Ch. 5 Normal Probability Distributions

Ch. 5 Normal Probability Distributions

5.1 Introduction to Normal Distributions and the Standard Normal Distribution

1 Find Areas Under the Standard Normal Curve

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Provide an appropriate response.

1) Find the area under the standard normal curve between z = 0 and z = 3.

A) 0.4987 B) 0.9987 C) 0.0010 D) 0.4641

2) Find the area under the standard normal curve to the right of z = 1.

A) 0.1587 B) 0.8413 C) 0.1397 D) 0.5398

3) Find the area under the standard normal curve between z = 1 and z = 2.

A) 0.1359 B) 0.8413 C) 0.5398 D) 0.2139

4) Find the area under the standard normal curve to the left of z = 1.5.

A) 0.9332 B) 0.0668 C) 0.5199 D) 0.7612

5) Find the area under the standard normal curve between z = -1.5 and z = 2.5.

A) 0.9270 B) 0.7182 C) 0.6312 D) 0.9831

6) Find the area under the standard normal curve to the right of z = -1.25.

A) 0.8944 B) 0.5843 C) 0.6978 D) 0.7193

7) Find the area under the standard normal curve to the left of z = 1.25.

A) 0.8944 B) 0.1056 C) 0.2318 D) 0.7682

8) Find the area under the standard normal curve between z = 1.5 and z = 2.5.

A) 0.0606 B) 0.9938 C) 0.9332 D) 0.9816

9) Find the area under the standard normal curve between z = -1.25 and z = 1.25.

A) 0.7888 B) 0.8817 C) 0.6412 D) 0.2112

10) Find the sum of the areas under the standard normal curve to the left of z = -1.25 and to the right of z = 1.25.

A) 0.2112 B) 0.7888 C) 0.1056 D) 0.3944

Page 98

11) Find the area of the indicated region under the standard normal curve.

A) 0.0968 B) 0.9032 C) 0.4032 D) 0.0823


A) 0.6562 B) 1.309 C) 0.3438 D) 0.309


A) 0.9032 B) 0.0968 C) 0.0823 D) 0.9177


A) 0.1504 B) 0.1292 C) 0.8489 D) 0.0212

2 Find Probabilities Using the Standard Normal Distribution



1) Use the standard normal distribution to find P(0 < z < 2.25).

A) 0.4878 B) 0.5122 C) 0.8817 D) 0.7888

2) Use the standard normal distribution to find P(-2.25 < z < 0).

A) 0.4878 B) 0.5122 C) 0.6831 D) 0.0122

Page 99

3) Use the standard normal distribution to find P(-2.25 < z < 1.25).

A) 0.8822 B) 0.0122 C) 0.4878 D) 0.8944

4) Use the standard normal distribution to find P(-2.50 < z < 1.50).

A) 0.9270 B) 0.8822 C) 0.6167 D) 0.5496

5) Use the standard normal distribution to find P(z < -2.33 or z > 2.33).

A) 0.0198 B) 0.9802 C) 0.7888 D) 0.0606

6) For the standard normal curve, find the z-score that corresponds to the third quartile.

A) 0.67 B) -0.67 C) 0.77 D) -0.23

7) For the standard normal curve, find the z-score that corresponds to the first quartile.

A) -0.67 B) 0.67 C) 0.77 D) -0.23

8) For the standard normal curve, find the z-score that corresponds to the first decile.

A) -1.28 B) 1.28 C) -2.33 D) 0.16

5.2 Normal Distributions: Finding Probabilities

1 Find Probabilities for Normally Distributed Variables


Provide an appropriate response. Use the Standard Normal Table to find the probability.

1) IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. An individualʹs IQscore is found to be 110. Find the z-score corresponding to this value.

A) 0.67 B) -0.67 C) 1.33 D) -1.33


A) -0.67 B) 0.67 C) 1.33 D) -1.33


A) 1.33 B) -0.67 C) 0.67 D) -1.33

4) IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the IQ scorethat corresponds to a z-score of 1.96.

