet, half of the roller coasters have

a height between 140 and 220, and the remaining quarter

have a height between 220 and 240.

11.2 Monitoring Progress (pp. 594 –596)

1.

5, 14, 16, 16, 20, 22, 28, 30

lower half ⎧⎪⎪⎪⎨⎪⎪⎪⎩

least value

fi rst quartile

median

⎧⎪⎪⎪⎨⎪⎪⎪⎩

third quartile

greatest value

upper half

Least value: 5

First quartile: 14 + 16 —

2 = 30

— 2 = 15

Median: 16 + 20 —

2 = 36

— 2 = 18

Third quartile: 22 + 28 —

2 = 50

— 2 = 25

Greatest value: 30

2 6 10 14 18 22 26 30 34Points

median

least value

fi rst quartile

median

third quartile

greatest value


Chapter 11

2. The range is 30 − 18 = 12 years. This means that the ages

vary by no more than 12 years.

IQR = Q3 − Q1 = 29 − 23 = 6

The interquartile range is 6 years. This means that the middle

half of the ages vary by no more than 6 years.

3. 25% of the ages are between 18 and 23, 50% of the ages

are between 23 and 29, and 25% of the ages are between

29 and 30.

4. For both shops, the right whisker is longer than the left

whisker, and most of the data are on the left side of the

plot. So, both distributions are skewed right. The range and

interquartile range of the prices at Shop A are greater than

the range and interquartile range at Shop B. So, the surfboard

prices are more spread out at Shop A.

11.2 Exercises (pp. 597–598)


1. In order to fi nd the fi rst quartile of the data set, order the

data, fi nd the median of the data set, and then fi nd the

median of the lower half.

2. The one that is different is “Find the difference of the

greatest value and the least value of the data set.” This range

is 21 − 1 = 23. The other three ask for the interquartile

range, which is 20 − 11 = 9.


3. The least value is 3.

4. The greatest value is 14.

5. The third quartile is 11.

6. The fi rst quartile is 6.

7. The median is 8.

8. The range is 14 − 3 = 11.

9.

0, 2, 3, 4, 4, 5, 5, 6


least value

fi rst quartile

median

⎧⎪⎪⎪⎨⎪⎪⎪⎩

third quartile

greatest value

upper half

Least value: 0


2 = 5 —

2 = 2.5

Median: 4 + 4 —

2 = 8 —

2 = 4


2 = 10

— 2 = 5

Greatest value: 6

0 1 2 3 4 5 6 7Hours

10.

16, 17, 18, 20, 21, 22, 23, 25


least value

fi rst quartile

median

⎧⎪⎪⎪⎨⎪⎪⎪⎩third

quartilegreatest

value

upper half

Least value: 16


2 = 35

— 2 = 17.5

Median: 20 + 21 —

2 = 41

— 2 = 20.5


2 = 45

— 2 = 22.5

Greatest value: 25

16 18 20 22 24 26Length(inches)

11.

−4, −3, −3, −2, 0, 0, 1, 2, 2, 5, 6

least value

fi rst quartile

lower half ⎧⎪⎪ ⎪⎪⎨⎪⎪ ⎪⎪⎩

median

upper half ⎧⎪⎪ ⎪⎪⎨⎪⎪ ⎪⎪⎩

third quartile

greatest value

Least value: −4

First quartile: −3

Median: 0

Third quartile: 2

Greatest value: 6

−4 −2 0 2 4 6Elevation(feet)

12.

95, 95, 105, 110, 114, 124, 124, 190, 300

least value

fi rst quartile


median

upper half ⎧⎪⎪⎪⎨⎪⎪⎪⎩

third quartile

greatest value

Least value: 95


2 = 200

— 2 = 100

Median: 114

Third quartile: 124 +190 —

2 = 314

— 2 = 157

Greatest value: 300

90 130 170 210 250 290Price(dollars)


Chapter 11

13.

0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 7

least value

fi rst quartile

median

lower half ⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩ ⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩ upper half

third quartile

greatest value

Least value: 0


2 = 2 —

2 = 1

Median: 2 + 2 —

2 = 4 —

2 = 2


2 = 7 —

2 = 3.5

Greatest value: 7

0 1 2 3 4 5 6 7 8Hours

14.

6, 7, 8, 8, 9, 10, 10, 12, 12, 13, 14, 14, 17, 21, 22

lower half ⎧⎪⎪ ⎪ ⎪ ⎪⎪⎨⎪⎪ ⎪ ⎪ ⎪⎪⎩

least value

fi rst quartile

median⎧⎪⎪ ⎪ ⎪ ⎪⎪⎨⎪⎪ ⎪ ⎪ ⎪⎪⎩ upper half

third quartile

greatest value

Least value: 6

First quartile: 8

Median: 12

Third quartile: 14

Greatest value: 22

4 8 12 16 20 24Length(inches)

15. a. The range is 18.25 − 8.75 = 9.5. This means that the

prices of the entrées vary by no more than $9.50.

b. 25% of the entrées cost between $8.75 and $10.50, 50%

of the entrées cost between $10.50 and $14.75, and 25%

of the entrées cost between $14.75 and $18.25.

c. The interquartile range is 14.75 − 10.5 = 4.25. This

means that the middle half of the prices vary by no more

than $4.25.

d. The right whisker is longer than the left whisker. So, the

data above Q3 is more spread out than the data below Q1.

16. a. The range is 17 − 0 = 17 runs. This means that the number

of runs vary by no more than 17. The interquartile range

is 9 − 2 = 7 runs. This means that the middle half of the

number of runs vary by no more than 7.

b. 25% of the number of runs scored are between 0 and 2,

50% of the number of runs scored are between 2 and 9, and

25% of the number of runs scored are between 9 and 17.

c. The box is wider between Q2 and Q3 because

Q2 − Q1 = 4 − 2 = 2 and Q3 − Q2 = 9 − 4 = 5.

So, the data are more spread out between Q2 and Q3.

17. a. For Sales Rep A, the whisker lengths are equal, and the

median is in the middle of the plot. So, the distribution

for Sales Rep A is symmetric. For Sales Rep B, the right

whisker is longer than the left whisker, and most of the

data are on the left side of the plot. So, the distribution for

Sales Rep B is skewed right.

b. The range and interquartile range of the monthly car

sales for Sales Rep B are greater than the range and

interquartile range for Sales Rep A. So, the monthly car

sales for Sales Rep B are more spread out.

c. The least value is 4, and it belongs to Sales Rep B. So, Sales

Rep B had the single worst sales month during the year.

18. When a distribution is skewed left, most of the data are not

on the left side of the plot. The distribution is skewed left.

So, most of the data are on the right side of the plot.

19. The range must be greater than the interquartile range. So,

the range is 36, and the interquartile range is 12.

20. a. Because 11 is the greatest value, it is always true that the

data set contains the value 11.

b. The data set may not contain the value 6. Sample answer: The data could be 2, 4, 5, 5, 8, 12.

c. The right whisker is longer, and most of the data are

on the left side of the plot. So, it is always true that the

distribution is skewed right.

d. The mean is not always the same as the median. So, the

mean may not be 5.

21. a. For both brands, the right whiskers are longer than the

left whiskers, and most of the data are on the left side of

the plots. So, the distributions for both brands are skewed

right.

b. For Brand A, the range of the upper 75% of battery lives

is about 7 − 3.5 = 3.5 hours.

For Brand B, the range of the upper 75% of battery lives

is about 5.75 − 3.25 = 2.5 hours.

c. The interquartile range of Brand A is about

4.75 − 3.5 = 1.25 hours, and the interquartile range

of Brand B is about 4.25 − 3.25 = 1 hour. So, the

interquartile range of Brand A is greater.

d. The range of Brand A is greater. So, Brand A probably has

a greater standard deviation.

e. You should buy Brand A because 75% of the battery lives

are greater than 3.5 hours.

22. The data set must have 3 as its least value and 18 as its

greatest value. Also, the fi rst quartile must be 4, the median

must be 10.5, and the third quartile must be 15.

Sample answer: A data set that fi ts this criteria and could be

represented by the box-and-whisker plot shown: 3, 4, 4, 10,

11, 14, 16, 18.


Chapter 11

23. Yes, it is possible for the box-and-whisker plots to be

different even though they have the same median, the same

interquartile range, and the same range.

Sample answer:

2 6 10 14 18 22

B

A

These two plots each have a median of 10, a range of 16, and

an interquartile range of 6.


24. The zeros of f are p = −9 and q = 3.

The x-coordinate of the vertex is −9 + 3 —

2 = −6

— 2

= −3.

f (x) = −2(x + 9)(x − 3)

f (−3) = −2(−3 + 9)(−3 − 3)

= −2(6)(−6)

= 72

So, plot the x-intercepts (−9, 0) and (3, 0) as well as the

vertex (−3, 72). Also, because a < 0, the parabola opens

down. Draw a smooth curve through the points.

x

y

10

20

30

40

50

42 6−4−6 −2

25. The zeros are p = 5 and q = −5.

The x-coordinate of the vertex is 5 + (−5) —

2 = 0 —

2 = 0.

y = 3(x − 5)(x + 5)

= 3(0 − 5)(0 + 5)

= 3(−5)(5)

= −75


vertex (0, −75). Also, because a > 0, the parabola opens up.

Draw a smooth curve through the points.

x

y10

−20

−30

−40

−50

−60

−80

421 3−2−4−3 −1

26. y = 4x2 − 16x − 48

y = 4(x2 − 4x − 12)

y = 4(x − 6)(x + 2)

So, the zeros are p = 6 and q = −2.


