Chapter 3 Solutions 3.1 (a) 87% of the girls her daughter’s age weigh the same or less than she does

and 67% of girls her daughter’s age are her height or shorter. (b) According to the Los Angeles Times, the speed limits on California highways

are such that 85% of the vehicle speeds on those stretches of road are no greater than the speed limit.

3.2 (a) Peter’s time was better than 80% of his previous race times that season, but

it was better than only 50% of the racers at the league championship meet. (b) Although Molly’s league swim meet time was better than only 50% of her

meet times for the season, it was better than 80% of the other participants in the league swim meet.

3.3 (a) Francine’s bone density is 1.45 standard deviations below the mean hip bone

density for 25-year-old women of 956 grams/cm2. (b)

−1.45 =948 − 956

standard deviation⇒ standard devation = -8

-1.45= 5.517. The standard

deviation of the reference population is 5.517 grams/cm2. 3.4 (a) For their respective ages, Louise has the healthier bone hip density since her

bone density is 0.5 standard deviations above the mean bone hip density for her age group while Francine’s was 1.45 standard deviations below the average for her age group.

(b) 0.50 =948 − 944

standard deviation⇒ standard deviation = 4

0.5= 8. The result is logical since

we might have expected the variability in bone density values to increase as women age.

3.5 (a) Since 23 of the salaries are less than or equal to Lidge’s salary, his salary is

at the 2329

= 79.3 percentile. Roughly 79% of the team had salaries less than

or equal to Lidge’s salary.

(b) z =6,350,000 − 3,388,617

3,767,484= 0.79. Lidge’s salary was 0.79 standard deviations

above the mean salary of $3,388,617. 3.6 Madson’s salary in 2008 was $1,400,000. Since 9 of the salaries were less

than or equal to Madson’s salary, his salary percentile is 9

29= 31.0. Madson’s

Chapter 3 33

z-score is z =1,400,000 − 3,388,617

3,767,484= −0.53. Madson had a low salary compared

to the rest of the team since his percentile is less than 50% and his z-score is negative, indicating a salary below the mean team salary.

3.7 (a) The total area under the curve must equal 1. Since the base of the rectangle

is 1, the height must also be 1. (b) The balance point (mean) is 0.5. (c) The median is also 0.5 . It is the same as the mean since the density curve

is symmetric. (d) (0.4)(1) = 0.4. 40%. 3.8 (a)

The height of the curve is ½ so that the total area of the curve is 1

(2*1/2=1).

(b) 1.4 −1

2= 0.2. 20% of the observations lie between 1 and 1.4.

(c) Q1 is the value such that 25% of the curve is less than this value:

0.25 =Q1

2⇒ Q1 = 0.5. The median and third quartile are found similarly:

0.5 =Median

2⇒ Median =1; 0.75 = Q3

2⇒ Q3 =1.5.

3.9 Answers will vary. One possibility:

The distribution is skewed to the right so that the median < mean. The median is approximately 3.5 and the mean is approximately 4.2.

Chapter 3 34

3.10 Answers will vary. One possibility:

The distribution is well approximated by a uniform distribution between 0 and

9 having mean=median=4.5. 3.11 (a) The overall shape of the distribution is symmetric and mound-shaped.

Hence, the mean and median are equal and are at point A. (b) The overall shape of the distribution is skewed to the left. The mean is at A

and the median is at B. 3.12

3.13 (a) The distribution is skewed to the right, thus the mean is to the right of the

median. Point “A” clearly does not mark the 50th percentile, so the median must be at “B” and the mean at “C”.

(b) The distribution is symmetric so the mean and the median both mark the 50th percentile of the distribution and are located at “A”.

(c) The distribution is skewed to the left, thus the mean is to the left of the median. Point “C” clearly does not mark the 50th percentile so the median must be at “B” and the mean at “A”.

3.14 It actually means that 90% of similar men have cholesterol levels that are

less than or equal to Martin’s level. Martin’s cholesterol is comparatively high, not low.