A) 129.4 B) 115.6 C) 122.4 D) 132.1


A) 0.91 B) -0.91 C) 1.10 D) -1.10

Page 100


A) 0.35 B) -0.35 C) 2.83 D) -2.83

7) The lengths of pregnancies of humans are normally distributed with a mean of 268 days and a standarddeviation of 15 days. Find the probability of a pregnancy lasting more than 300 days.

A) 0.0166 B) 0.9834 C) 0.2375 D) 0.3189

8) The lengths of pregnancies of humans are normally distributed with a mean of 268 days and a standarddeviation of 15 days. Find the probability of a pregnancy lasting less than 250 days.

A) 0.1151 B) 0.1591 C) 0.0606 D) 0.0066

9) The distribution of cholesterol levels in teenage boys is approximately normal with μ = 170 and σ = 30 (Source:U.S. National Center for Health Statistics). Levels above 200 warrant attention. Find the probability that ateenage boy has a cholesterol level greater than 200.

A) 0.1587 B) 0.8413 C) 0.3419 D) 0.2138

10) The distribution of cholesterol levels in teenage boys is approximately normal with μ = 170 and σ = 30 (Source:U.S. National Center for Health Statistics). Levels above 200 warrant attention. Find the probability that ateenage boy has a cholesterol level greater than 225.

A) 0.0336 B) 0.0606 C) 0.0718 D) 0.0012

11) An airline knows from experience that the distribution of the number of suitcases that get lost each week on acertain route is approximately normal with μ = 15.5 and σ = 3.6. What is the probability that during a givenweek the airline will lose less than 20 suitcases?

A) 0.8944 B) 0.3944 C) 0.1056 D) 0.4040

12) An airline knows from experience that the distribution of the number of suitcases that get lost each week on acertain route is approximately normal with μ = 15.5 and σ = 3.6. What is the probability that during a givenweek the airline will lose more than 20 suitcases?

A) 0.1056 B) 0.3944 C) 0.4040 D) 0.8944

13) An airline knows from experience that the distribution of the number of suitcases that get lost each week on acertain route is approximately normal with μ = 15.5 and σ = 3.6. What is the probability that during a givenweek the airline will lose between 10 and 20 suitcases?

A) 0.8314 B) 0.3944 C) 0.1056 D) 0.4040

14) Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $32,000 and a standard deviation of $3000. If a teacher is selected at random, find the probability thathe or she makes more than $36,000.

A) 0.0918 B) 0.9082 C) 0.1056 D) 0.4040

15) Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $32,000 and a standard deviation of $3000. If a teacher is selected at random, find the probability thathe or she makes less than $28,000.

A) 0.0918 B) 0.9981 C) 0.2113 D) 0.9827

Page 101

16) Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. The cheerleaders for a local professional basketball team must be between 65.5 and 68.0inches. If a woman is randomly selected, what is the probability that her height is between 65.5 and 68.0inches?

A) 0.1844 B) 0.9608 C) 0.7881 D) 0.3112

17) The lengths of pregnancies of humans are normally distributed with a mean of 268 days and a standarddeviation of 15 days. A baby is premature if it is born three weeks early. What percent of babies are bornprematurely?

A) 8.08% B) 6.81% C) 9.21% D) 10.31%

18) The distribution of cholesterol levels in teenage boys is approximately normal with μ = 170 and σ = 30 (Source:U.S. National Center for Health Statistics). Levels above 200 warrant attention. What percent of teenage boyshave levels between 170 and 225?

A) 3.36% B) 6.06% C) 46.64% D) 56.13%

19) Assume that blood pressure readings are normally distributed with μ = 120 and σ = 8. A blood pressurereading of 145 or more may require medical attention. What percent of people have a blood pressure readinggreater than 145?

A) 0.09% B) 99.91% C) 6.06% D) 11.09%

20) Assume that the heights of American men are normally distributed with a mean of 69.0 inches and a standarddeviation of 2.8 inches. The U.S. Marine Corps requires that men have heights between 64 and 78 inches. Findthe percent of men meeting these height requirements.

A) 96.26% B) 3.67% C) 99.93% D) 31.12%

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

21) Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. The U.S. Army requires that the heights of women be between 58 and 80 inches. If awoman is randomly selected, what is the probability that her height is between 58 and 80 inches?

2 Interpret Normal Distributions


Provide an appropriate response. Use the Standard Normal Table to find the probability.