2 = 4 —

2 = 2.

y = 4x2 − 16x − 48

= 4(2)2 − 16(2) − 48

= 4(4) − 32 − 48

= 16 − 32 − 48

= −16 − 48

= −64


vertex (2, −64). Also, because a > 0, the parabola opens up.

Draw a smooth curve through the points.

x

y10

−50

−60

−70

421 3 5−1

27. h(x) = −x2 + 5x + 14

h(x) = − ( x2 − 5x − 14 )

h(x) = −(x − 7)(x + 2)

The zeros are p = 7 and q = −2.


2 = 5 —

2 = 2.5.

h(x) = −x2 + 5x + 14

= −(2.5)2 + 5(2.5) + 14

= −6.25 + 12.5 + 14

= 6.25 + 14

= 20.25


vertex (2.5, 20.25). Also, because a < 0, the parabola opens

down. Draw a smooth curve through the points.

x

y

8

4

2

6

14

16

18

20

421 3 5 6−1


Chapter 11


1. a. Sample answer: About 745 + 1075 + 1080 + 915 + 660 = 4475 of the

data values are within 2 inches of the mean of 40 inches.

There are a total of 5738 data values.

So, about 4475 — 5738 ≈ 0.780, or 78.0%, of the data values that

are within 1 standard deviation of the mean.

b. Sample answer: About 193 + 410 + 4475 + 375 + 97 = 5550 of the

data values are within 4 inches of the mean of 40 inches.

There are a total of 5738 data values.

So, about 5550 — 5738 ≈ 0.967, or 96.7%, of the data values that

are within 2 standard deviations of the mean.

c. Sample answer: About 5 + 90 + 5550 + 45 + 10 = 5700 of the data

values are within 6 inches of the mean of 40 inches. There

are a total of 5738 data values. So, about 5700 — 5738 ≈ 0.993,

or 99.3%, of the data values that are within 3 standard

deviations of the mean.

2. a. The set of adult female heights has a smaller standard

deviation. This means that the heights of adult females do

not vary as far from the mean.

b. Sample answer: About 19 + 32 + 25 + 32 + 30 + 26 + 23 = 187 of

the adult males surveyed have a height that is between 67

inches and 73 inches. A total of 250 males were surveyed.

So, about 187 — 250 = 0.748, or about 75%, of the heights are

between 67 inches and 73 inches.

3. Sample answer: You can characterize the basic shape of a

distribution by making a histogram of the data and drawing a

curve through the tops of the bars.

4. a. Most of the data are clustered around the mean in

the center of the data, and the data on the right of the

distribution are approximately a mirror image of the data

on the left of the distribution.

b. The curve connecting the tops of the bars looks like a bell.

c. Sample answer: Two other real-life examples of

symmetric distributions are standardized test scores and

the length of daylight each day over a year.


1.

0

5

10

15

20

25

30

35

40

1–10 11–20 21–30

Number of pounds

Freq

uen

cy

31–40 41–50 51–60

Aluminum Cans Collected

The tail of the graph extends to the left, and most of the data

are on the right. So, the distribution is skewed left.

2. a. Email Attachments Sent Frequency

1–20 2

21–40 3

41–60 9

61–80 10

81–100 4

101–120 2

0

2

4

6

8

10

12

1–20 21–40 41–60

Number of attachments

Freq

uen

cy

61–80 81–100 101–120

Email Attachments Sent

b. Because the data on the right of the distribution are

approximately a mirror image of the data on the left of

the distribution, the distribution is symmetric. So, use the

mean to describe the center and the standard deviation to

describe the variation.

3. Because most of the data are on the left of the distribution

for football players and the tail of the graph extends to the

right, the distribution is skewed right. So, the median and

fi ve-number summary best represent the distribution for

professional football players.

Because the data on the right of the distribution are

approximately a mirror image of the data on the left, the

distribution for company employees is symmetric. So, the

mean and standard deviation best represent the distribution

for company employees.

The median number of years of experience for professional

football players is probably in the 3–5 or 6–8 interval, and

the mean number of years of experience for the company

employees is probably in the 9–11 or 12–14 interval. So, a

typical company employee is more likely to have more years

of experience than a typical professional football player.

The data for the company employees are more variable than

the data for the professional football players. This means

that the number of years of experience tends to differ more

from one company employee to the next.

4. The mean is greater than the median because the distribution

is skewed right.

5. If 50 more women are surveyed, you would expect about

0.68 ⋅ 50 = 34 of the women to own between 10 and

18 pairs of shoes.


Chapter 11

11.3 Exercises (pp. 604–606)


1. In a symmetric distribution, the data are evenly distributed

to the left and right of the highest bar. In a distribution that is

skewed left, most of the data are on the right. In a distribution

that is skewed right, most of the data are on the left.

2. The mean and standard deviation best describe a symmetric

distribution, and the median and fi ve-number summary best

describe skewed distributions.


3.

0

2

4

6

8

10

12

14

16

18

20

1–2 3–4 5–6

Number of volunteer hours

Freq

uen

cy

7–8 9–10 11–12 13–14

Monthly Student Volunteer Hours

Most of the data are in the center and the data on the right

are approximately a mirror image of the data on the left. So,

the distribution is symmetric.

4.

0

5

10

15

20

25

30

35

40

45

50

0–3 4–7 8–11

Hours online

Freq

uen

cy

12–15 16–19 20–23 24–27

Weekly Online Hours

Most of the data are on the right and the tail of the graph

extends to the left. So, the distribution is skewed left.

5. Most of the data are in the fi rst two stems and the tail of

the data extends to the higher values. So, the distribution is

skewed right.

6. Most of the data are in the center and the data in the top half

are approximately a mirror image of the data in the bottom

half. So, the distribution is symmetric.

7. Most of the data are in the center and the data on the right

are approximately a mirror image of the data on the left,

which means the distribution is symmetric. So, the mean and

standard deviation best represent the data.

8. Most of the data are on the right and the tail of the graph

extends to the left. So, the distribution is skewed left, which

means that the median and fi ve-number summary best

represent the data.

9. a. ATM Withdrawals

(dollars)Frequency

26–50 7

51–75 5

76–100 5

101–125 3

126–150 2

151–175 1

176–200 1

0

1

2

3

4

5

6

7

8

26–50 51–75 76–100

Amount of money (dollars)

Freq

uen

cy

101–125 126–150 151–175 176–200

ATM Withdrawals

b. Most of the data are on the left and the tail of the graph

extends to the right. So, the distribution is skewed right,

which means that the median and fi ve-number summary

best represent the data.

c. Most of the data values are on the left side of the graph,

which represent withdrawals of less than $150. So, most

people were charged a fee for their withdrawals.


Chapter 11

10. a. IQ Scores Frequency

151–166 3

167–182 8

183–198 4

199–214 1

215–230 2

0

1

2

3

4

5

6

7

8

151–166 167–182 183–198

IQ score

Freq

uen

cy

199–214 215–230

IQ Scores

b. Most of the data are on the left and the tail of the graph

extends to the right. So, the distribution is skewed right,

which means that the median and fi ve-number summary

best represent the data.

c. As you include more and more IQ scores in the data set,

the shape of the distribution will become more symmetric.

11. When most of the data are on the right, the distribution is

skewed left, not right.



12. Use the standard deviation when the distribution of the data

is symmetric, not skewed.

Because the distribution is skewed, use the fi ve-number

summary to describe the variation of the data.

13. A stem-and-leaf plot requires that you list every data item,

but a histogram only displays the frequency of data items in

each range. So, a histogram would be more appropriate for a

large data set.

14. For a symmetric distribution, the mean is closest to the

center. So, it is the most appropriate measure to use for

describing the center of the data. Also for a symmetric

distribution, the variation is approximately the same on each

side of the center. So, the standard deviation is the most

appropriate measure for describing the variation of the data.

For a skewed distribution, the median is closest to the center

of the data. So, it is the most appropriate measure to use

for describing the center. Also for a skewed distribution,

the variation is different on each side. So, the fi ve-number

summary gives the most accurate description of the variation.

15. Because most of the data are on the right of the distribution

for Town A and the tail of the graph extends to the left, the

distribution is skewed left. So, the median and fi ve-number

summary best represent the center and distribution for Town A.

Because the data on the right of the distribution for Town

B are approximately a mirror image of the data on the left

of the distribution, the distribution is symmetric. So, the

mean and standard deviation best represent the center and

distribution for Town B.

The median of the Town A data set is in the 70–79 interval,

while the mean of the Town B data set is probably in the

60–69 interval. So, on a typical day, it is likely that the

temperature in Town A is higher than the temperature in

Town B.

The data for Town B is more variable than the data for Town

A. This means that the daily high temperature tends to differ

more for Town B.


Chapter 11

16.

0

1

2

3

4

5

6

7

8

9

10

11

12

13

8–10 11–13 14–16

Price range (dollars)

Res

tau

ran

t B

17–19 20–22 23–25

0

1

2

3

4

5

6

7

8

9

10

11

12

13R

esta

ura

nt

A

Freq

uen

cyEntrée Prices

Because most of the data are on the left of the distribution

for Restaurant A and the tail of the graph extends to the

right, the distribution is skewed right. So, the median

and fi ve-number summary best represent the center and

distribution for Restaurant A.

Because most of the data are on the right of the distribution

for Restaurant B and the tail of the graph extends to the left,

the distribution is skewed left. So, the median and fi ve-

number summary best represent the center and distribution

for Restaurant B.

The median of the Restaurant A data set is in the

14–16 interval, while the median of the Restaurant B data

set is between the 17–19 and 20–22 intervals. So, entrée

prices from Restaurant B are typically higher.

Because most of the prices for Restaurant A seem

to be concentrated in the two middle bars, the data

for Restaurant B is more spread out than the data for

Restaurant A. This means that the entrée prices for

Restaurant B tend to vary more.