Chapter 3 35

3.15 (a)

(b) The mean of the 50 observations is calculated to be 5.598 percent. The

median is 5.45 percent. The standard deviation is 1.391 percent. The distribution of unemployment rates is fairly symmetric, with a center of about 5.5 percent and a spread of 5.6 percent.

(c) In the ordered data set, Illinois’s unemployment rate is 40th. Thus, the percentile for Illinois is 40/50=.8. 80% of the unemployment rates are less than or equal to Illinois’s unemployment rate. Illinois has a pretty high unemployment rate compared to the remaining 49 states.

(d) Since .4*50=20, the state whose unemployment rate is 20th among the ordered observations represents the 40th percentile. Texas represents the 40th percentile with an unemployment rate of 5.1%. The z-score for this

state is z =5.1− 5.598

1.391= −0.36.

3.16 (a) The unemployment rate for Illinois was 2.4% (6.9-4.5) higher in September,

2008 than in December, 2000. (b) The percentile for Illinois (its unemployment rate compared to the rest of the

country) did not change much from December of 2000 to September of 2008, it actually dropped slightly. It was at the 86th percentile in December of 2000 (43/50=0.86) versus the 80th percentile in September of 2008.

(c) The z-score for Illinois in December of 2000 was z =4.5 − 3.47

1=1.03. The

unemployment rate for Illinois was slightly higher in December of 2000 than in September of 2008 relative to the rest of the country. It was 1.03 standard deviations above the mean in December of 2000 whereas it was

0.94 z =6.9 − 5.598

1.391= 0.94

⎛ ⎝ ⎜

⎞ ⎠ ⎟ .standard deviations above the mean in September

of 2008.

Chapter 3 36

3.17 (a) The only way to obtain a z-score of 0 is if the x value equals the mean.

Thus, the mean is 170 cm. To find the standard deviation, use the fact that a z-score of 1 corresponds to a height of 177.5 cm. Then

1=177.5 −170

standard deviation⇒ standard deviation = 7.5 . Thus, the standard deviation of

the height distribution for 15-year-old-males is 7.5 cm.

(b) Since 2.5 =x −170

7.5, x = 2.5 * 7.5 +170 =188.75 cm. A height of 188.75 cm has a

z-score of 188.75 cm. 3.18 (a) Since a z-score of 0 corresponds to a height of 162.5, the mean of the

distribution of heights for 16-year-old females is 162.5. Since a z-score of 1 corresponds to a height of 169, we have

1=169 −162.5

standard deviation⇒ standard deviation = 6.5. The standard deviation of the

height distribution for 16-year-old females is 6.5 cm.

(b) Since −1.5 =x −162.5

6.5, x = −1.5 * 6.5 +162.5 =152.75 cm. A height of 152.75 cm

has a z-score of -1.5. 3.19

(a) z =175 −170

7.5= 0.67. Paul is slightly taller than average for his age. His height

is 0.67 standard deviations above the average male height for his age. (b) 75% of boys Paul’s age are shorter than or equal to Paul’s height. 3.20

(a) z =170 −162.5

6.5=1.15. Miranda is taller than average for her age. Her height is

1.15 standard deviations above the average female height for her age. (b) Among 16-year-old girls, 88% are Miranda’s height, 170 cm, or shorter. (c) Relative to her peers, Miranda is taller than Paul since her height is at the

88th percentile and Paul’s is only at the 75th. 3.21 111 ± 2(11) = (89, 133).

3.22 z100 = 100−111

11= −1. Since 68% of the scores are within one standard

deviation of the mean, 34% lie between the mean and z = −1. Thus, 50% + 34% = 84% are larger than 100. In our sample, 62 of the 74 scores are greater than 100 so the sample percentage, 62/74=84%, is the same.

3.23 144 has a z-score of 3 (111 + 3(11) = 144). We would expect half of 0.3%, or 0.15% to be above 144 (since 99.7% are within three standard deviations of the mean). It is not surprising that none of the students in the sample

Chapter 3 37

had an IQ score this large as we would only expect a value this large, on average, about 1.5 times out of every thousand.

