Ch. 7 The Normal Probability Distributionsite.iugaza.edu.ps/mriffi/files/2018/02/Statistics... ·...

Ch. 7 The Normal Probability Distribution

7.1 Properties of the Normal Distribution

1 Use the uniform probability distribution.

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

Provide an appropriate response.

1) High temperatures in a certain city for the month of August follow a uniform distribution over the interval 63°F

to 90°F. What is the probability that a randomly selected August day has a high temperature that exceeded

68°F?

A) 0.8148 B) 0.1852 C) 0.4444 D) 0.037

2) High temperatures in a certain city for the month of August follow a uniform distribution over the interval 61°F

to 91°F. Find the high temperature which 90% of the August days exceed.

A) 64°F B) 88°F C) 71°F D) 91°F

3) The diameter of ball bearings produced in a manufacturing process can be explained using a uniform

distribution over the interval 3.5 to 5.5 millimeters. What is the probability that a randomly selected ball

bearing has a diameter greater than 4.4 millimeters?

A) 0.55 B) 0.8 C) 2 D) 0.4889

4) Suppose x is a uniform random variable with c = 40 and d = 90. Find the probability that a randomly selected

observation exceeds 75.

A) 0.3 B) 0.7 C) 0.1 D) 0.9

5) Suppose x is a uniform random variable with c = 10 and d = 70. Find the probability that a randomly selected

observation is between 13 and 65.

A) 0.87 B) 0.13 C) 0.5 D) 0.8

6) A machine is set to pump cleanser into a process at the rate of 6 gallons per minute. Upon inspection, it is

learned that the machine actually pumps cleanser at a rate described by the uniform distribution over the

interval 5.5 to 8.5 gallons per minute. Find the probability that between 6.0 gallons and 7.0 gallons are pumped

during a randomly selected minute.

A) 0.33 B) 0.67 C) 0 D) 1

7) Suppose a uniform random variable can be used to describe the outcome of an experiment with the outcomes

ranging from 30 to 80. What is the probability that this experiment results in an outcome less than 40?

A) 0.2 B) 0.09 C) 0.29 D) 1

8) A random number generator is set top generate integer random numbers between 1 and 10 inclusive following

a uniform distribution. What is the probability of the random number generator generating a 7?

A) 0.7 B) 0 C) 0.07 D) 0.5

9) True or False: In a uniform probability distribution, any random variable is just as likely as any other random

variable to occur, provided the random variables belong to the distribution.

A) True B) False

2 Graph a normal curve.



10) Compare a graph of the normal density function with mean of 0 and standard deviation of 1 with a graph of a

normal density function with mean equal to 4 and standard deviation of 1. The graphs would

A) Have the same height but one would be shifted 4 units to the right.

B) Have the same height but one would be shifter 4 units to the left.

C) Have no horizontal displacement but one would be steeper that the other.

D) Have no horizontal displacement but one would be flatter than the other.

11) Compare a graph of the normal density function with mean of 0 and standard deviation of 1 with a graph of a

normal density function with mean equal to 0 and standard deviation of 0.5. The graphs would

A) Have no horizontal displacement but one would be steeper that the other.

B) Have no horizontal displacement but one would be flatter than the other.

C) Have the same height but one would be shifted 4 units to the right.

D) Have the same height but one would be shifter 4 units to the left.

12) Draw a normal curve with μ = 140 and σ = 10. Label the mean and the inflection points.

A)

130 140 150

B)

120 140 160

C)

135 140 145

D)

130 140 150

3 State the properties of the normal curve.

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


13) You are performing a study about the weight of preschoolers. A previous study found the weights to be

normally distributed with a mean of 30 and a standard deviation of 4. You randomly sample 30 preschool

children and find their weights to be as follows.

25 25 26 26.5 27 27 27.5 28 28 28.5

29 29 30 30 30.5 31 31 32 32.5 32.5

33 33 34 34.5 35 35 37 37 38 38

a) Draw a histogram to display the data. Is it reasonable to assume that the weights are normally

distributed? Why?

b) Find the mean and standard deviation of your sample.

c) Is there a high probability that the mean and standard deviation of your sample are consistent

with those found in previous studies? Explain your reasoning.


14) The graph of a normal curve is given. Use the graph to identify the value of μ and σ.