17. Sample answer: The salaries of employees at a large

company tend to have a distribution that is skewed right.

18. Sample answer: The ages of residents and staff at a

retirement home tend to have a distribution that is

skewed left.

19. a.

200 500 800 1100 1400 1700 2000 2300Numberof songs

Sophomores

Freshmen

For freshmen, the whiskers are the same size, and the

median is in the center of the plot. So, the distribution is

symmetric. For sophomores, the whisker and box to the

left of the median are larger than the whisker and box

to the right of the median. So, most of the data is on the

right, and the distribution is skewed left.

b. The centers and spreads of the two data sets are quite

different from each other. The median for sophomores is

nearly twice the median for freshmen, and the mean for

sophomores is signifi cantly higher than that for freshmen.

So, the number of songs downloaded by freshmen tends

to be less than the number of songs downloaded by

sophomores. Also, there is more variability in the number

songs downloaded by sophomores.

c. Assuming the symmetric distribution is bell-shaped,

you know about 68% of the data lie within 1 standard

deviation of the mean. Because the mean is 1150, and

the standard deviation is 420, the interval from

1150 − 420 = 730 to 1150 + 420 = 1570 represents

about 68% of the data. So, you would expect about

0.68 ⋅ 45 ≈ 31 of the freshmen surveyed to have between

730 and 1570 songs downloaded on their MP3 players.

d. Assuming the symmetric distribution is bell-shaped,

you know about 95% of the data lie within 2 standard

deviation of the mean. Because the mean is 1150, and

the standard deviation is 420, the interval from

1150 − 2(420) = 310 to 1150 + 2(420) = 1990

represents about 95% of the data. So, you would expect

about 0.95 ⋅ 100 = 95 of the freshmen surveyed to have

between 310 and 1990 songs downloaded on their

MP3 players.

20. a.

200 500 800 1100 1400 1700 2000 2300

Numberof songs

Sophomores

Freshmen

The mean and median of this data set are less than

the mean and median, respectively, of the data set for

sophomores. So, the number of songs downloaded by

freshmen still tends to be less than the number of songs

downloaded by sophomores, but this data set has much

more variability than the previous data set for freshmen.

b. Because the whisker and box to the left of the median are

longer than the median and box to the right of the median,

the data are more concentrated on the right side of the

plot and the distribution is skewed left. So, the median is

greater than the mean for this group of freshmen.

21. The distribution will still be symmetric because the data

will all be related proportionally. Because this is a data

transformation using multiplication, the measures of center

and distribution will each be doubled.


Chapter 11

22. a. C; The data on the left are approximately a mirror image

of the data on the right. So, the distribution is symmetric.

This matches box-and-whisker plot C, which also shows

a symmetric distribution, because the whisker and box

to the right of the median are approximately the same size

as the whisker and box, respectively, to the left of

the median.

b. A; Most of the data are on the left of the histogram and

the tail extends to the right. So, the distribution is skewed

right. This matches box-and-whisker plot A, which also

shows a distribution that is skewed right, because most of

the data are concentrated on the left side of the plot.

c. B; Most of the data are on the right of the histogram, and

the tail extends to the left. So, the distribution is skewed

left. This matches box-and-whisker plot B, which also

shows a distribution that is skewed left, because most of

the data are concentrated on the right side of the plot.

23. a. Waiting Times Frequency

0–9 2

10–19 4

20–29 7

30–39 4

40–49 3

0

1

2

3

4

5

6

7

8

0–9 10–19 20–29

Time (minutes)

Freq

uen

cy

30–39 40–49

Restaurant Waiting Times

The data on the right side of the graph are approximately

a mirror image of the data on the left side. So, the

distribution is approximately symmetric.

b. Waiting Times Frequency

0–4 1

5–9 1

10–14 1

15–19 3

20–24 3

25–29 4

30–34 2

35–39 2

40–44 3

45–49 0

0

1

2

3

4

5

6

0–4

5–9

10–1

4

15–1

9

20–2

4

25–2

9

30–3

4

35–3

9

40–4

4

45–4

9

Time (minutes)

Freq

uen

cy

Restaurant Waiting Times

This graph shows that there are more data on the right

side of the graph, and that a tail extends to the left. So,

when the number of intervals is increased, the distribution

becomes skewed left.

c. More intervals show the spread of the data more

accurately. So, the histogram in part (b) best represents

the data.

24. Sample answer: The age at which people receive diplomas or

degrees is a bimodal distribution, because peaks in frequency

occur both at the typical high school graduation age and

again at the typical college graduation age.


25. x + 6 ≥ 0 −6 −6

x ≥ −6

So, the domain of the function is x ≥ −6.

26. 2x ≥ 0

2x

— 2 ≥ 0 —

2

x ≥ 0 So, the domain of the function is x ≥ 0.

27. x − 7 ≥ 0

+7 +7

x ≥ 7 So, the domain of the function is x ≥ 7.


Chapter 11

11.1–11.3 What Did You Learn? (p. 607)

1. In Exercise 15, the outlier is much less than the rest of

the data values, and it decreases the mean. In Exercise 16,

the outlier is much greater than the rest of the data values,

and it increases the mean. When a number is much less, it

decreases the total of the data items, and when a data value

is much greater, it increases the total. Because the mean

depends on the sum of the data values, it makes sense that

an outlier that is much less will decrease the mean and an

outlier that is much greater will increase the mean.

2. Sample answer: The residents would be much older than the

staff, and the residents would signifi cantly outnumber the

staff. So, most data values would be on the right.

11.1–11.3 Quiz (p. 608)

1. a. Mean: — x = 3

1 —

2 + 3 + 5 + 3

1 —

2 + 2

1 —

2 +

1 —

2

——— 6 = 18

— 6 = 3

Median: 1 — 2 , 2

1 —

2 , 3, 3

1 — 2 , 3

1 —

2 , 5

The mean of the two middle numbers is

3 + 3 1 —

2

— 2 = 6 1 —

2 ÷ 2 = 13

— 2 ⋅ 1 —

2 = 13

— 4

= 3 1 — 4 .

Mode: 1 — 2 , 2

1 —

2 , 3, 3

1 — 2 , 3

1 — 2 , 5

The data value that occurs most often is 3 1 —

2 .

The mean is 3, the median is 3 1 —

4 , and the mode is 3

1 —

2 . The

median best represents the data, because it is in the center

of the data, and it is the average of the mean and the mode.

2. a. Mean:

— x = 1000 + 1200 + 2568 + . . . + 1328 + 1000 + 1100

————— 9

= 11,834 —

9 = 1314 8 —

9

Median: 1000, 1000, 1100, 1180, 1191, 1200, 1267,

1328, 2568

The middle number is 1191.

Mode: 1000, 1000, 1100, 1180, 1191, 1200, 1267, 1328, 2568


The mean is 1314 8 —

9 , the median is 1191, and the mode is

1000. Because the data have an outlier of 2568, which is

much greater than the other data values, the mean is greater

than most of the data. Also, the mode is less than most of the

data. So, the median best represents the data.

3. Female students: 2, 3, 4, 4, 6

So, the range is 6 − 2 = 4 absences.

— x = 6 + 2 + 4 + 3 + 4 ——

5 = 19

— 5 = 3.8

x — x x − — x (x − — x )2

6 3.8 2.2 4.84

2 3.8 −1.8 3.24

4 3.8 0.2 0.04

3 3.8 −0.8 0.64

4 3.8 0.2 0.04

4.84 + 3.25 + 0.04 + 0.64 + 0.04 ———

5 = 8.8

— 5 = 1.76

√—

1.76 ≈ 1.3


Male students: 3, 5, 6, 6, 9

So, the range is 9 − 3 = 6 absences.

— x = 5 + 3 + 6 + 6 + 9 ——

5 = 29

— 5 = 5.8

x — x x − — x (x − — x )2

5 5.8 −0.8 0.64

3 5.8 −2.8 7.84

6 5.8 0.2 0.04

6 5.8 0.2 0.04

9 5.8 3.2 10.24

0.64 + 7.84 + 0.04 + 0.04 + 10.24 ————

5 = 18.8

— 5 = 3.76

√—

3.76 ≈ 1.9


The data have a greater range and standard deviation for the

male students. So, the numbers of absences are more spread

out for the male students.


Chapter 11

4. Juniors: 10, 14, 15, 15, 18, 19, 20, 21

So, the range is 21 − 10 = 11 points.

— x = 19 + 15 + 20 + 10 + 14 + 21 + 18 + 15 ————

8

= 132 —

8 = 16.5

x — x x − — x (x − — x )2

19 16.5 2.5 6.25

15 16.5 −1.5 2.25

20 16.5 3.5 12.25

10 16.5 −6.5 42.25

14 16.5 −2.5 6.25

21 16.5 4.5 20.25

18 16.5 1.5 2.25

15 16.5 −1.5 2.25

6.25 + 2.25 + 12.25 + 42.25 + 6.25 + 20.25 + 2.25 + 2.25

—————— 8

= 94 —

8 ≈ 11.75

√—

11.75 ≈ 3.4


Seniors: 15, 19, 19, 22, 26, 29, 30, 32

So, the range is 32 − 15 = 17 points.

— x = 22 + 19 + 29 + 32 + 15 + 26 + 30 + 19 ————

8

= 192 —

8 = 24

x — x x − — x (x − — x )2

22 24 −2 4

19 24 −5 25

29 24 5 25

32 24 8 64

15 24 −9 81

26 24 2 4

30 24 6 36

19 24 −5 25

4 + 25 + 25 + 64 + 81 + 4 + 36 + 25 ————

8 = 264

— 8 = 33

√—

33 ≈ 5.7


The data have a greater range and standard deviation for the

seniors. So, the numbers of points scored are more spread

out for the seniors.