1) The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15days. Out of 50 pregnancies, how many would you expect to last less than 250 days?

2) The distribution of cholesterol levels in teenage boys is approximately normal with μ = 170 and σ = 30. Levelsabove 200 warrant attention. If 95 teenage boys are examined, how many would you expect to have cholesterollevels greater than 225?

3) An airline knows from experience that the distribution of the number of suitcases that get lost each week on acertain route is approximately normal with μ = 15.5 and σ = 3.6. In one year, how many weeks would youexpect the airline to lose between 10 and 20 suitcases?

Page 102

4) Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. The U.S. Army requires that the heights of women be between 58 and 80 inches. If 200women want to enlist in the U.S. Army, how many would you expect to meet the height requirements?

5) Assume that the heights of men are normally distributed with a mean of 69.0 inches and a standard deviationof 2.8 inches. The U.S. Marine Corps requires that the heights of men be between 64 and 78 inches. If 500 menwant to enlist in the U.S. Marine Corps, how many would you not expect to meet the height requirements?

5.3 Normal Distributions: Finding Values

1 Find a z-score Given the Area Under the Normal Curve



1) Find the z-score that corresponds to the given area under the standard normal curve.




Page 103


5) Find the z-scores for which 90% of the distributionʹs area lies between -z and z.

A) (-1.645, 1.645) B) (-2.33, 2.33) C) (-1.96, 1.96) D) (-0.99, 0.99)

6) Find the z-scores for which 98% of the distributionʹs area lies between -z and z.

A) (-2.33, 2.33) B) (-1.645, 1.645) C) (-1.96, 1.96) D) (-0.99, 0.99)

7) Find the z-score for which 70% of the distributionʹs area lies to its right.

A) -0.53 B) -0.98 C) -0.81 D) -0.47

8) Find the z-score that is greater than the mean and for which 70% of the distributionʹs area lies to its left.

A) 0.53 B) 0.98 C) 0.81 D) 0.47

9) Use a standard normal table to find the z-score that corresponds to the cumulative area of 0.0188.

A) -2.08 B) 2.08 C) 2.1 D) -2.1

10) Find the z-score that has 84.85% of the distributionʹs area to its right.

A) 1.03 B) -1.03 C) 0.39 D) -0.39

11) Find the z-score for which 99% of the distributionʹs area lies between -z and z.

A) (-2.575, 2.575) B) (-1.28, 1.28) C) (-1.96, 1.96) D) (-1.645, 1.645)

2 Find a z-score Given a Percentile



1) For the standard normal curve, find the z-score that corresponds to the 90th percentile.

A) 1.28 B) 0.28 C) 1.52 D) 2.81

2) For the standard normal curve, find the z-score that corresponds to the 30th percentile.

A) -0.53 B) -0.98 C) -0.47 D) -0.12

3) For the standard normal curve, find the z-score that corresponds to the 7th decile.

A) 0.53 B) 0.98 C) 0.47 D) 0.12

4) Use a standard normal table to find the z-score that corresponds to the 98th percentile.

A) 2.055 B) 1.405 C) 2.33 D) 0.25

Page 104

3 Use Normal Distributions to Answer Questions



1) IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the x -score thatcorresponds to a z-score of 2.33.

A) 134.95 B) 125.95 C) 139.55 D) 142.35

2) IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the x -score thatcorresponds to a z-score of -1.645.

A) 75.3 B) 79.1 C) 82.3 D) 91.0

3) The scores on a mathematics exam have a mean of 65 and a standard deviation of 5. Find the x-value thatcorresponds to the z-score -1.28.

A) 58.6 B) 71.4 C) 60.0 D) 63.7


4) A mathematics professor gives two different tests to two sections of his college algebra courses. The first classhas a mean of 56 with a standard deviation of 9 while the second class has a mean of 75 with a standarddeviation of 15. A student from the first class scores a 62 on the test while a student from the second classscores an 83 on the test. Compare the scores.


5) Compare the scores: a score of 75 on a test with a mean of 65 and a standard deviation of 8 and a score of 75 ona test with a mean of 70 and a standard deviation of 4.

A) The two scores are statistically the same.