3 6 9 12 15 18 21

A) μ = 12, σ = 3 B) μ = 3, σ = 12 C) μ = 12, σ = 9 D) μ = 9, σ = 12

15) The normal density curve is symmetric about

A) Its mean

B) The horizontal axis

C) An inflection point

D) A point located one standard deviation from the mean

16) The highest point on the graph of the normal density curve is located at

A) its mean B) an inflection point C) μ + σ D) μ + 3σ

17) Approximately ____% of the area under the normal curve is between μ - 2σ and μ + 2σ.

A) 95 B) 99.7 C) 68 D) 50

Determine whether the graph can represent a normal curve. If it cannot, explain why.

18)

A) The graph cannot represent a normal density function because it increases as x becomes very large or

very small.

B) The graph cannot represent a normal density function because it takes negative values for some values of

x.

C) The graph cannot represent a normal density function because the area under the graph is greater than 1.

D) The graph can represent a normal density function.

19)

A) The graph can represent a normal density function.

B) The graph cannot represent a normal density function because it has no inflection points.

C) The graph cannot represent a normal density function because as x increases without bound, the graph

takes negative values.

D) The graph cannot represent a normal density function because the area under the graph is greater than 1.

20)

A) The graph cannot represent a normal density function because the graph takes negative values for some

values of x.

B) The graph cannot represent a normal density function because the area under the graph is less than 1.

C) The graph cannot represent a normal density function because it is not symmetric.


21)

A) The graph cannot represent a normal density function because it does not approach the horizontal axis as

x increases or decreases without bound.

B) The graph cannot represent a normal density function because it is not bell shaped.

C) The graph can represent a normal density function.

D) The graph cannot represent a normal density function because it has no inflection points.

22)

A) The graph can represent a normal density function.

B) The graph cannot represent a normal density function because it has no inflection points.

C) The graph cannot represent a normal density function because its maximum value is too small.

D) The graph cannot represent a normal density function because the area under the graph is less than 1.

23)

A) The graph cannot represent a normal density function because it is not symmetric.

B) The graph cannot represent a normal density function because as x increases without bound, the graph

takes negative values.

C) The graph cannot represent a normal density function because it is bimodal.


24)

A) The graph cannot represent a normal density function because it is bimodal.

B) The graph cannot represent a normal density function because a normal density curve should approach

but not reach the horizontal axis as x increases and decreases without bound.

C) The graph can represent a normal density function.

D) A and B are both true.

4 Explain the role of area in the normal density function.



25) The analytic scores on a standardized aptitude test are known to be normally distributed with mean μ = 610

and standard deviation σ = 115.

(a) Draw a normal curve with the parameters labeled.

(b) Shade the region that represents the proportion of test takers who scored less than 725.

(c) Suppose the area under the normal curve to the left of X = 725 is 0.8413. Provide two interpretations of this

result.

26) The weight of 2-year old hyraxes is known to be normally distributed with mean

μ = 2200 grams and standard deviation σ = 365 grams


(b) Shade the region that represents the proportion of hyraxes who weighed more than 2930 grams.

(c) Suppose the area under the normal curve to the left of X = 2930 is 0.0228. Provide two interpretations of this

result.

27) The average mpg (miles per gallon) of a new model of motorcycle is known to be normally distributed with

mean μ = 27.4 mpg and standard deviation σ = 2.9 mpg.


(b) Shade the region that represents the proportion of mpgs between 29.3 and 23.7.

(c) Suppose the area under the normal curve to between X = 29.3 and X = 23.7 is 0.6419. Provide two

interpretations of this result.


28) True or False: The area under the normal curve drawn with regard to the population parameters is the same as

the proportion of the population that has these characteristics.

A) True B) False

29) True or False: The area under the normal curve drawn with regard to the population parameters is the same as

the probability that a randomly selected individual of a population has these characteristics.

A) True B) False

30) True or False: The proportion of the population that has certain characteristics is the same as the probability

that a randomly selected individual of the population has these same characteristics.

A) True B) False

7.2 Applications of the Normal Distribution

1 Find and interpret the area under a normal curve.



1) 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

2) 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

3) 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

4) 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

5) 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

6) 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

7) 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

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

Determine the area under the standard normal curve that lies between:

11) z = 1 and z = 2

A) 0.1359 B) 0.8641 C) 0.0006 D) 0.0008

12) z = 0.8 and z = 1.4

A) 0.1311 B) 0.7881 C) 0.9192 D) 0.2119

13) z = -0.1 and z = 0.1A) 0.0796 B) 0.4602 C) 0.5398 D) 0.5

14) z = -2 and z = -0.1A) 0.4374 B) 0.4602 C) 0.5398 D) 0.0228

Find the indicated probability.