5.

15, 20, 25, 30, 40, 55, 60, 70


least value

fi rst quartile

median

⎧⎪⎪⎪⎨⎪⎪⎪⎩

third quartile

greatest value

upper half

Least value: 15


2 = 45

— 2 = 22.5

Median: 30 + 40 —

2 = 70

— 2 = 35


2 = 115

— 2 = 57.5

Greatest value: 70

15 25 35 45 55 65Age

6.

20, 20, 20, 30, 35, 40, 40, 40, 50, 50, 60, 60

lower half ⎧⎪⎪ ⎪ ⎪ ⎪⎪⎨⎪⎪ ⎪ ⎪ ⎪⎪⎩

least value

fi rst quartile

median third quartile

greatest value

⎧⎪⎪ ⎪ ⎪ ⎪⎪⎨⎪⎪ ⎪ ⎪ ⎪⎪⎩ upper half

Least value: 20


2 = 50

— 2 = 25

Median: 40 + 40 —

2 = 80

— 2 = 40


2 = 100

— 2 = 50

Greatest value: 60

20 30 40 50 60Minutes

7.

0

2

4

6

8

10

12

14

16

18

0–2 3–5 6–8

Score

Freq

uen

cy

9–11 12–14

Quiz Scores




Chapter 11

8. a. Mean:

— x = 98 + 119 + 95 + 211 + 130 + 98 + 100 + 125 —————

8

= 976 —

8 = 122

Median: 95, 98, 98, 100, 119, 125, 130, 211

The mean of the two middle numbers is

100 + 119

— 2 =

219 —

2 = 109.5.

Mode: 95, 98, 98, 100, 119, 125, 130, 211


The range is 211 − 95 = 116.

x — x x − — x (x − — x )2

98 122 −24 576

119 122 −3 9

95 122 −27 729

211 122 89 7921

130 122 8 64

98 122 −24 576

100 122 −22 484

125 122 3 9

576 + 9 + 729 + 7921 + 64 + 576 + 484 + 9

————— 8

= 10,368 —

8 = 1296

√—

1296 = 36

So, the mean is $122, the median is $109.50, the mode is

$98, the range is $116, and the standard deviation is $36.

b. The price of $211 is signifi cantly greater than the other

prices. So, it is the outlier.

Mean: — x = 98 + 119 + 95 + 130 + 98 + 100 + 125 ————

7

= 765 —

7 ≈ 109.29

Median: 95, 98, 98, 100, 119, 125, 130


Mode: 95, 98, 98, 100, 119, 125, 130

The data value that occurs most often is still 98.

So, the outlier increases the mean by about

$122 − $109.29 = $12.71 and it increases the median by

$109.50 − $100 = $9.50, but the outlier does not affect

the mode.

c.

95, 98, 98, 100, 119, 125, 130, 211


least value

fi rst quartile

median

⎧⎪⎪⎪⎨⎪⎪⎪⎩third

quartilegreatest

value

upper half

Least value: 95


2 = 196

— 2 = 98

Median: 100 + 119 —

2 = 219

— 2 = 109.5


2 = 255

— 2 = 127.5

Greatest value: 211

80 100 120 140 160 180 200 220Price

The interquartile range is 127.5 − 98 = 29.5. This means

that the middle half of the prices vary by no more than

$29.50.

Because most of the data items are on the left side of the

plot and the right whisker is longer than the left whisker,

the distribution is skewed right.

d. Mean: — x = 122(1 − 0.05) = 122(0.95) = 115.9

Median: 109.5(0.95) = 104.025

Mode: 98(1 − 0.05) = 98(0.95) = 93.1

Range: 116(1 − 0.05) = 116(0.95) = 110.2

Standard deviation: 36(1 − 0.05) = 36(0.95) = 34.2

9. a. Time (minutes) Frequency

3–5 2

6–8 4

9–11 10

12–14 3

15–17 1

0

2

4

6

8

10

12

3–5 6–8 9–11

Time (minutes)

Freq

uen

cy

12–14 15–17

Presentation Times

b. Because the data on the right are approximately a mirror

image of the data on the left, the distribution is symmetric.

So, the mean and standard deviation best represent

the data.

c. Because most of the data are in the 9–11 range or close to

it, most of the presentations are about the right length.


Chapter 11


1. a.

Beginning of

season

T-Shirt Size

S M L XL XXL Total

Col

or

blue/white 5 6 7 6 5 29

blue/gold 5 6 7 6 5 29

red/white 5 6 7 6 5 29

black/white 5 6 7 6 5 29

black/gold 5 6 7 6 5 29

Total 25 30 35 30 25 145

End of season

T-Shirt Size

S M L XL XXL Total

Col

or

blue/white 5 4 1 0 2 12

blue/gold 3 6 5 2 0 16

red/white 4 2 4 1 3 14

black/white 3 4 1 2 1 11

black/gold 5 2 3 0 2 12

Total 20 18 14 5 8 65

b. Sample answer: Next season, for each shirt category, you

could order the quantity equal to the difference between

the number at the beginning of this season and the number

at the end of this season. From the table, you can see that

you should order fewer t-shirts that are sizes small and

medium, and large especially. You may even want to order

a couple more of the t-shirts that were sold out by the end

of the season, such as the blue/white, size XL.

2. a. Sample answer:

Gender

Males Females Total

Hou

rs

0 hours per week 145 145 290

1–7 hours per week 110 115 225

8+ hours per week 70 115 185

Total 325 375 700

b. Sample answer: More females work 8+ hours per week

than males. Approximately equal numbers of males and

females work 1–7 hours per week and 0 hours per week.

3. Sample answer: Rows and columns represent different

categories, each entry represents the number in both

categories.


1.

Tablet Computer

Yes No Total

Cel

l P

hone Yes 34 124 158

No 18 67 85

Total 52 191 243

So, 52 students own a tablet, and 191 students do not own a

tablet. Also, 158 students own a cell phone, and 85 students

do not own a cell phone. In total, 243 students were

surveyed.

2.

Summer Job

Yes No Total

Gen

der Male 57 18 75

Female 45 12 57

Total 102 30 132

3.

Summer Job

Yes No Total

Gen

der Male 0.43 0.14 0.57

Female 0.34 0.09 0.43

Total 0.77 0.23 1

So, about 0.23, or 23%, of the students are not getting a

summer job.

4.

Major in Medical Field

Yes No

Cla

ss Junior 0.36 0.64

Senior 0.30 0.70

So, given that a student is a senior, the conditional relative

frequency that he or she is planning to major in a medical

fi eld is about 0.30, or 30%.


Chapter 11

5.

Tablet Computer

Yes No

Cel

l P

hone Yes

34 —

52 ≈ 0.65

124 —

191 ≈ 0.65

No 18

— 52

≈ 0.35 67

— 191

≈ 0.35

The conditional relative frequencies for students who

have cell phones are about the same whether they have a

tablet computer or not. Similarly, the conditional relative

frequencies for students who do not have cell phones are

about the same whether they have a tablet computer or not.

So, there is not an association between owning a tablet

computer and owning a cell phone.

11.4 Exercises (pp. 614–616)


1. Each entry in a two-way table is called a joint frequency.

2. It is appropriate to use a two-way table to organize data

when data are collected from one source and belong to two

different categories.

3. A marginal relative frequency is the sum of the joint relative

frequencies in a row or column. A conditional relative

frequency is the ratio of a joint relative frequency to the

marginal relative frequency.

4. In order to fi nd conditional relative frequencies, you can

use the marginal relative frequency of each row or of

each column. Or, you can fi nd the ratio of a joint relative

frequency to the corresponding marginal relative frequency.


Table for Exercises 5–8

Buy Lunch at School

Yes No Total

Cla

ss Freshmen 92 86 178

Sophomore 116 52 168

Total 208 138 346

5. A total of 178 freshmen were surveyed.

6. A total of 168 sophomores were surveyed.

7. A total of 208 students buy lunch at school.

8. A total of 138 students do not buy lunch at school.

9.

Set Academic Goals

Yes No Total

Gen

der Male 64 168 232

Female 54 142 196

Total 118 310 428

So, 232 male students and 196 female students were asked if

they set academic goals. A total of 118 students set academic

goals, and 310 students have not. A total of 428 students

were asked if they set academic goals.

10.

Cat

Yes No Total

Dog

Yes 104 208 312

No 186 98 284

Total 290 306 596

So, 312 people have a dog, while 284 people do not. Also,

290 people have a cat, while 306 people do not. A total of

596 people were asked whether they have a cat or dog.

11.

Participate in Spirit Week

Yes No Undecided Total

Cla

ss Freshmen 112 56 54 222

Sophomore 92 68 32 192

Total 204 124 86 414

So, 222 freshmen and 192 sophomores responded to the

survey. A total of 204 students say they will participate

in school spirit week, 124 students say they will not, and

86 students are undecided. A total of 414 students were

surveyed.

12.

Type of Degree

Associate’s Bachelor’s Master’s Total

Gen

der Male 58 126 42 226

Female 62 118 48 228

Total 120 244 90 454

So, 226 males and 228 females responded to the survey.

A total of 120 seniors plan to receive an associate’s

degree, 244 seniors plan to receive a bachelor’s degree,

and 90 seniors plan to receive a master’s degree. A total of

454 seniors were surveyed.


Chapter 11

13.

Traveled on an Airplane

Yes No Total

Cla

ss Freshman 90 62 152

Sophomore 184 16 200

Total 274 78 352

14.

Plan to Attend School Dance

Yes No Total

Gen

der Male 38 46 84

Female 88 24 112

Total 126 70 196

15.