B) A score of 75 with a mean of 65 and a standard deviation of 8 is better.

C) A score of 75 with a mean of 70 and a standard deviation of 4 is better.

D) You cannot determine which score is better from the given information.

6) Compare the scores: a score of 88 on a test with a mean of 79 and a score of 78 on a test with a mean of 70.

A) You cannot determine which score is better from the given information.

B) The two scores are statistically the same.

C) A score of 75 with a mean of 65 and a standard deviation of 8 is better.

D) A score of 75 with a mean of 70 and a standard deviation of 4 is better.

7) Compare the scores: a score of 220 on a test with a mean of 200 and a standard deviation of 21 and a score of 90on a test with a mean of 80 and a standard deviation of 8.

A) A score of 90 with a mean of 80 and a standard deviation of 8 is better.


C) The two scores are statistically the same.


Page 105

8) Two high school students took equivalent language tests, one in German and one in French. The student takingthe German test, for which the mean was 66 and the standard deviation was 8, scored an 82, while the studenttaking the French test, for which the mean was 27 and the standard deviation was 5, scored a 35. Compare thescores.

A) A score of 82 with a mean of 66 and a standard deviation of 8 is better.




9) SAT scores have a mean of 1026 and a standard deviation of 209. ACT scores have a mean of 20.8 and astandard deviation of 4.8. A student takes both tests while a junior and scores 1130 on the SAT and 25 on theACT. Compare the scores.

A) A score of 25 on the ACT test was better.

B) A score of 1130 on the SAT test was better.



10) SAT scores have a mean of 1026 and a standard deviation of 209. ACT scores have a mean of 20.8 and astandard deviation of 4.8. A student takes both tests while a junior and scores 860 on the SAT and 16 on theACT. Compare the scores.

A) A score of 860 on the SAT test was better.

B) A score of 16 on the ACT test was better.




11) Assume that blood pressure readings are normally distributed with μ = 124 and σ = 9. A researcher wishes toselect people for a study but wants to exclude the top and bottom 10 percent. What would be the upper andlower readings to qualify people to participate in the study?

12) Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $35,000 and a standard deviation of $4000. What is the cutoff salary for teachers in the top 10%?


13) Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $34,000 and a standard deviation of $2000. What is the cutoff salary for teachers in the bottom 10%?

A) $31,440 B) $36,560 C) $30,710 D) $37,290


14) The times for completing one circuit of a bicycle course are normally distributed with a mean of 82.1 minutesand a standard deviation of 4.8 minutes. An association wants to sponsor a race but will cut the bottom 25% ofriders. In a trial run, what should be the cutoff time?

Page 106

15) Assume that the heights of men are normally distributed with a mean of 68.9 inches and a standard deviationof 1.7 inches. If the top 5 percent and bottom 5 percent are excluded for an experiment, what are the cutoffheights to be eligible for this experiment? Round your answers to one decimal place.


16) Assume that the heights of women are normally distributed with a mean of 64.9 inches and a standarddeviation of 1.6 inches. Find Q3, the third quartile that separates the bottom 75% from the top 25%.

A) 66.0 B) 63.8 C) 66.7 D) 66.9

17) The body temperatures of adults are normally distributed with a mean of 98.6 ° F and a standard deviation of0.44° F. What temperature represents the 95th percentile?

A) 99.32° F B) 97.88° F C) 99.16° F D) 99.46° F

18) In a certain normal distribution, find the standard deviation σ when μ = 50 and 10.56% of the area lies to theright of 55.

A) 4 B) 2 C) 3 D) 5

19) In a certain normal distribution, find the mean μwhen σ = 5 and 5.48% of the area lies to the left of 78.

A) 86 B) 70 C) 94 D) 62


20) In a certain normal distribution, 6.3% of the area lies to the left of 36 and 6.3% of the area lies to the right of 42.Find the mean μ and the standard deviation σ.


21) A tire company finds the lifespan for one brand of its tires is normally distributed with a mean of 47,500 milesand a standard deviation of 3000 miles. If the manufacturer is willing to replace no more than 10% of the tires,what should be the approximate number of miles for a warranty?

