Content check — Statistics, Chapter 6: The Normal Distribution

6-1 §6.1 source ↗

A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________.

Answer

ounces of water in a bottle

6-3 §6.1 source ↗

X ~ N(1, 2)

σ = _______

Answer

2

6-5 §6.1 source ↗

X ~ N(–4, 1)

What is the median?

Answer

–4

6-7 §6.1 source ↗

X ~ N(–2, 1)

μ = _______

Answer

–2

6-9 §6.1 source ↗

What does standardizing a normal distribution do to the mean?

Answer

The mean becomes zero.

6-11 §6.1 source ↗

What is the z-score of x = 12, if it is two standard deviations to the right of the mean?

Answer

z = 2

6-13 §6.1 source ↗

What is the z-score of x = –2, if it is 2.78 standard deviations to the right of the mean?

Answer

z = 2.78

6-15 §6.1 source ↗

Suppose X ~ N(2, 6). What value of x has a z-score of three?

Answer

x = 20

6-17 §6.1 source ↗

Suppose X ~ N(9, 5). What value of x has a z-score of –0.5?

Answer

x = 6.5

6-19 §6.1 source ↗

Suppose X ~ N(4, 2). What value of x is 1.5 standard deviations to the left of the mean?

Answer

x = 1

6-21 §6.1 source ↗

Suppose X ~ N(8, 9). What value of x is 0.67 standard deviations to the left of the mean?

Answer

x = 1.97

6-23 §6.1 source ↗

Suppose X ~ N(12, 6). What is the z-score of x = 2?

Answer

z = –1.67

6-25 §6.1 source ↗

Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z-score of x = 5.5?

Answer

z ≈ –0.33

6-27 §6.1 source ↗

In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean.

Answer

0.67, right

6-29 §6.1 source ↗

In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean.

Answer

3.14, left

6-31 §6.1 source ↗

About what percent of x values from a normal distribution lie within one standard deviation, left and right, of the mean of that distribution?

Answer

about 68 percent

6-33 §6.1 source ↗

About what percent of x values lie between the second and third standard deviations, both sides?

Answer

about 4 percent

6-35 §6.1 source ↗

Suppose X ~ N(–3, 1). Between what x values does 95.45 percent of the data lie? The range of x values is centered at the mean of the distribution (i.e., –3).

Answer

between –5 and –1

6-37 §6.1 source ↗

About what percent of x values lie between the mean and three standard deviations?

Answer

about 50 percent

6-39 §6.1 source ↗

About what percent of x values lie between the first and second standard deviations from the mean, both sides?

Answer

about 27 percent

6-41 §6.1 source ↗

Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts.

Define the random variable X in words. X = _______________.

Answer

The lifetime of a Sunshine CD player measured in years

6-43 §6.2 source ↗

How would you represent the area to the left of one in a probability statement?

A graph showing a bell shaped curve of normal distribution with a vertical line to the right of center labeled with a 3. The axes are unlabeled. — Figure 6.12 in the book ↗

Answer

P(x < 1)

6-45 §6.2 source ↗

Is P(x < 1) equal to P(x ≤ 1)? Why or why not?

Answer

Yes, because they are the same in a continuous distribution: P(x = 1) = 0

6-47 §6.2 source ↗

What is the area to the right of three?

Answer

1 – P(x < 3) or P(x > 3)

6-49 §6.2 source ↗

If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x?

Answer

1 – 0.543 = 0.457

6-51 §6.2 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

Find the probability that x < 30.

Answer

0.0013

6-53 §6.2 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

Find the 60^th percentile.

Answer

56.03

6-55 §6.2 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

X ~ N(–3, 4)

Find the probability that x is between one and four.

Answer

0.1186

6-57 §6.2 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

Use the following information to answer the next three exercises: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.

a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.

b. P(0 < x < ____________) = ___________. Use zero for the minimum value of x.

Empty normal distribution curve. — Figure 6.16 in the book ↗

Answer

a. Check student’s solution

b. 3, 0.1979

6-59 §6.2 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

Find the 70^th percentile of the distribution for the time a CD player lasts.

a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70 percent.

b. P(x < k) = __________. Therefore, k = _________.

Answer

a. Check student’s solution

b. 0.70, 4.78 years

6-61 §6.1 source ↗

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

What is the z-score for a patient who takes 10 days to recover?

a. 1.5

b. 0.2

c. 2.2

d. 7.3

Answer

c

6-63 §6.1 source ↗

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

The heights of the 430 basketball players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with a mean, µ = 79 inches, and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences:

a. 77 inches

b. 85 inches

c. If a player reported his height had a z-score of 3.5, would you believe him? Explain your answer.