15) Assume that the random variable X is normally distributed, with mean μ = 70 and standard deviation σ = 8.

Compute the probability P(X < 80).

A) 0.8944 B) 0.8849 C) 0.1056 D) 0.9015


Compute the probability P(X > 116).

A) 0.2119 B) 0.1977 C) 0.7881 D) 0.2420


Compute the probability P(37 < X < 85).

A) 0.8914 B) 0.8819 C) 0.8944 D) 0.7888


18) A physical fitness association is including the mile run in its secondary-school fitness test. The time for this

event for boys in secondary school is known to possess a normal distribution with a mean of 440 seconds and a

standard deviation of 40 seconds. Find the probability that a randomly selected boy in secondary school can

run the mile in less than 348 seconds.

A) 0.0107 B) 0.4893 C) 0.9893 D) 0.5107



standard deviation of 50 seconds. Find the probability that a randomly selected boy in secondary school will

take longer than 325 seconds to run the mile.

A) 0.9893 B) 0.4893 C) 0.0107 D) 0.5107

20) Suppose a brewery has a filling machine that fills 12 ounce bottles of beer. It is known that the amount of beer

poured by this filling machine follows a normal distribution with a mean of 12.18 ounces and a standard

deviation of 0.04 ounce. Find the probability that the bottle contains fewer than 12.08 ounces of beer.

A) 0.0062 B) 0.4938 C) 0.9938 D) 0.5062


poured by this filling machine follows a normal distribution with a mean of 12.14 onces and a standard

deviation of 0.04 ounce. Find the probability that the bottle contains more than 12.14 ounces of beer.

A) 0.5 B) 1 C) 0 D) 0.4



deviation of 0.04 ounce. Find the probability that the bottle contains between 12.00 and 12.06 ounces.

A) 0.1525 B) 0.8351 C) 0.1649 D) 0.8475

23) The length of time it takes college students to find a parking spot in the library parking lot follows a normal

distribution with a mean of 6.5 minutes and a standard deviation of 1 minute. Find the probability that a

randomly selected college student will find a parking spot in the library parking lot in less than 6.0 minutes.

A) 0.3085 B) 0.1915 C) 0.3551 D) 0.2674


distribution with a mean of 3.0 minutes and a standard deviation of 1 minute. Find the probability that a

randomly selected college student will take between 1.5 and 4.0 minutes to find a parking spot in the library

lot.

A) 0.7745 B) 0.4938 C) 0.0919 D) 0.2255

25) The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a normal distribution with

a mean of 12.39 ounces and a standard deviation of 0.26 ounce. The cans only hold 12.65 ounces of soda. Every

can that has more than 12.65 ounces of soda poured into it causes a spill and the can needs to go through a

special cleaning process before it can be sold. What is the probability a randomly selected can will need to go

through this process?

A) 0.1587 B) 0.3413 C) 0.8413 D) 0.6587

26) A new phone system was installed last year to help reduce the expense of personal calls that were being made

by employees. Before the new system was installed, the amount being spent on personal calls followed a

normal distribution with an average of $500 per month and a standard deviation of $50 per month. Refer to

such expenses as PCEʹs (personal call expenses). Using the distribution above, what is the probability that a

randomly selected month had a PCE of between $375.00 and $590.00?

A) 0.9579 B) 0.0421 C) 0.9999 D) 0.0001


by employees. Before the new system was installed, the amount being spent on personal calls follows a normal

distribution with an average of $700 per month and a standard deviation of $50 per month. Refer to such

expenses as PCEʹs (personal call expenses). Find the probability that a randomly selected month had a PCE that

falls below $550.

A) 0.0013 B) 0.7857 C) 0.2143 D) 0.9987

28) The tread life of a particular brand of tire is a random variable best described by a normal distribution with a

mean of 60,000 miles and a standard deviation of 2800 miles. What is the probability a particular tire of this

brand will last longer than 57,200 miles?