A) 43,660 B) 51,340 C) 42,565 D) 52,435

5.4 Sampling Distributions and the Central Limit Theorem

1 Interpret Sampling Distributions



1) The distribution of room and board expenses per year at a four-year college is normally distributed with amean of $5850 and standard deviation of $1125. Random samples of size 20 are drawn from this populationand the mean of each sample is determined. Which of the following mean expenses would be consideredunusual?

A) $5180 B) $6350 C) $6180 D) none of these

Page 107

2 Find Probabilities



1) The lengths of pregnancies are normally distributed with a mean of 267 days and a standard deviation of 15days. If 36 women are randomly selected, find the probability that they have a mean pregnancy between 267days and 269 days.

A) 0.2881 B) 0.7881 C) 0.2119 D) 0.5517

2) The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15days. If 64 women are randomly selected, find the probability that they have a mean pregnancy between 266days and 268 days.

A) 0.3577 B) 0.7881 C) 0.2881 D) 0.5517

3) Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. If 100 women are randomly selected, find the probability that they have a mean heightgreater than 63.0 inches.

A) 0.9918 B) 0.0082 C) 0.2881 D) 0.8989

4) Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standarddeviation of 2.5 inches. If 75 women are randomly selected, find the probability that they have a mean heightbetween 63 and 65 inches.

A) 0.9811 B) 0.3071 C) 0.0188 D) 0.2119

5) Assume that the heights of men are normally distributed with a mean of 67.9 inches and a standard deviationof 2.8 inches. If 64 men are randomly selected, find the probability that they have a mean height greater than68.9 inches.

A) 0.0021 B) 0.8188 C) 0.9005 D) 9.9671

6) Assume that the heights of men are normally distributed with a mean of 69.0 inches and a standard deviationof 2.8 inches. If 64 men are randomly selected, find the probability that they have a mean height between 68and 70 inches.

A) 0.9958 B) 0.9979 C) 0.0021 D) 0.9015


7) The body temperatures of adults are normally distributed with a mean of 98.6 ° F and a standard deviation of0.60° F. If 25 adults are randomly selected, find the probability that their mean body temperature is less than99° F.


8) The body temperatures of adults are normally distributed with a mean of 98.6 ° F and a standard deviation of0.60° F. If 36 adults are randomly selected, find the probability that their mean body temperature is greaterthan 98.4° F.

A) 0.9772 B) 0.0228 C) 0.8188 D) 0.9360

Page 108


9) Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $32,000 and a standard deviation of $3000. If 100 teachers are randomly selected, find the probabilitythat their mean salary is less than $32,500.


10) Assume that the salaries of elementary school teachers in the United States are normally distributed with amean of $32,000 and a standard deviation of $3000. If 100 teachers are randomly selected, find the probabilitythat their mean salary is greater than $32,500.

A) 0.0475 B) 0.9525 C) 0.3312 D) 0.1312

11) Assume that blood pressure readings are normally distributed with a mean of 116 and a standard deviation of4.8. If 36 people are randomly selected, find the probability that their mean blood pressure will be less than 118.

A) 0.9938 B) 0.0062 C) 0.8819 D) 0.8615

12) Assume that blood pressure readings are normally distributed with a mean of 120 and a standard deviation of8. If 100 people are randomly selected, find the probability that their mean blood pressure will be greater than122.

A) 0.0062 B) 0.9938 C) 0.8819 D) 0.8615

13) The average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22pounds (Source: American Dietetic Association). If a sample of 50 individuals is randomly selected, find theprobability that the mean of the sample will be less than 200 pounds.

A) 0.9015 B) 0.0985 C) 0.7613 D) 0.8815

14) The average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22pounds (Source: American Dietetic Association). If a sample of 50 individuals is randomly selected, find theprobability that the mean of the sample will be greater than 200 pounds.

A) 0.0985 B) 0.7613 C) 0.8815 D) 0.9015

15) A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standarddeviation of 0.2 ounce. If a sample of 9 cups is selected, find the probability that the mean of the sample will beless than 12.1 ounces. Find the probability if the sample is just 1 cup.