Spanish Class

Yes No Total

Fre

nch

Cla

ss Yes 45 54 99

No 64 82 146

Total 109 136 245

16.

Condition

New Used Total

Ori

gin Domestic 36 28 64

Foreign 19 15 34

Total 55 43 98

17.

Exercise Preference

Aerobic Anaerobic Total

Gen

der Male

88 —

350 ≈ 0.25

104 —

350 ≈ 0.30 0.55

Female 96

— 350

≈ 0.27 62

— 350

≈ 0.18 0.45

Total 0.52 0.48 1

18.

Meat

Turkey Ham Total

Bre

ad White 452

— 1348

≈ 0.34 146

— 1348

≈ 0.11 0.55

Wheat 328

— 1348

≈ 0.24 422

— 1348

≈ 0.31 0.45

Total 0.58 0.42 1

19. Of the students surveyed, about 52% prefer aerobic exercise

and about 30% are males who prefer anaerobic exercise.

20. Of the sandwiches included in the survey, 55% are on wheat

bread and 34% are turkey on white bread.

21. One hundred eighty-seven is only the number of freshmen

who participated in the fundraiser.

A total of 187 + 85 = 272 freshmen responded to

the survey.

22. The table entries do not give the ratio of joint relative

frequencies to the total number of values. So, the two-way

table does not show the joint relative frequencies.

Each table entry is the ratio of the joint frequency to a

marginal frequency. So, the two-way table shows conditional

relative frequencies.

23.

Menu

Potluck Catered Total

Mea

l Lunch 36 48 84

Dinner 44 72 116

Total 80 120 200

Menu

Potluck Catered

Mea

l Lunch 36

— 84

≈ 0.43 48

— 84

≈ 0.57

Dinner 44

— 116

≈ 0.38 72

— 116

≈ 0.62

So, given that an employee prefers a lunch event, the

conditional relative frequency that he or she prefers a catered

event is about 0.57, or 57%.

24.

Type

Tiger Hawk Dragon Total

Styl

e Realistic 67 74 51 192

Cartoon 58 18 24 100

Total 125 92 75 292

Type

Tiger Hawk Dragon

Styl

e Realistic 67

— 125

≈ 0.54 74

— 92

≈ 0.80 51

— 75

= 0.68

Cartoon 58

— 125

≈ 0.46 18

— 92

≈ 0.20 24

— 75

= 0.32

So, given that a student prefers a hawk mascot, the

conditional relative frequency that he or she prefers a cartoon

mascot is about 20%.


Chapter 11

25.

Live on Campus

Yes No Total

Hav

e a

Car Yes 25% 10% 35%

No 60% 5% 65%

Total 85% 15% 100%

Live on Campus

Yes No

Hav

e a

Car Yes

0.25 —

0.35 ≈ 0.71

0.10 —

0.35 ≈ 0.29

No 0.60

— 0.65

≈ 0.92 0.05

— 0.65

≈ 0.08

Of the students who have a car, about 71% live on campus.

Of the students who do not have a car, about 92% live on

campus. So, a student who does not have a car is more likely

to live on campus than a student who does not have a car. So,

there is an association between living on campus and having

a car at college.

26.

Participate in a Sport

Yes No Total

Wat

ch

Spor

ts o

n T

V

Yes 34% 36% 70%

No 14% 16% 30%

Total 48% 52% 100%

Participate in a Sport

Yes No

Wat

ch

Spor

ts o

n T

V

Yes 0.34

— 0.48

≈ 0.71 0.36

— 0.52

≈ 0.69

No 0.14

— 0.48

≈ 0.29 0.16

— 0.52

≈ 0.31

Of the students who participate in a sport, about 71% watch

sports on TV. Of the students who do not participate in a

sport, about 69% watch sports on TV. So, whether a student

participates in a sport does not seem to mean that he or she

is signifi cantly more or less likely to watch sports on TV. So,

there is not an association between participating in a sport

and watching sports on TV.

27. Age

21–30 31–40 41–50 51–60 61–70 Total

Ski Yes 87 93 68 37 20 305

No 165 195 148 117 125 750

Total 252 288 216 154 145 1055

Age

21–30 31–40 41–50 51–60 61–70

Ski Yes

87 —

252 ≈ 0.35

93 —

228 ≈ 0.32

68 —

216 ≈ 0.31

37 —

154 ≈ 0.24

20 —

145 ≈ 0.14

No 165

— 252

≈ 0.65 195

— 288

≈ 0.68 148

— 216

≈ 0.69 117

— 154

≈ 0.76 125

— 145

≈ 0.86

Based on this sample, about 35% of adults ages 21–30

participate in recreational skiing, and about 14% of adults

ages 61–70 participate in recreational skiing. In general, the

table shows that as adults get older, they are less likely to

participate in recreational skiing. So, there is an association

between age and participation in recreational skiing.

28.

Type of Degree

Associate’s Bachelor’s Master’sG

ende

r Male 58

— 226

≈ 0.26 126

— 226

≈ 0.56 42

— 226

≈ 0.19

Female 62

— 228

≈ 0.27 118

— 228

≈ 0.52 48

— 228

≈ 0.21

The percentages for each type of degree are similar between

genders. So, there is not an association between gender and

type of degree.

29. Venn diagrams can be used to display the information in a

two-way table that has two rows and two columns.

30. a.

Genre

Comedy Action Horror

Gen

der Male 30 50 20

Female 40 30 20

b. Sample answer: The graph is preferred because it displays

the data visually.


Chapter 11

31. a.

Seat Location

Main Floor Balcony Total

Tic

ket

Typ

e Adult 2x + y x + 2y 3x + 3y

Child x − 40 3x − y − 80 4x − y − 120

Total 800 1009 1809

b. (2x + y) + (x − 40) = 800

3x + y − 40 = 800

+40 +40

3x + y = 840

(x + 2y) + (3x − y − 80) = 1009

4x + y − 80 = 1009

+80 +80

4x + y = 1089

(3x + 3y) + (4x − y − 120) = 1809

7x + 2y − 120 = 1809

+120 +120

7x + 2y = 1929

4x + y = 1089

−(3x + y = 840)

x = 249

3x + y = 840

3(249) + y = 840

747 + y = 840

−747 −747

y = 93

Check: 3x + y = 840 4x + y = 1089

3(249) + 93 =?

840 4(249) + 93 =?

1089

747 + 93 =?

840 996 + 93 =?

1089

840 = 840 ✓ 1089 = 1089 ✓

7x + 2y = 1929

7(249) + 2(93) =?

1929

1743 + 186 =?

1929

1929 = 1929 ✓

So, x = 249 and y = 93.

c. 3x + 3y = 3(249) + 3(93)

= 747 + 279

= 1029

So, about 1029 —

1809 ≈ 0.57, or 57% of the tickets are adult

tickets.

d. 4x − y − 120 = 4(249) − 93 − 120

= 996 − 93 − 120

= 903 − 120

= 783

3x − y − 80 = 3(249) − 93 − 80

= 747 − 93 − 80

= 574

So, about 574 —

783 ≈ 0.73, or 73% of the child tickets are

balcony tickets.

32. “One-way tables” display data collected from one source

that belong to one category. “Two-way tables” display

data collected from one source that belong to two different

categories. It is possible to have a “Three-way table” that

displays data collected from one source that belong to three

different categories.

Sample answer:

Gender and Socio-Economic Status

Male Female

Type of Program Low Middle High Low Middle High

General 7 10 4 9 10 5

Academic 4 22 21 15 22 21

Vocation 4 15 4 8 16 3


33.

x 0 1 2 3 4

y 144 24 4 2 —

3

1 —

9

+1 +1 +1 +1

× 1 — 6 ×

1 — 6 ×

1 — 6 ×

1 — 6

As x increases by 1, y is multiplied by a common factor of 1 —

6 .

So, the table represents an exponential function.

34.

x −1 0 1 2 3

y 3 0 −1 0 3

+1 +1 +1 +1

−3 −1 +1 +3

+2 +2 +2

As x increases by 1, y has a second-level common difference

of 2. So, the table represents a quadratic function.


Chapter 11


1. a. Sample answer:

Birds: 307

— 3962

⋅ 360° ≈ 27.9°

Mammals: 2746

— 3962

⋅ 360° ≈ 249.5°

Amphibians: 145

— 3962

⋅ 360° ≈ 13.2°

Reptiles: 75 —

3962 ⋅ 360° ≈ 6.8°

Unknown: 689

— 3962

⋅ 360° ≈ 62.6°

Mammals: 2746

Road Kill

Amphibians: 145Reptiles: 75

Birds: 307 Unknown: 689

The circle graph shows the types of animals as parts

of a whole.

b. Sample answer:

19921994

19961998

20002002

20042006

20082010

20120

20

40

60

80

100

120

140

0

Year

Black Bear Road Kill

Nu

mb

er o

f b

lack

bea

rs

x

y

The scatter plot shows the relationship between two sets

of numerical data.

c. Sample answer:

Raccoon Road Kill Weights

Stem Leaf

9 4 5

10

11 0

12 4 9

13 4 6 9

14 0 5 8 8

15 2 7

16 8

17 0 2 3 5

18 5 5 6 7

19 0 1 4

20 4

21 3 5 5 5

22

23

24

25 4

Key: 9 ∣ 4 = 9.4 pounds

The stem-and-leaf plot shows how the raccoon weights

are distributed.

d. Sample answer:

Animals Killed by Motor Vehicles

Racoon

Skunk

Ground sq

uirrel

Opossum

Deer

Gray s

quirrel

Cottonta

il

Barn o

wl

Jack

rabbit

Gopher sn

ake

200

400

600

800

1000

1200

1400

1600

1800 1693

1372

845763 761 715 629

486 466363

0

The bar graph shows the data in each specifi c category.

