Content check — Business 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%

6-33 §6.1 source ↗

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

Answer

about 4%

6-35 §6.1 source ↗

Suppose X ~ N(–3, 1). Between what x values does 95.45% 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%

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%

6-41 §6.1 source ↗

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

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

Answer

The lifetime of a wearable fitness device measured in years.

6-43 §6.3 source ↗

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

A bell-shaped curve representing a normal distribution is shown, with a vertical line indicating x=1 on the horizontal axis. — Figure 6.13 in the book ↗

Answer

P(x < 1)

6-45 §6.3 source ↗

Is P(x < 1) equal to P(x ≤ 1)? Why?

Answer

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

6-47 §6.3 source ↗

What is the area to the right of three?

An illustration of a bell curve, or normal distribution, with a vertical line segment marking the value 3 on the x-axis, which is labeled 'x'. — Figure 6.16 in the book ↗

Answer

1 – P(x < 3) or P(x > 3)

6-49 §6.3 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 §6.3 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 §6.3 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

6-55 §6.3 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

Use the following information to answer the next three exercise: The life of wearable fitness devices is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A wearable fitness device is guaranteed for three years. We are interested in the length of time a wearable fitness device lasts. Find the probability that a wearable fitness device 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 6.17 in the book ↗

Answer

a. Answers may vary.

b. 3, 0.1979

6-57 §6.3 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

An experiment with a probability of success given as 0.40 is repeated 100 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at least 45 successes.

Answer

0.154

6-58 §6.3 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

An experiment with a probability of success given as 0.30 is repeated 90 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at least 22 successes.

Answer

0.874

6-59 §6.3 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

An experiment with a probability of success given as 0.40 is repeated 100 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have from 35 to 45 successes.

Answer

0.693

6-60 §6.3 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

An experiment with a probability of success given as 0.30 is repeated 90 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have from 26 to 30 successes.

Answer

0.346

6-61 §6.3 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

An experiment with a probability of success given as 0.40 is repeated 100 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at most 34 successes.

Answer

0.110

6-62 §6.3 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

An experiment with a probability of success given as 0.30 is repeated 90 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at most 34 successes.

Answer

0.946

6-63 §6.3 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

A multiple choice test has a probability any question will be guesses correctly of 0.25. There are 100 questions, and a student guesses at all of them. Use the normal distribution to approximate the binomial distribution, and determine the probability at least 30, but no more than 32, questions will be guessed correctly.

Answer

0.071

6-64 §6.3 source ↗

Use the following information to answer the next four exercises:

X ~ N(54, 8)

A multiple choice test has a probability any question will be guesses correctly of 0.25. There are 100 questions, and a student guesses at all of them. Use the normal distribution to approximate the binomial distribution, and determine the probability at least 24, but no more than 28, questions will be guessed correctly.

Answer

0.347

6-66 §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 ten days to recover?

a. 1.5

b. 0.2

c. 2.2

d. 7.3

Answer

c

6-68 §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 National Basketball Association players were listed on team rosters at the start of a recent season. The heights of basketball players have an approximate normal distribution with 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 an NBA 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) = 92.615 inches, which is taller than 7 feet, 8 inches. There are very few NBA players this tall so the answer is no, not likely.

6-70 §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.

6-72 §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.

During a certain year, 1,475,623 students heading to college took the SAT. 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. During a different year, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT 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 they took?

Answer

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

a. X = 720 : $z =$ $\frac{720 - 520}{115} =$ $1.74$ . 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).

6-75 §6.3 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

6-77 §6.3 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 over 72 inches? Explain why or why not, and justify your answer numerically.

d. The middle 40% 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

6-79 §6.3 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 America 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 = percent of fat calories.

a. X ~ _____(_____,_____)

b. Find the probability that the percent 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% of their calories as fat is 0.3446.

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

6-81 §6.3 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 the 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 percent of the children spend over ten 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 ten hours a day unsupervised is less than 0.0001.

e. 2.21 hours

6-83 §6.3 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

6-85 §6.3 source ↗

Table 6.1 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 6.1 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 ten customers has been 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. Answers may vary.

c. Answers may vary.

d. Answers may vary.

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 normal curve as well.

6-88 §6.3 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.

An automotive factory can build an average of 6,000 cars and trucks a week. Generally, 10% 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 are 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. $z =$ $\pm 1 : x_{1} =$ $µ +$ $zσ =$ $10 +$ $1(3) =$ $13$ and $x 2 =$ $µ - zσ =$ $10 - 1(3) =$ $7.68%$ of the defective cars will fall between seven and 13.

ii. $z =$ $\pm 2 : x_{1} =$ $µ +$ $zσ =$ $10 +$ $2(3) =$ $16$ and $x 2 =$ $µ - zσ =$ $10 - 2(3) =$ $4. 95 %$ of the defective cars will fall between four and 16

iii. $z =$ $\pm 3 : x_{1} =$ $µ +$ $zσ =$ $10 +$ $3(3) =$ $19$ and $x 2 =$ $µ - zσ =$ $10 - 3(3) =$ $1. 99.7%$ of the defective cars will fall between one and 19.

6-90 §6.3 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 $1 scratch off lotto ticket will be a winner one out of five times. Out of a shipment of n = 190 lotto tickets, find the probability for the lotto tickets that there are

a. somewhere between 34 and 54 prizes.

b. somewhere between 54 and 64 prizes.

c. more than 64 prizes.

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) = 0.7641

b. For this problem: P(54 < x < 64) = 0.0018

c. For this problem: P(x > 64) = 0.0000012 (approximately 0)

6-92 §6.3 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 hospital has 49 births in a year. It is considered equally likely that a birth be a boy as it is the birth be a girl.

a. What is the mean?

b. What is the standard deviation?

c. Can this binomial distribution be approximated with a normal distribution?

d. If so, use the normal distribution to find the probability that at least 23 of the 49 births were boys.

Answer

a. 24.5

b. 3.5

c. Yes

d. 0.67

6-93 §6.3 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.

Historically, a final exam in a course is passed with a probability of 0.9. The exam is given to a group of 70 students.

a. What is the mean of the binomial distribution?

b. What is the standard deviation?

c. Can this binomial distribution be approximate with a normal distribution?

d. If so, use the normal distribution to find the probability that at least 60 of the students pass the exam?

Answer

a. 63

b. 2.5

c. Yes

d. 0.88

6-94 §6.3 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 tree in an orchard has 200 oranges. Of the oranges, 40 are not ripe. Use the normal distribution to approximate the binomial distribution, and determine the probability a box containing 35 oranges has at most two oranges that are not ripe.

Answer

0.02

6-95 §6.3 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 a large city one in ten fire hydrants are in need of repair. If a crew examines 100 fire hydrants in a week, what is the probability they will find nine of fewer fire hydrants that need repair? Use the normal distribution to approximate the binomial distribution.

Answer

0.37

6-96 §6.3 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.

On an assembly line it is determined 85% of the assembled products have no defects. If one day 50 items are assembled, what is the probability at least 4 and no more than 8 are defective. Use the normal distribution to approximate the binomial distribution.

Answer

0.50