3.24 The mean is 10 and the standard deviation is about 2. The entire range of

scores is about 16 – 4 = 12. Since the range of values in a normal distribution is about six standard deviations, we have 12/6 = 2 for σ.

3.25 (a) 336 ± 3(3) = (327, 345).

(b) z339 = 339−336

3= 1 ⇒ 16% of horse pregnancies last longer than 339 days.

This is because 68% are within one standard deviation of the mean, so 32% are more than one standard deviation from the mean and half of those are greater than 339.

3.26 (a) Since 2.1 is 1 standard deviation above the mean, 50% + 34% = 84% of

the cartons weigh less than 2.1 pounds (50% below the mean of 2 pounds and 34% between the mean and one standard deviation above the mean).

(b) Since 1.8 is 2 standard deviations below the mean, 2.5% of the cartons weigh less than 1.8 pounds (5% of the weights are more than 2 standard deviations from the mean with 2.5% being at least 2 standard deviations below and 2.5% being at least 2 standard deviations above).

(c) Since 1.9 is 1 standard deviation below the mean, 84% of the cartons weigh more than 1.9 pounds (50% above the mean of 2 pounds and 34% between the mean and 1.9 pounds).

3.27 (a)

The area to the left of z=-0.37 is 0.3557.

(b) The area to the right of z=-0.37

is 1-0.3557=0.6443.

Chapter 3 38

(c)

The area to the left of z=2.15 is

0.9842. (d) The area to the right of z=2.15

is 1-0.9842=0.0158. 3.28 (a) The area to the left of z=-1.58

is 0.0571. (b) The area to the right of z=1.58

is 1-0.9429=0.0571. (c) The area to the right of z=-0.46

is 1-0.3228=0.6772.

Chapter 3 39

(d) The area to the left of z=0.93 is

0.8238. 3.29 (a) The area between z= -1.33 and

z=1.65 is 0.9505-0.0918 =0.8587.

(b) The area between z =0.50 and

z=1.79 is 0.9633-0.6915 =0.2718.

3.30 (a) The area between z=-2.05 and

z=0.78 is 0.7823-0.0202 =0.7621.

(b) The area between z=-1.11 and

z=-0.32 is 0.3745-0.1335 = 0.241.

Chapter 3 40

3.31 (a) The area to the left of z is 0.20. The value of z with 20% of the

area to the left is z=-0.84. (b) The value of z with 55% (1-.45) of the observations less than z The area to the is approximately z=0.13. right of z is 0.45. 3.32 (a) The area to the left of z is 0.63. The value of z with 63% of the

observations less than z is 0.33. (b) The value of z with 25% (1-.75) of the observations less than z is The area to the -0.67. right of z is 0.75. 3.33 Step 1: State the Problem. Let x=the distance that Tiger’s ball travels. The

variable x has a Normal distribution with μ=304 and σ=8. Assuming Tiger is safe hitting his driver as long as the ball lands before the creek, we want to find the probability that x ≤ 317. Step 2: Standardize and draw a picture. For x=317, we have

z =x − μ

σ=

317 − 3048

=1.63.

So, x ≤ 317 corresponds to z ≤ 1.63 under the standard Normal curve.

Chapter 3 41

Step 3: Use the table. From Table A, we see that the proportion of

observations less than 1.63 is 0.9484. The probability that Tiger is safe hitting his driver is about 0.95. Step 4: Conclusion. Since the probability that Tiger’s drive misses the creek is about 0.95, we can conclude that Tiger is safe hitting his driver.

3.34 Step 1: State the problem and draw a picture. Let x=the score on the

Wechsler Adult Intelligence Scale of a randomly selected adult aged 20 to 34. The variable x has a Normal distribution with μ=110 and σ=25. We want to find the score that separates the top 25% from the bottom 75% of scores.