2. You can display data using different plots and graphs to

make it easy to interpret and draw conclusions from the data.

3. Answers will vary. Examples of each data display can easily

be found by typing the name of the data display into a browser

or search engine. Remember to make note of the source.


Chapter 11


1. Even though telephone numbers are numerical, they are

not counts or measurements, and it does not make sense to

compare them numerically. So, the data are qualitative.

2. Ages are a numerical measurement. So, the data are

quantitative.

3. Lengths of videos are numerical measurements. So, the data

are quantitative.

4. Types of fl owers are nonnumerical entries that can be

separated into different categories. So, the data are

qualitative.

5. Sample answer:

10

20

30

40

50

60

70

0

Eye color

Eye Color Survey

Nu

mb

er o

f st

ud

ents

63

37

25

103 2

Brown Blue Hazel Green Gray Amber

6. Sample answer:

Interstate A Interstate B

9 5

8 8 8 8 8 7 5 5 5 2 6 5 6 7 9 9

7 5 5 5 2 1 0 7 0 0 1 1 2 3 5 7 8 9 9

2 0 8 0 1 4 4

Key: 2 | 6 | 5 = 62 mi/h on Interstate A, 65 mi/h on

Interstate B

7. Sample answer:

2,000

4,000

6,000

8,000

10,000

12,000

14,000

16,000

18,000

20,000

0

Academic year

Tuition, Room, and Boardat College and Universities

Ave

rag

e co

st (

do

llars

)

2007–2008 2008–2009 2009–2010 2010–2011

2008 2009 2010 2011 2012 20130

1

2

3

4

5

6

7

8

9

10

0

Year

Mean Hourly Wage for Employeesat Fast-Food Restaurant

Wag

e (d

olla

rs p

er h

ou

r)

11.5 Exercises (pp. 621–622)


1. Sample answer: A line graph is misleading if the numbers on

the scale are unevenly spaced or if the scale does not begin at

zero and there is no break.

2. The data set that does not belong with the other three is

“breeds of dogs at a pet store” because this data set is

qualitative, and the other three are quantitative.


3. Brands of cars are nonnumerical entries that can be separated

into different categories. So, the data are qualitative.

4. Weights are numerical measurements. So, the data are

quantitative.

5. Budgets are numerical measurements. So, the data are

quantitative.

6. File formats are nonnumerical entries that can be separated

into different categories. So, the data are qualitative.

7. Shoe sizes are numerical measurements. So, the data are

quantitative.

8. Street addresses have a numerical component, but they are

not counts or measurements, and it does not make sense

to compare or measure them. Instead, they are labels that

can be separated into different categories. So, the data

are qualitative.


Chapter 11

9. Sample answer: An appropriate data display for the number

of students in a marching band each year is a line graph,

because it will display numerical data over time.

10. Sample answer: An appropriate data display for a

comparison of students’ grades in two different classes is a

back-to-back stem-and-leaf plot, because it allows you to

compare data values from two different sources.

11. Sample answer: An appropriate data display for the favorite

sports of students in your class is a bar graph, because you

want to show the data in each specifi c category.

12. Sample answer: An appropriate data display for the

distribution of teachers by age is a histogram, because it

shows frequencies of data values in equally-sized intervals.

13. Sample answer:

21, 21, 22, 22, 23, 23, 24, 24, 24, 24, 25,

26, 26, 27, 27, 28, 28, 28, 29, 30, 30, 32, 32

least value

fi rst quartile

lower half ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

median

upper half ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

third quartile

greatest value

Least value: 21

First quartile: 23

Median: 26

Third quartile: 28

Greatest value: 32

18, 20, 20, 22, 23, 23, 24, 24, 24, 24,

25, 25, 26, 27, 27, 27, 27, 28, 29, 32, 36

least value

fi rst quartile

lower half ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

median

upper half ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

third quartile

greatest value

Least value: 18


2 = 46

— 2 = 23

Median: 25


2 = 54

— 2 = 23

Greatest value: 36

18 20 22 24 26 28 30 32 34 36Age

2011 Women (Japan)

Ages of World Cup Winners

2010 Men (Spain)

A double box-and-whisker plot shows the distributions

of the data.

14. Sample answer:

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

0

Month

Precipitation

Ave

rag

e p

reci

pit

atio

n (

inch

es)

Jan

uar

y

Feb

ruar

y

Mar

ch

Ap

ril

May

Jun

e

July

Au

gu

st

Sep

tem

ber

Oct

ob

er

No

vem

ber

Dec

emb

er

A line graph shows how the data change over time.

15. Sample answer:

Grades (out of 100) on a Test

Stem Leaf

5 2 3 9

6 2 3

7 4 5 7

8 0 1 3 4 5 7 8 9

9 5 6 7

10 0

Key: 5 ∣ 3 = 53 out of 100

A stem-and-leaf plot shows how the grades are distributed.


Chapter 11

16. Sample answer:

White: 25

— 100

⋅ 360° = 90°

Red: 12

— 100

⋅ 360° = 43.2°

Yellow: 1 —

100 ⋅ 360° = 3.6°

Black: 21

— 100

⋅ 360° = 75.6°

Green: 3 —

100 ⋅ 360° = 10.8°

Silver/gray: 27

— 100

⋅ 360° = 97.2°

Blue: 6 —

100 ⋅ 360° = 21.6°

Brown/beige: 5 —

100 ⋅ 360° = 18°

Color of Cars that Drive by

Your House

White: 25

Red: 12

Yellow: 1Black: 21

Green: 3

Silver/Gray: 27

Blue: 6Brown/Beige: 5

A circle graph shows the data as parts of a whole.

17. Sample answer:

Age

Frequency for 2010 Men’s World Cup Winner (Spain)

Frequency for 2011 Women’s World

Cup Winner (Japan)

18–21 2 3

22–25 9 9

26–29 8 7

30–33 4 1

34–37 0 1

2

4

6

8

10

12

14

0

Ages of World Cup Winners

2010

Men

’s W

orl

dC

up

win

ner

(Sp

ain

)

2

4

6

8

10

12

14

0

Age

2011

Wo

men

’s W

orl

dC

up

win

ner

(Ja

pan

)

18–21 22–25 26–29 30–33 34–37

18. Sample answer:

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

0

Month

Precipitation

Ave

rag

e p

reci

pit

atio

n (

inch

es)

Januar

y

Febru

ary

Mar

chApril

May

June

July

August

Septe

mber

October

Novem

ber

Decem

ber

19. Sample answer:

Score Frequency

51–60 3

61–70 2

71–80 4

81–90 7

91–100 4

Grades (out of 100) on a Test

2

4

6

8

10

0

Score

Freq

uen

cy

51–60 61–70 71–80 81–90 91–100

20. Sample answer:

Colors of Cars that Drive byYour House

4

8

12

16

20

24

28

0

Color

Nu

mb

er o

f ca

rs

White Red Yellow Black Green Silver/Gray

Blue Brown/Beige

21. The scale on the vertical axis has very small increments.

So, someone might believe that the annual sales more than

tripled from 2010 to 2013.

22. The interval for the third bar is greater than the other

two bars and the vertical scale does not have a break. So,

someone might believe that the frequency for each 30-minute

interval is increasing.


Chapter 11

23. The increments on the scale are not equal. So, someone

might believe that the temperatures are evenly distributed.

24. The increments on the vertical axis are not equal. So,

someone might believe that the decay rate is linear.

25. Sample answer:

Annual Sales

10

20

30

40

50

60

70

80

90

100

110

0

Year

Sale

s (m

illio

ns

of

do

llars

)

2010 2011 2012 2013

26. Sample answer:

Minutes Frequency

0–29 12

30–59 15

60–89 13

90–119 6

Bicycling

3

6

9

12

15

18

0

Minutes

Freq

uen

cy

0–29 30–59 60–89 90–119

27. The data are quantitative, not qualitative. A bar graph is most

appropriate for qualitative data. So, Classmate B is correct

that the data would be better displayed in a histogram.

28. The units on the vertical axis are missing. Sample answer: The units may be hundreds of stores opened, in which case,

the number of stores being opened might be decreasing, but

their sales may be increasing as new stores are opened.

29. a. The total of the numbers in the circle graph is 120,

when only 100 students were asked. So, the number

of responses does not match the number of students

surveyed, but this is not clearly conveyed by the graph.

b. Sample answer: A bar graph could be used because a bar

graph shows data in specifi c categories.

30. Sample answer: Doughnut display

31. Sample answer: A dot plot clearly shows the mode of a data

set, because the mode(s) will have the most dots.


32. The input −5 has two different outputs of −1 and 1. So, the

relation is not a function.

33. Each input has exactly one output. So, the relation is a

function.

11.4–11.5 What Did You Learn? (p. 623)

1. Because more sophomores said yes than no, your sophomore

friend is more likely to have said yes.

2. Sample answer: Business decisions are based on data. So,

accurate graphs help to make correct decisions.

Chapter 11 Review (pp. 624–626)

1. New mean:

— x = 3.5 + 4.0 + 4.4 + 3.9 + . . . + 4.5 + 2.0 + 5.0 + 4.0 —————

11

= 44 —

11 = 4

New median: 2.0, 3.5, 3.9, 4.0, 4.0, 4.1, 4.3, 4.3, 4.4, 4.5, 5.0

The middle number is 4.1.

New mode: 2.0, 3.5, 3.9, 4.0, 4.0, 4.1, 4.3, 4.3, 4.4, 4.5, 5.0

The data values that occur most often are 4.0

and 4.3.