Answer

a. Use the z-score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average.

b. Use the z-score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average.

c. Height = 79 + 3.5(3.89) = 90.67 inches, which is over 7.7 feet tall. There are very few NBA players this tall; so, the answer is no, not likely.

6-65 §6.1 source ↗

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

Kyle’s doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure, given in millimeters, of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. If X = a systolic blood pressure score, then X ~ N (125, 14).

a. Which answer(s) is/are correct?

i. Kyle’s systolic blood pressure is 175.

ii. Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age.

iii. Kyle’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age.

iv. Kyles’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men.

b. Calculate Kyle’s blood pressure.

Answer

a. iv

b. Kyle’s blood pressure is equal to 125 + (1.75)(14) = 149.5.

6-67 §6.1 source ↗

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

In 2005, 1,475,623 students heading to college took the SAT exam. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard deviation σ = 115.

a. Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence.

b. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score?

c. For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternative to the SAT math test, and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test that each person took?

Answer

Let X = an SAT math score and Y = an ACT math score.

a. X = 720 $\frac{720 – 520}{15}$ = 1.74 The exam score of 720 is 1.74 standard deviations above the mean of 520.

b. z = 1.5
The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520.

c. $\frac{X – μ}{σ}$ = $\frac{700 – 514}{117}$ ≈ 1.59, the z-score for the SAT. $\frac{Y – μ}{σ}$ = $\frac{30 – 21}{5.3}$ ≈ 1.70, the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better—has the higher z-score).

6-69 §6.2 source ↗

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

The 90^th percentile for recovery times is –

a. 8.89

b. 7.07

c. 7.99

d. 4.32

Answer

c

6-71 §6.2 source ↗

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

Find the probability that it takes at least eight minutes to find a parking space.

a. 0.0001

b. 0.9270

c. 0.1862

d. 0.0668

Answer

d

6-73 §6.2 source ↗

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.

a. X ~ _____ (_____ ,_____)

b. Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement.

c. Would you expect to meet many Asian adult males taller than 72 inches? Explain why or why not, and numerically justify your answer.

d. The middle 40 percent of heights fall between what two values? Sketch the graph, and write the probability statement.

Answer

a. X ~ N(66, 2.5)

b. 0.5404

c. No, the probability that an Asian male is over 72 inches tall is 0.0082.

6-75 §6.2 source ↗

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

The percent of fat calories that a person in the United States consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percentage of fat calories.

a. X ~ _____ (_____, _____)

b. Find the probability that the percentage of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined.

c. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement.

Answer

a. X ~ N(36, 10)

b. The probability that a person consumes more than 40 percent of their calories as fat is 0.3446.

c. Approximately 25 percent of people consume less than 29.26 percent of their calories as fat.

6-77 §6.2 source ↗

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time that child spends alone per day.

a. In words, define the random variable X.

b. X ~ _____ (_____, _____)

c. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement.

d. What percentage of the children spend more than 10 hours per day unsupervised?

e. Seventy percent of the children spend at least how long per day unsupervised?

Answer

a. X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.

b. X ~ N(3, 1.5)

c. The probability that the child spends less than one hour a day unsupervised is 0.0918.

d. The probability that a child spends over 10 hours a day unsupervised is less than 0.0001.

e. 2.21 hours

6-79 §6.2 source ↗

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days.

a. In words, define the random variable X.

b. X ~ _____ (_____, _____)

c. If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement.

d. Sixty percent of all trials of this type are completed within how many days?

Answer

a. X = the distribution of the number of days a particular type of criminal trial will take

b. X ~ N(21, 7)

c. The probability that a randomly selected trial will last more than 24 days is 0.3336.

d. 22.77

6-81 §6.2 source ↗

Table 6.3 in the book ↗

0.50	4.25	5	6	7.25
1.75	4.25	5.25	6	7.25
2	4.25	5.25	6.25	7.25
2.25	4.25	5.5	6.25	7.75
2.25	4.5	5.5	6.5	8
2.5	4.75	5.5	6.5	8.25
2.75	4.75	5.75	6.5	9.5
3.25	4.75	5.75	6.75	9.5
3.75	5	6	6.75	9.75
3.75	5	6	6.75	10.75

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. Table 6.3 displays the ordered real data, in minutes.

a. Calculate the sample mean and the sample standard deviation.

b. Construct a histogram.

c. Draw a smooth curve through the midpoints of the tops of the bars.

d. In words, describe the shape of your histogram and smooth curve.

e. Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by X ~ _____ (_____, _____)

f. Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes.

g. Determine the cumulative relative frequency for waiting less than 6.1 minutes.

h. Why aren’t the answers to part f and part g exactly the same?

i. Why are the answers to part f and part g as close as they are?

j. If only 10 customers were surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion.