A) 0.9332; 0.6915 B) 0.9332; 0.1915 C) 0.4332; 0.1915 D) 0.4332; 0.6915

16) A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standarddeviation of 0.2 ounce. If a sample of 9 cups is selected, find the probability that the mean of the sample will begreater than 12.1 ounces. Find the probability if the sample is just 1 cup.

A) 0.0668; 0.3085 B) 0.0668; 0.8085 C) 0.5668; 0.8085 D) 0.5668; 0.3085

Page 109

3 Determine if the Finite Correction Factor Should be Used



1) A study of 800 homeowners in a certain area showed that the average value of the homes is $182,000 and thestandard deviation is $15,000. If 50 homes are for sale, find the probability that the mean value of these homesis less than $185,000. Remember: check to see if the finite correction factor applies.

2) A study of 800 homeowners in a certain area showed that the average value of the homes is $182,000 and thestandard deviation is $15,000. If 50 homes are for sale, find the probability that the mean value of these homesis greater than $185,000. Remember: check to see if the finite correction factor applies.

4 Concepts



1) A soda machine dispenses normally distributed amounts of soda with a mean of 20 ounces and a standarddeviation of 0.2 ounce. Are you more likely to randomly select one bottle with more than 20.3 ounces or areyou more likely to select a sample of eight bottles with a mean amount of more than 20.3 ounces? Explain.

2) A soda machine dispenses normally distributed amounts of soda with a mean of 20 ounces and a standarddeviation of 0.2 ounce. Are you more likely to randomly select one bottle with an amount between 19.8 ouncesand 20.2 ounces or are you more likely to select a sample of eight bottles with a mean amount between 19.8ounces and 20.2 ounces? Explain.

5.5 Normal Approximations to Binomial Distributions

1 Decide if the Normal Distribution Can be Used to Approximate the Binomial Distribution



1) In a recent survey, 90% of the community favored building a police substation in their neighborhood. Yourandomly select 16 citizens and ask each if he or she thinks the community needs a police substation.Decide whether you can use the normal distribution to approximate the binomial distribution. If so, find themean and standard deviation. If not, explain why.

2) A recent survey found that 71% of all adults over 50 wear glasses for driving. You randomly select 30 adultsover 50, and ask if he or she wears glasses. Decide whether you can use the normal distribution to approximatethe binomial distribution. If so, find the mean and standard deviation, If not, explain why.

3) According to government data, the probability than an adult was never married is 16%. You randomly select 48adults and ask if he or she was ever married. Decide whether you can use the normal distribution toapproximate the binomial distribution, If so, find the mean and standard deviation, If not, explain why.

4) Decide if it is appropriate to use the normal distribution to approximate the random variable x for a binomialexperiment with sample size of n = 7 and probability of success p = 0.4.

5) Decide if it is appropriate to use the normal distribution to approximate the random variable x for a binomialexperiment with sample size of n = 27 and probability of success p = 0.5.

Page 110

6) Decide if it is appropriate to use the normal distribution to approximate the random variable x for a binomialexperiment with sample size of n = 25 and probability of success p = 0.2

2 Find the Correction for Continuity



1) Ten percent of the population is left-handed. A class of 100 students is selected. Convert the binomialprobability P(x > 12) to a normal probability by using the correction for continuity.

A) P(x > 12.5) B) P(x < 11.5) C) P(x ≤ 12.5) D) P(x ≥ 11.5)

2) Ten percent of the population is left-handed. A class of 9950 students is selected. Convert the binomialprobability P(x < 14) to a normal probability by using the correction for continuity.

A) P(x < 13.5) B) P(x ≥ 13.5) C) P(x > 14.5) D) P(x < 14.5)

3) Ten percent of the population is left-handed. A class of 100 students is selected. Convert the binomialprobability P(x < 12) to a normal probability by using the correction for continuity.

A) P(x < 11.5) B) P(x > 12.5) C) P(x ≤ 12.5) D) P(x ≥ 11.5)

4) Ten percent of the population is left-handed. A class of 7650 students is selected. Convert the binomialprobability P(x ≤ 12) to a normal probability by using the correction for continuity.

A) P(x ≤ 12.5) B) P(x ≥ 11.5) C) P(x > 12.5) D) P(x < 11.5)

3 Find a Probability Using the Appropriate Distribution



1) The failure rate in a statistics class is 30%. In a class of 50 students, find the probability that exactly five studentswill fail. Use the normal distribution to approximate the binomial distribution.