The new mean is 4, which is the same as the old mean. The

new median is 4.1, which is less than the old median. The

data set now has two modes, 4.0 and 4.3. Because 4.0 is

equal to the old mean, the mean stays the same. Because 4.0

is less than the old median, the median decreases. Because

you ran 4.0 miles on another day as well and the old mode

occurred twice, 4.0 becomes an additional mode.

2. New mean:

— x = 3.5 + 4.0 + 4.4 + 3.9 + . . . + 4.5 + 2.0 + 5.0 + 10.0 —————

11

= 50 —

11 ≈ 4.5

New median: 2.0, 3.5, 3.9, 4.0, 4.1, 4.3, 4.3, 4.4, 4.5, 5.0, 11.0

The middle number is 4.3.

New mode: 2.0, 3.5, 3.9, 4.0, 4.1, 4.3, 4.3, 4.4, 4.5, 5.0, 11.0

The data value that occurs most often is 4.3.

The new mean is about 4.5, which is greater than the old

mean. The new median is 4.3, which is greater than the old

median. The mode is still 4.3. Because 10.0 is greater than

the old mean, the mean increases. Because 10.0 is greater

than the median, the median increases. Because this is the

only day you have run 11.0 miles, the mode stays the same.


Chapter 11

3. Mean:

— x = 11 + 0 + 10 + 3 + (−9) + 10 + 3 + (−2) + 10 —————

9

= 36 —

9 = 4

Median: −9, −2, 0, 3, 3, 10, 10, 10, 11


Mode: −9, −2, 0, 3, 3, 10, 10, 10, 11


The mean is 4, the median is 3, and the mode is 10.

4. Mean:

— x = 0 + 0 + 1 + 1 + 1 + 1 + 2 + 2 + 4 + 5 ————

10

= 17 —

10 = 1.7

Median: 0, 0, 1, 1, 1, 1, 2, 2, 4, 5

The middle number is 1 + 1 —

2 = 2 —

2 = 1.

Mode: 0, 0, 1, 1, 1, 1, 2, 2, 4, 5


The mean is 1.7, the median is 1, and the mode is 1.

5. Player A: 174, 185, 190, 194, 200, 203, 205, 210, 219, 230

So, the range for Player A is 230 − 174 = 56.

— x = 205 + 185 + 210 + . . . + 219 + 203 + 230 ————

10

= 2010 —

10 = 201

x — x x − — x (x − — x )2

205 201 4 16

185 201 −16 256

210 201 9 81

174 201 −27 729

194 201 −7 49

190 201 −11 121

200 201 −1 1

219 201 18 324

203 201 2 4

230 201 29 841

16 + 256 + 81 + 729 + 49 + 121 + 1 + 324 + 4 + 841

————— 10

= 2422 —

10 = 242.2

So, the standard deviation for Player A is about

√—

242.2 ≈ 15.56.

Player B: 154, 168, 172, 181, 192, 204, 228, 235, 235, 240

So, the range for Player B is 240 − 154 = 86.

— x = 228 + 172 + 154 + . . . + 240 + 235 + 192 ————

10

= 2010 —

10 = 201

x — x x − — x (x − — x )2

228 201 27 729

172 201 −29 841

154 201 −47 2209

235 201 34 1156

168 201 −33 1089

205 201 4 16

181 201 −20 400

240 201 39 1521

235 201 34 1156

192 201 −9 81

729 + 841 + 2209 + . . . + 1521 + 1156 + 81

———— 10

= 9198 —

10 = 919.8

So, the standard deviation for Player B is about

√—

919.8 ≈ 30.33.

The data for Player B have a greater range and standard

deviation. So, Player B’s bowling scores are more spread out.


Chapter 11

6. Store A: 140, 150, 160, 180, 190, 200, 250, 250

So, the range for Store A is 250 − 140 = $110.

— x = 140 + 200 + 150 + 250 + 180 + 250 + 190 + 160 —————

8

= 1520 —

8 = 190

x — x x − — x (x − — x )2

140 190 −50 2500

200 190 10 100

150 190 −40 1600

250 190 60 3600

180 190 −10 100

250 190 60 3600

190 190 0 0

160 190 −30 900

2500 + 100 + 1600 + 3600 + 100 + 3600 + 0 + 900

————— 8

= 12,400 —

8 = 1550

So, the standard deviation for Store A is about

√—

1550 ≈ $39.37.

Store B: 160, 190, 190, 225, 240, 260, 285, 310

So, the range for Player B is 310 − 160 = $150.

— x = 225 + 260 + 190 + 160 + 310 + 190 + 285 ————

8

= 1860 —

8 = 232.5

x — x x − — x (x − — x )2

225 232.5 −7.5 56.25

260 232.5 27.5 756.25

190 232.5 −42.5 1806.25

160 232.5 −72.5 5256.25

310 232.5 77.5 6006.25

190 232.5 −42.5 1806.25

285 232.5 52.5 2756.25

240 232.5 7.5 56.25

56.25 + 756.25 + . . . + 2756.25 + 56.25

———— 8

= 18,500

— 8 = 2312.5

So, the standard deviation for Store B is about

√—

2312.5 ≈ $48.09.

The data for Store B have a greater range and standard

deviation. So, Store B’s prices are more spread out.

7. Mean: — x = 109 + 25 = 134

Median: 104 + 25 = 129

Mode: 96 + 25 = 121

Range: 45

Standard deviation: 3.6

8. Mean: — x = 109(0.6) = 65.4

Median: 104(0.6) = 62.4

Mode: 96(0.6) = 57.6

Range: 45(0.6) = 27

Standard deviation: 3.6(0.6) = 2.16

9. Ordered data: 14, 14, 16, 17, 17, 18, 20, 21, 22

Least value: 14


2 = 30

— 2 = 15

Median: 17


2 = 41

— 2 = 20.5

Greatest value: 22

14 16 18 20 22Age

Because most of the data items are on the left side of the

plot and the right whisker and box are longer than the left

whisker and box, the distribution is skewed right.

10. Ordered data: 120, 160, 180, 200, 200, 210, 230, 230

Least value: 120


2 = 340

— 2 = 170

Median: 200 + 200 —

2 = 400

— 2 = 200


2 = 440

— 2 = 220

Greatest value: 230

110 130 150 170 190 210 230Mass(kilograms)

Because most of the data items are on the right side of the

plot, and the left whisker and box are longer than the right

whisker and box, the distribution is skewed left.


Chapter 11

11. a. Amounts of Money in Pocket

2

4

6

8

10

12

0

Freq

uen

cy

0–0.99 1–1.99 2–2.99 3–3.99 4–4.99 5–5.99

Most of the data are on the left and the tail of the graph

extends to the right. So, the distribution is skewed right.

b. Because the distribution is skewed, use the median to

describe the center and the fi ve-number summary to

describe the variation.

c. The adults typically have more money in their pockets and

the amounts vary more.

12. a.

Food Court

Like Dislike Total

Adults 21

— 200

= 0.105 79

— 200

= 0.395 0.5

Teenagers 96

— 200

= 0.48 4 —

200 = 0.02 0.5

Total 0.585 0.415 1

Shop

pers

b.

Food Court

Like Dislike

Adults 0.105

— 0.585

≈ 0.18 0.395

— 0.415

≈ 0.95

Teenagers 0.48

— 0.585

≈ 0.82 0.02

— 0.415

≈ 0.05Shop

pers

13. Sample answer:

Perfect Attendance

Class

10

20

30

40

50

60

70

80

90

0

Nu

mb

er o

f st

ud

ents

Freshman Sophomore Junior Senior

A bar graph shows data in specifi c categories.

14. Heights are numerical measurements. So, the data are

quantitative.

15. Grade levels are numerical, but they are labels and you

cannot measure them. So, the data are qualitative.

Chapter 11 Test (p. 627)

1. Because the data on the right of the distribution are

approximately a mirror image of the data on the left, the

distribution is symmetric. So, use the mean to describe

the center and use the standard deviation to describe the

variation.

2. Because most of the data are on the left and the tail of the

graph extends to the right, the distribution is skewed right.

So, use the median to describe the center and use the

fi ve-number summary to describe the variation.

3. Because most of the data are on the right and the tail of the

graph extends to the left, the distribution is skewed left.

So, use the median to describe the center and use the fi ve-

number summary to describe the variation.

4. a. The statement is always true because the marginal

frequencies in the “total” row and the “total” column each

represent the total population.

b. The statement is sometimes true because the distribution

of the data may not be symmetric.

c. The statement is sometimes true because qualitative data

are not typically numerical, but sometimes it is for data

such as jersey numbers or telephone numbers, which are

labels, not counts or measurements.


Chapter 11

5. Mean: — x = 15.5 + 7.8 + 18.9 + . . . + 9.75 + 12.25 + 21.7 —————

8

= 120 —

8 = 15

Median: 7.8, 9.75, 10.6, 12.25, 15.5, 18.9, 21.7, 23.5

The middle number is 12.25 + 15.5 ——

2 = 27.75

— 2 = 13.875.

Mode: 7.8, 9.75, 10.6, 12.25, 15.5, 18.9, 21.7, 23.5

The data each occur once.

Range: 23.5 − 7.8 = 15.7

Standard deviation

x — x x − — x (x − — x )2

15.5 15 0.5 0.25

7.8 15 −7.2 51.84

18.9 15 3.9 15.21

23.5 15 8.5 72.25

10.6 15 −4.4 19.36

9.75 15 −5.25 27.5625

12.25 15 −2.75 7.5625

21.7 15 6.7 44.89

0.25 + 51.84 + 15.21 + . . . + 27.5625 + 7.5625 + 44.89

————— 8

= 238.925 —

8 = 29.865625

√—

238.925

— 8 ≈ 5.46

The mean is $15.00, the median is $13.88, the data set has

no mode, the range is $15.70, and the standard deviation is

$5.46.