Area = 0.25 Step 2: Use the table. From Table A, we see that the value of z with an

area of 0.75 to its left is 0.67. Step 3: “Unstandardize”. We know that the standardized value of the unknown x is z=0.67. We find x by solving the following equation:

0.67 =x −110

25⇒ x = 0.67 * 25 +110 =126.75.

Step 4: Conclusion. A score of 127 puts a person in the top 25% of all scores.

3.35 Step 1: State the Problem. Let x=a randomly selected female’s score on the

SAT Math test. The variable x has a Normal distribution with μ=500 and σ=111. We want to find the probability that x > 533, the male mean. Step 2: Standardize and draw a picture. For x=533, we have

z =x − μ

σ=

533 − 500111

= 0.30.

So, x > 533 corresponds to z > 0.30 under the standard Normal curve.

Chapter 3 42

Step 3: Use the table. From Table A, we see that the proportion of

observations less than 0.30 is 0.6179. Therefore, the proportion of observations greater than 0.30 is 1-0.6179 = 0.3821. Step 4: Conclusion. Approximately 38% of females scored higher than the male mean.

3.36 (a) Step 1: State the problem and draw a picture. Let x=the score on the SAT

Math test of a randomly selected male. The variable x has a Normal distribution with μ=533 and σ=118. We want to find the score that separates the top 15% from the bottom 85% of scores.

Area = 0.85

Step 2: Use the table. From Table A, we see that the value of z with an

area of 0.85 to its left is 1.04. Step 3: “Unstandardize”. We know that the standardized value of the unknown x is z=1.04. We find x by solving the following equation:

1.04 =x − 533

118⇒ x =1.04 *118 + 533 = 655.72

Step 4: Conclusion. The 85th percentile of the SAT Math score distribution for males is 655.7.

(b) Step 1: State the Problem. Let x=a randomly selected female’s score on the SAT Math test. The variable x has a Normal distribution with μ=500 and σ=111. We want to determine the percentile corresponding to a value of 655.7. Step 2: Standardize and draw a picture. For x=655.7, we have

z =x − μ

σ=

655.7 − 500111

=1.40.

To find the percentile corresponding to x = 655.7, we find the area under the standard Normal curve corresponding to z < 1.40.

Chapter 3 43

Step 3: Use the table. From Table A, we see that the proportion of observations less than 1.40 is 0.9192. Step 4: Conclusion. The 85th percentile for the SAT Math score distribution for males corresponds to the 92nd percentile for the SAT Math score distribution for females.

3.37 (a) Step 1: State the Problem. Let x=the weight of a randomly selected bag of

potatoes from the shipment. The variable x has a Normal distribution with μ=10 and σ=0.5. We want to determine the percent of bags weighing less than 10.25 pounds. Step 2: Standardize and draw a picture. For x=10.25, we have

z =x − μ

σ=

10.25 −100.5

= 0.5.

To find the percent of bags weighing less than 10.25 pounds, we find the area under the standard Normal curve corresponding to z < 0.5.

Step 3: Use the table. From Table A, we see that the percent of

observations less than 0.5 is 0.6915. Step 4: Conclusion. Approximately 69% of the bags in the shipment weighed less than 10.25 pounds.

(b) Step 1: State the Problem. Let x=the weight of a randomly selected bag of potatoes from the shipment. The variable x has a Normal distribution with μ=10 and σ=0.5. We want to determine the percent of bags weighing between 9.5 and 10.25 pounds. Step 2: Standardize and draw a picture. From part (a), we know that the percent of bags weighing less than 10.25 pounds corresponds to P(z<0.5). For x=9.5, we have

z =x − μ

σ=

9.5 −100.5

= −1.

Thus, the area we would like to find is the area under the standard Normal curve between -1 and 0.5.

To calculate this area, we need to find the percent of bags weighing less than

10.25 pounds and subtract from this the percent of bags weighing less than

Chapter 3 44

9.5 pounds. From part (a), we have P(x<10.25)=P(z<0.5) = 0.6915. To find the percent of bags weighing less than 9.5 pounds, we find the area under the standard Normal curve corresponding to z < -1.