2) A local motel has 100 rooms. The occupancy rate for the winter months is 60%. Find the probability that in agiven winter month at least 70 rooms will be rented. Use the normal distribution to approximate the binomialdistribution.

3) A local motel has 100 rooms. The occupancy rate for the winter months is 60%. Find the probability that in agiven winter month fewer than 70 rooms will be rented. Use the normal distribution to approximate thebinomial distribution.


4) A student answers all 48 questions on a multiple-choice test by guessing. Each question has four possibleanswers, only one of which is correct. Find the probability that the student gets exactly 15 correct answers. Usethe normal distribution to approximate the binomial distribution.

A) 0.0823 B) 0.8577 C) 0.7967 D) 0.0606

Page 111

5) If the probability of a newborn child being female is 0.5, find the probability that in 100 births, 55 or more willbe female. Use the normal distribution to approximate the binomial distribution.

A) 0.1841 B) 0.7967 C) 0.8159 D) 0.0606

6) An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150advanced reservations, find the probability that there will be fewer than 20 no-shows.

A) 0.2451 B) 0.7549 C) 0.7967 D) 0.3187

7) Find the probability that in 200 tosses of a fair six-sided die, a five will be obtained at least 40 times.

A) 0.1210 B) 0.8810 C) 0.0871 D) 0.3875

8) Find the probability that in 200 tosses of a fair six-sided die, a five will be obtained at most 40 times.

A) 0.9131 B) 0.1190 C) 0.8810 D) 0.0853

9) A telemarketer found that there was a 1% chance of a sale from his phone solicitations. Find the probability ofgetting 5 or more sales for 1000 telephone calls.

A) 0.9599 B) 0.0401 C) 0.8810 D) 0.0871


10) The author of a statistics book has trouble deciding whether to use the words ʺheʺ or ʺsheʺ in the bookʹsexamples. To solve the problem, the author flips a coin each time the problem arises. If a head shows, theauthor uses ʺheʺ and if a tail shows, the author uses ʺsheʺ. If this problem occurs 100 times in the book, what isthe probability that ʺsheʺ will be used 58 times?

4 Concepts



1) Match the binomial probability P(x < 23) with the correct statement.

A) P(there are fewer than 23 successes) B) P(there are at most 23 successes)

C) P(there are more than 23 successes) D) P(there are at least 23 successes)

2) Match the binomial probability P(x < 65) with the correct statement.

A) P(there are fewer than 65 successes) B) P(there are at most 65 successes)

C) P(there are more than 65 successes) D) P(there are at least 65 successes)

3) Match the binomial probability P(x ≤ 23) with the correct statement.

A) P(there are at most 23 successes) B) P(there are at least 23 successes)

C) P(there are more than 23 successes) D) P(there are fewer than 23 successes)

4) Match the binomial probability P(x > 23) with the correct statement.

A) P(there are more than 23 successes) B) P(there are at least 23 successes)

C) P(there are at most 23 successes) D) P(there are fewer than 23 successes)

Page 112

5) Ten percent of the population is left-handed. In a class of 100 students, write the binomial probability for thestatement ʺThere are at least 12 left-handed students in the class.ʺ

A) P(x ≥ 12) B) P(x < 12) C) P(x > 12) D) P(x ≤ 12)

6) Ten percent of the population is left-handed. In a class of 100 students, write the binomial probability for thestatement ʺThere are more than 12 left-handed students in the class.ʺ

A) P(x > 12) B) P(x < 12) C) P(x ≤ 12) D) P(x = 12)

7) Ten percent of the population is left-handed. In a class of 100 students, write the binomial probability for thestatement ʺThere are at most 12 left-handed students in the class.ʺ

A) P(x ≤ 12) B) P(x < 12) C) P(x = 12) D) P(x ≥ 12)

8) Ten percent of the population is left-handed. In a class of 104 students, write the binomial probability for thestatement ʺThere are at most 23 left-handed students in the class.ʺ

A) P(x ≤ 23) B) P(x < 23) C) P(x > 23) D) P(x ≥ 23)