6. Mean: — x = 15(1 − 0.20) = 15(0.8) = 12

Median: 13.875(1 − 0.20) = 13.875(0.8) = 11.1

Mode: Each data value occurs once.

Range: 15.7(1 − 0.20) = 15.7(0.8) = 12.56

Standard deviation: 5.46(1 − 0.20) = 5.46(0.8) = 4.37

The mean is $12.00, the median is $11.10, the data set has

no mode, the range is $12.56, and the standard deviation is

$4.37.

7. Most of the data items have the same stem of 1. So, a

histogram would best represent the data because it would

show the distribution better.

8. a.

Brand A:

8.5, 11.5, 13.5, 13.5, 14.5, 15.5, 16.25, 16.75, 18.5, 20.75

least value

fi rst quartile

lower half ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

median

upper half ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

third quartile

greatest value

Least value: 8.5

First quartile: 13.5

Median: 14.5 + 15.5 —

2 = 30

— 2 = 15

Third quartile: 16.75

Greatest value: 20.75

Brand B:

7.0, 8.5, 8.5, 9.0, 9.0, 9.5, 9.75, 10.25, 10.5, 12.5

least value

fi rst quartile

lower half ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

median

upper half ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

third quartile

greatest value

Least value: 7.0

First quartile: 8.5

Median: 9.0 + 9.5 —

2 = 18.5

— 2 = 9.25

Third quartile: 10.25

Greatest value: 12.5

6 8 10 12 14 16 18 20 22Battery life(hours)

Brand A

Brand B

b. For brand A, the box and the whisker on the right are

each approximately the same length as the box and the

whisker, respectively, on the left. So, the distribution is

approximately symmetric. For brand B, the box and the

whisker on the right are each longer than the box and the

whisker, respectively, on the left. So, the data are more

concentrated on the left, and the distribution is skewed

right.

c. The range, 20.75 − 8.5 = 12.25, and interquartile range,

16.75 − 13.5 = 3.25, of brand A are greater than the

range, 12.5 − 7.0 = 5.5, and interquartile range,

10.25 − 8.5 = 1.75, respectively, of brand B. So, brand

A’s battery lives are more spread out.


Chapter 11

d. Because the distribution of the data for brand A is

approximately symmetric, the mean best represents the

center, and because the distribution of the data for brand B is

skewed right, the median best represents the center. The mean

of the battery lives for brand A is — x = 149.25

— 10

= 14.925,

while the median of the battery lives for Brand B is 9.25. So,

the battery life for brand A is typically longer than the battery

life for brand B.

Also, by looking at the box-and-whisker plot, you can see

that the data for brand A is more variable than the data for

brand B. This means that the battery life for brand A tends

to differ more from one battery to the next.

9. Sample answer:

Exercise

Preferred method of exercise

5

10

15

20

25

30

35

0Nu

mb

er o

f st

ud

ents

Walking Jogging Biking Swimming Liftingweights

Dancing

A bar graph shows data in specifi c categories.

10. a.

Attending Field Trip

Yes No Total

Male 92 29 121

Female 119 31 150

Total 211 60 271

Gen

der

So, 211 students are attending the fi eld trip, 60 students

are not attending the fi eld trip, 121 males responded to

the survey, 150 females responded to the survey, and

271 students were surveyed.

b.

Attending Field Trip

Yes No

Male 92

— 121

≈ 0.760 29

— 121

≈ 0.240

Female 119

— 150

≈ 0.793 31

— 150

≈ 0.217Gen

der

So, about 79.3% of females are attending the class fi eld trip.

Chapter 11 Standards Assessment (pp. 628–629)

1.

Cell Phones

Yes No Total

Male 27 12 39

Female 31 17 48

Total 58 29 87

Gen

der

Cell Phones

Yes No

Male 27

— 39

≈ 0.69 12

— 39

≈ 0.31

Female 31

— 48

≈ 0.65 17

— 48

≈ 0.35Gen

der

So, about 69% of males and about 65% of females have cell

phones. So, your friend is correct that a greater percent of

males in your grade have cell phones than females.

2. a. Because f (x) is quadratic and h(x) is linear, eventually f (x)

increases more rapidly than h(x). So, when x > 2, f (x) > h(x).

b. Because g(x) is exponential and f (x) is quadratic,

eventually g(x) increases more rapidly than f (x). So, when

x > 5, g(x) > f (x).

c. Because g(x) is exponential and h(x) is linear, eventually

g(x) increases more rapidly than h(x). So, when x > 3,

g(x) > h(x).

3. a. The box and the whisker on the right are each longer than

the box and the whisker, respectively, on the left. So, the

data are more concentrated on the left and the distribution

is skewed right.

b. The box and the whisker on the right are each approximately

the same length as the box and the whisker, respectively, on

the left. So, the distribution is approximately symmetric.

c. Most of the data are on the right and the tail of the graph


d. Most of the data are on the left and the tail of the graph

extends to the right. So, the distribution is skewed right.


Chapter 11

4. The slope of the equation is

m = y2 − y1

— x2 − x1 = −5 − 4

— 1− (−2)

= −5−4 —

1 + 2 = −9

— 3 , or −3.

y = mx + b

y = −3x + b

4 = −3(−2) + b

4 = 6 + b

−6 −6

−2 = b

Check y = −3x − 2 y = −3x − 2

−5 =?

−3(1) − 2 4 =?

−3(−2) − 2

−5 =?

−3 − 2 4 =?

6 − 2

−5 = −5 ✓ 4 = 4 ✓

y = 2x2 − x − 6 y = 2x2 − x − 6

−5 =?

2(1)2 − 1 − 6 4 =?

2(−2)2 − (−2) − 6

−5 =?

2(1) − 1 − 6 4 =?

2(4) + 2 − 6

−5 =?

2 − 1 − 6 4 =?

8 + 2 − 6

−5 =?

1 − 6 4 =?

10 − 6

−5 = −5 ✓ 4 = 4 ✓

So, the equation is y = −3x − 2.

5. y = −3x2, for x ≥ 0

x = −3y2

x — −3

= −3y2

— −3

− 1 —

3 x = y2

± √— − 1 —

3 x = √

— y2

± √— − 1 —

3 x = y, for y ≥ 0

So, y = −3x2 and y = √— − 1 —

3 x are inverse functions.

y = 1 — 2 x + 2

x = 1 — 2 y + 2

x − 2 = 1 — 2 y + 2 − 2

x − 2 = 1 —

2 y

2(x − 2) = 2 ( 1 — 2 y )

2(x) − 2(2) = y

2x − 4 = y

So, y = 1 — 2 x + 2 and y = 2x − 4 are inverse functions.

y = −x + 7

x = −y + 7

x − 7 = −y + 7 − 7

x − 7 = −y

−1(x − 7) = −1(−y)

−1(x) − 1(−7) = y

−x + 7 = y

So, y = −x + 7 and y = −x + 7 are inverse functions.

y = x2 − 5, for x ≥ 0

x = y2 − 5 x + 5 = y2 − 5 + 5

x + 5 = y2

± √—

x + 5 = √—

y2

± √—

x + 5 = y, for y ≥ 0

So, y = x2 − 5 and y = √—

x + 5 are inverse functions.

6. C; Q3 − Q1 = 7 − 4 = 3

So, the middle half of the presentation lengths vary by no

more than 3 minutes.


Chapter 11

7. a. — x = 36 + 48 + 42 + 63 + 28 + x + 57 + 52

———— 8 = 45.5

x + 326

— 8 = 45.5

8 ⋅ x + 326 —

8 = 8 ⋅ 45.5

x + 326 = 364

−326 −326

x = 38

So, when x = 38, the mean of the scores is 45.5.

b. 28, 36, 42, x, 48, 52, 57, 63

The mean of the two middle numbers is:

x + 48

— 2 = 47

2 ⋅ x + 48 —

2 = 2 ⋅ 47

x + 48 = 94

−48 −48

x = 46

So, when x = 46, the median of the scores is 47.

c. 20, 36, 42, 48, 52, 57, 63, x

When x = 63, the mode of the scores is 63.

d. 28, 36, 42, 48, 52, 57, 63, x

x − 28 = 71

+28 +28

x = 99

So, when x = 99, the range of the scores is 71.

8. Graph the function

y =

x2 + 4x + 7, if x ≤ −1

1 —

2 x + 2, if x > −1

x −4 −3 −2 −1 −1 0 1 2 3 4

y 7 4 3 4 1.5 2 2.5 3 3.5 4

closed

dot

opendot

x

y

4

6

8

42−2−4

So, the range of the function is y > 1.5. Of the choices, the

numbers that are in the range of the function are 2, 2 1 —

2 , 3, 3

1 —

2 ,

and 4.

9. a. The value of D(2) appears to be about 500 feet. This

means that the traveler has walked about 500 feet from the

starting point after 2 minutes.

b. From the graph D(15) = 3500. So, x = 15. It takes the

traveler 15 minutes to travel 3500 feet.

c. There is no change in y, the travelers distance between

(4, 1000) and (12, 1000). So, the traveler is waiting for the

shuttle bus for 12 − 4 = 8 minutes.

d. The traveler is on the shuttle bus between (12, 1000)

and (13, 3000). So, the traveler rides

3000 − 1000 = 2000 feet.

e. Before riding the shuttle bus, the traveler walks

1000 − 0 = 1000 feet. After riding the shuttle bus, the

traveler walks 3500 − 3000 = 500 feet. So, the traveler

walks a total of 1000 + 500 = 1500 feet.

⎧⎨⎩

chapter 11€¦ · 7.175 6 1.2 √ — 7.175 — 6 ≈ 1.09 the standard deviation of the old data...

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