Step 3: Use the table. From Table A, we see that the percent of

observations less than -1 is 0.1587. Thus, the percent of bags weighing between 9.5 and 10.25 pounds is 0.6915-0.1587=0.5328. Step 4: Conclusion. Approximately 53% of the bags in the shipment weighed between 9.5 and 10.25 pounds.

3.38 We want to find μ so that 40% of the distribution is less than 60 seconds.

Area = 0.4 The z value with an area of 0.4 to the left is -0.25. We can then find μ as

follows:

−0.25 =60 − μ

20⇒ μ = 60 + 0.25 * 20 = 65.

The mean of the distribution is 65 seconds.

Chapter 3 45

3.39

The picture shows that 68% of women are between 62.5 and 67.5 inches tall, 95% are between 60 and 70 inches tall, and 99.7% are between 57.5 and 72.5 inches tall.

3.40 The mean is 28 and the standard deviation is about 1. Most of the values are

between 25 and 31 which, in a normal distribution, would be about six standard deviations. Since the range is about six, the standard deviation must be about 1.

3.41 In a normal distribution, 68% of the values are within one standard deviation

of the mean. Since the distribution is symmetric, 34% of these values must be below the mean. That leaves 16% of values that fall more than one standard deviation below the mean. Since the percentile rank of an observation is the percent of observations below it, the percentile rank of an observation one standard deviation below the mean is 16. Similarly, 95% of the observations are within two standard deviations of the mean, leaving 97.5% below a value that is two standard deviations above the mean.

3.42 (a) The area to the right of z=-1.81

is 1- 0.0351=0.9649. (b)

The area to the left of z=2.29 is 0.9890.

57.5 60 62.5 65 67.5 70 72.5

68%

99.7%

95%

Chapter 3 46

(c) The area between z=-1.81 and

z= -0.47 is 0.3192-0.0351 = 0.2841.

(d) The area between z=-1.02 and

z = 0.65 is 0.7422–0.1539 = 0.5883.

3.43 (a) The area to the left of z is 32%. The value of z with an area of

0.32 to the left is z=-0.47. (b) The area to the right The value of z with an area of of z is 0.4. 0.6 (1-0.4) to the left is z=0.25. (c) The area to the left of z is 0.10. The value of z with an area of

0.10 to the left is z=-1.28.

Chapter 3 47

(d)

The area to the of z is 0.83. The value of z with an area of

0.83 to the left is z=0.95.

3.44 z =820 −1017

211= − 0.93. From the calculator [normalcdf(−1000,−0.93)] we

see that about 17.6% of the scores were less than 820. 3.45 (a) A score of 130 is two standard deviations above the mean. From the

calculator [normalcdf(2,100)] we see that about 2.28% of children had very superior scores in 1932.

(b) Now we have a distribution with a mean of 120 and a standard deviation of

15. z =130 −120

15= 0.67 ⇒ about 25.1% [normalcdf(0.67,100)] would

have very superior scores. 3.46 We want to find the score such that 98% of the scores are less than or equal

to its value.

Area to the left of z is 0.98. The z value with an area of 0.98 to the left is 2.05. To find the

corresponding IQ score, we solve the following:

2.05 =x −100

15⇒ x = 2.05 *15 +100 =130.75.

3.47 (a) $215 and $395 represent one standard deviation above and below the mean

of $305. Thus, about 68% spent an amount in this range.

(b) z =287 − 305

90= − 0.2 ⇒ about 42% [normalcdf(−100,-.2)] spent less than

$287 on textbooks.

Chapter 3 48

3.48 (a) Joey’s score on the national standardized math test was as high or higher

than 97% of students who took the test. His score on the reading portion of the test was as high or higher than 72% of students who took the test.