Answer

a. mean = 5.51, s = 2.15

b. Check student's solution.

c. Check student's solution.

d. Check student's solution.

e. X ~ N(5.51, 2.15)

f. 0.6029

g. The cumulative frequency for less than 6.1 minutes is 0.64.

h. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one.

i. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30.

j. The approximation would have been less accurate, because the smaller sample size means that the data does not fit a normal curve as well.

6-83 §6.2 source ↗

Table 6.4 in the book ↗

40,000	40,000	45,050	45,500	46,249	48,134
49,133	50,071	50,096	50,466	50,832	51,100
51,500	51,900	52,000	52,132	52,200	52,530
52,692	53,864	54,000	55,000	55,000	55,000
55,000	55,000	55,000	55,082	57,000	58,008
59,680	60,000	60,000	60,492	60,580	62,380
62,872	64,035	65,000	65,050	65,647	66,000
66,161	67,428	68,349	68,976	69,372	70,107
70,585	71,594	72,000	72,922	73,379	74,500
75,025	76,212	78,000	80,000	80,000	82,300

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

Table 6.4 shows a sample of the maximum capacity—maximum number of spectators—of sports stadiums. The table does not include horse-racing or motor-racing stadiums.

a. Calculate the sample mean and the sample standard deviation for the maximum capacity of sports stadiums.

b. Construct a histogram.

c. Draw a smooth curve through the midpoints of the tops of the bars of the histogram.

d. In words, describe the shape of your histogram and smooth curve.

e. Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by X ~ _____ (_____, _____).

f. Use the distribution in part e to calculate the probability that the maximum capacity of sports stadiums is less than 67,000 spectators.

g. Determine the cumulative relative frequency that the maximum capacity of sports stadiums is less than 67,000 spectators. Hint—Order the data and count the sports stadiums that have a maximum capacity less than 67,000. Divide by the total number of sports stadiums in the sample.

h. Why aren’t the answers to part f and part g exactly the same?

Answer

1. mean = 60,136
s = 10,468

2. Answers will vary

3. Answers will vary

4. Answers will vary

5. X ~ N(60136, 10468)

6. 0.7440

7. The cumulative relative frequency is 43/60 = 0.717.

8. The answers for part f and part g are not the same because the normal distribution is only an approximation.

6-85 §6.2 source ↗

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10 percent of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68–95–99.7 empirical rule—one standard deviation, two standard deviations, and three standard deviations from the mean being referred to? Assume a normal distribution for the defective cars in the sample.

Answer

• n = 100; p = 0.1; q = 0.9

• μ = np = (100)(0.10) = 10

• σ = $\sqrt{n p q}$ = $\sqrt{(100)(0 .1)(0 .9)}$ = 3

i. z = ±1: x₁ = µ + zσ = 10 + 1(3) = 13 and x2 = µ – zσ = 10 – 1(3) = 7. 68 percent of the defective cars will fall between seven and 13

ii. z = ±2: x₁ = µ + zσ = 10 + 2(3) = 16 and x2 = µ – zσ = 10 – 2(3) = 4. 95 percent of the defective cars will fall between four and 16

iii. z = ±3: x₁ = µ + zσ = 10 + 3(3) = 19 and x2 = µ – zσ = 10 – 3(3) = 1. 99.7 percent of the defective cars will fall between one and 19

6-87 §6.2 source ↗

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.

A child playing a carnival game will be a winner one out of five times. If 190 games are played, what is the probability that there are

a. somewhere between 34 and 54 wins

b. somewhere between 54 and 64 wins

c. more than 64 wins

Answer

• n = 190; p = $\frac{1}{5}$ = 0.2; q = 0.8

• μ = np = (190)(0.2) = 38

• σ = $\sqrt{n p q}$ = $\sqrt{(190)(0 .2)(0 .8)}$ = 5.5136

a. For this problem: P(34 < x < 54) = normalcdf(34,54,48,5.5136) = 0.7641

b. For this problem: P(54 < x < 64) = normalcdf(54,64,48,5.5136) = 0.0018

c. For this problem: P(x > 64) = normalcdf(64,10⁹⁹,48,5.5136) = 0.0000012 (approximately 0)