9) Ten percent of the population is left-handed. In a class of 100 students, write the binomial probability for thestatement ʺThere are exactly 12 left-handed students in the class.ʺ

A) P(x = 12) B) P(x ≥ 12) C) P(x ≤ 12) D) P(x > 12)

Page 113

Ch. 5 Normal Probability DistributionsAnswer Key

5.1 Introduction to Normal Distributions and the Standard Normal Distribution1 Find Areas Under the Standard Normal Curve

1) A2) A3) A4) A5) A6) A7) A8) A9) A10) A11) A12) A13) A14) A

2 Find Probabilities Using the Standard Normal Distribution1) A2) A3) A4) A5) A6) A7) A8) A

5.2 Normal Distributions: Finding Probabilities1 Find Probabilities for Normally Distributed Variables

1) A2) A3) A4) A5) A6) A7) A8) A9) A10) A11) A12) A13) A14) A15) A16) A17) A18) A19) A20) A21) If x = 58, then z = -2.24 and P(x) = 0.0125. If x = 80, then z = 6.56 and P(x) = 0.9999.

P(58 < x < 80) = 0.9999 - 0.0125 = 0.9874.2 Interpret Normal Distributions

1) About 6 pregnancies

Page 114

2) About 3 teenage boys3) About 43 weeks4) About 197 women5) About 19 men

5.3 Normal Distributions: Finding Values1 Find a z-score Given the Area Under the Normal Curve

1) z = -0.582) z = -1.713) z = 0.424) z = 3.075) A6) A7) A8) A9) A10) A11) A

2 Find a z-score Given a Percentile1) A2) A3) A4) A

3 Use Normal Distributions to Answer Questions1) A2) A3) A4) z = (62 - 56)/9 = 0.667; z = (83 - 75)/15 = 0.533. The student with the score of 62 has the better score.5) A6) A7) A8) A9) A10) A11) (112.5, 135.5)12) x = μ + zσ = 35,000 + (1.28)(4000) = $40,12013) A14) x = μ + zσ = 82.1 + (0.675)(4.8) = 85.3415) 66.1 inches, 71.7 inches16) A17) A18) A19) A20) μ = 39, σ = 1.9621) A

5.4 Sampling Distributions and the Central Limit Theorem1 Interpret Sampling Distributions

1) A2 Find Probabilities

1) A2) A3) A4) A5) A6) A

Page 115

7) 0.99968) A9) 0.952510) A11) A12) A13) A14) A15) A16) A

3 Determine if the Finite Correction Factor Should be Used1) 50/800 = 0.0625 = 6.25%, hence the finite correction factor applies; P(x < 185,000) = 0.92792) 50/800 = 0.0625 = 6.25%, hence the finite correction factor applies; P(x > 185,000) = 1 - 0.9279 = 0.0721

4 Concepts1) It is more likely to select one bottle with more than 20.3 ounces because a large percentage of the data is now closer to

the mean.2) It is more likely to select a sample of eight bottles with an amount between 19.8 ounces and 20.2 ounces because a

large percentage of the data is now closer to the mean.5.5 Normal Approximations to Binomial Distributions1 Decide if the Normal Distribution Can be Used to Approximate the Binomial Distribution

1) cannot use normal distribution, nq = (16)(0.1) = 1.6 < 52) use normal distribution, μ = 21.3 and σ = 2.49.3) use normal distribution, μ = 7.68 and σ = 2.54.4) cannot use normal distribution5) can use normal distribution6) can use normal distribution

2 Find the Correction for Continuity1) A2) A3) A4) A

3 Find a Probability Using the Appropriate Distribution1) P(4.5 < X < 5.5) = P(-3.24 < z <-2.93) = 0.0017 - 0.0006 = 0.00112) P(x ≥ 69.5) = 0.02623) P(x < 69.5) = 0.97384) A5) A6) A7) A8) A9) A10) P(57.5 < x < 58.5) = P(1.50 < z < 1.70) = 0.9554 - 0.9332 = 0.0222

4 Concepts1) A2) A3) A4) A5) A6) A7) A8) A9) A

Page 116

Documents

Ch. 5 Normal Probability Distributions