(b) Only if we assume that the distribution of scores nationally follows a Normal distribution. If so, his scores can be found by determining the z values that represent the 97th and 72nd percentiles of the standard Normal distribution, 1.88 and 0.58, respectively.

3.49 (a) The area to the right of z=-2.25

is 1-0.0122=0.9878. (b) The area between z=-2.25 and

z=1.77 is 0.9616-0.0122 = 0.9494.

(c) The area to the left of z=2.89 is

0.9981. (d) The area between z=1.44 and

z=2.89 is 0.9981-0.9251 = 0.0730.

Chapter 3 49

3.50 (a) The area to the left of z is 0.70. The z value with an area of 0.7

to the left is 0.52. (b) The area to the right The z value with an area of z is 0.90. of 0.1 (1-0.9) to the left is -1.28. (c) The area to the left of z is 0.46. The z value with an area of 0.46

to the left is -0.10. 3.51

(a) z =2500 − 3668

511= −2.29. The z value that corresponds to a baby weighing less

than 2500 grams at birth is -2.29. The percent of babies weighing less than this is the area to the left, 0.01. 1% of babies will be identified as low birth weight.

(b) The z-values corresponding to the quartiles (25th percentile, median, 75th percentile) are -0.67, 0 and 0.67. The median is equal to the mean of the distribution, 3668. To find the x values corresponding to the 25th and 75th percentiles, we can solve the z equations for x as follows:

−0.67 =x − 3668

511⇒ x = −0.67 * 511+ 3668 = 3325.63;

0.67 =x − 3668

511⇒ x = 0.67 * 511+ 3668 = 4010.37.

The quartiles of the birth weight distribution are 3325.6, 3668 and 4010.4 grams.

Chapter 3 50

3.52 (a)

Based on the histogram, the distribution of coliform counts appears to be

approximately normal. (b) The mean is calculated to be 5.88 and the standard deviation is calculated to

be 2.026. (c) Observations between 3.854 and 7.906 are within one standard deviation of

the mean. 65 out of 100, or 65%, of the observations are within one standard deviation of the mean. Observations between 1.828 and 9.932 are within two standard deviations of the mean. 95 out of 100, or 95%, of the observations are within two standard deviations of the mean. Observations between -0.198 and 11.958 are within three standard deviations of the mean. All of the observations, or 100%, are within three standard deviations of the mean. Note how closely this follows the 68-95-99.7 rule.

3.53 (a) The overall shape of the distribution is bi-modal. Because of the symmetry

of the curve, the mean and the median are the same and both are equal to B.

(b) The overall shape of the distribution is skewed to the right. The mean is pulled toward the tail of the distribution. Accordingly, the median is at A and the mean is at B.

3.54 We want to find the head sizes that correspond to the 5th and 95th percentiles

of the distribution. The corresponding z values are -1.645 and 1.645. We solve for the x values as follows:

−1.645 =x − 22.8

1.1⇒ x = −1.645 *1.1+ 22.8 = 20.99

1.645 =x − 22.8

1.1⇒ x =1.645 *1.1+ 22.8 = 24.61.

Head sizes that are less than 20.99 inches or greater than 24.61 inches get custom-made helmets.

Chapter 3

51

3.55 (a) The middle 95% of all yearly returns lie within 2 standard deviations of the

mean or between 12-2*16.5=-21 and 12+2*16.5=45. The middle 95% of all yearly returns are between -21% and 45%.

(b) z =0 −1216.5

= −0.73.

The area to the left of z=-0.73 is 0.2327. The market is down for about 23% of the years.

(c) z =15 −12

16.5= 0.18 and z =

25 −1216.5

= 0.79.

The area between z =0.18 and z=0.79 is 0.7852-0.5714=0.2138. 3.56 (a) The density curve is skewed to the right. This makes sense because you

would expect most people to play the game in a relatively short amount of time with some taking much longer (they aren’t as good at the game or they must leave their computer for some reason).

(b) The author scored in the top 19%. Thus, the time it took the author to play the game was less than 81% of all of the game times for the week.