A) 0.8413 B) 0.1587 C) 0.2266 D) 0.7266


mean of 60,000 miles and a standard deviation of 1500 miles. What is the probability a certain tire of this brand

will last between 56,850 miles and 57,300 miles?

A) 0.0180 B) 0.9813 C) 0.4920 D) 0.4649


30) A firm believes the internal rate of return for its proposed investment can best be described by a normal

distribution with mean 47% and standard deviation 3%. What is the probability that the internal rate of return

for the investment will be at least 42.5%?

31) A firm believes the internal rate of return for its proposed investment can best be described by a normal

distribution with mean 25% and standard deviation 3%. What is the probability that the internal rate of return

for the investment exceeds 31%?

32) Farmers often sell fruits and vegetables at roadside stands during the summer. One such roadside stand has a

daily demand for tomatoes that is approximately normally distributed with a mean equal to 571 tomatoes per

day and a standard deviation equal to 30 tomatoes per day. If there are 529 tomatoes available to be sold at the

roadside stand at the beginning of a day, what is the probability that they will all be sold?


33) Given a distribution that follows a standard normal curve, what does the graph of the curve do as z increases

in the positive direction?

A) The graph of the curve approaches zero.

B) The graph of the curve approaches 1.

C) The graph of the curve approaches an inflection point.

D) The graph of the curve eventually intersects the horizontal axis.

2 Find the value of a normal random variable.


Find the indicated z-score.

34) Find the z-score for which the area under the standard normal curve to its left is 0.96

A) 1.75 B) 1.82 C) 1.03 D) -1.38

35) Find the z-score for which the area under the standard normal curve to its left is 0.40

A) -0.25 B) 0.25 C) 0.57 D) -0.57

36) Find the z-score for which the area under the standard normal curve to its left is 0.09.

A) -1.34 B) -1.39 C) -1.26 D) -1.45


A) -1.75 B) -1.89 C) -1.48 D) -1.63


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

39) Find the z-score for which the area under the standard normal curve to its right is 0.07.

A) 1.48 B) 1.39 C) 1.26 D) 1.45


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


A) 1.34 B) 1.39 C) 1.26 D) 1.45

42) Find the z-score having area 0.86 to its right under the standard normal curve; that is, find z 0.86 .

A) -1.08 B) 1.08 C) 0.8051 D) 0.5557

43) For a standard normal curve, find the z-score that separates the bottom 90% from the top 10%.

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


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


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

46) Determine the two z-scores that separate the middle 87.4% of the distribution from the area in the tails of the

standard normal distribution.

A) -1.53, 1.53 B) -1.39, 1.39 C) -1.46, 1.46 D) -1.45, 1.45

47) Determine the two z-scores that separate the middle 96% of the distribution from the area in the tails of the

standard normal distribution.

A) -2.05 and 2.05 B) -1.75 and 1.75 C) 0 and 2.05 D) -2.33 and 2.33

48) 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)

49) 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)

Find the value of zα.

50) z0.2

A) 0.84 B) -0.84 C) 0.58 D) 0.8

Find the indicated percentile.

51) Assume that the random variable X is normally distributed with mean μ = 70 and standard deviation σ = 8.

Find the 40th percentile for X.

A) 68 B) 72 C) 64.72 D) 66.4




standard deviation of 60 seconds. The fitness association wants to recognize the boys whose times are among

the top (or fastest) 10% with certificates of recognition. What time would the boys need to beat in order to earn

a certificate of recognition from the fitness association?

A) 393.2 sec B) 371.3 sec C) 546.8 sec D) 568.7 sec



standard deviation of 40 seconds. Between what times do we expect most (approximately 95%) of the boys to

run the mile?

A) between 371.6 and 528.4 sec B) between 384.2 and 515.824 sec

C) between 355 and 545 sec D) between 0 and 515.824 sec

54) The amount of corn chips dispensed into a 24-ounce bag by the dispensing machine has been identified as

possessing a normal distribution with a mean of 24.5 ounces and a standard deviation of 0.2 ounce. What chip

amount represents the 67th percentile for the bag weight distribution? Round to the nearest hundredth.

A) 24.59 oz B) 24.09 oz C) 24.63 oz D) 24.13 oz

55) Suppose a brewery has a filling machine that fills 12-ounce bottles of beer. It is known that the amount of beer


deviation of 0.04 ounce. The company is interested in reducing the amount of extra beer that is poured into the

12 ounce bottles. The company is seeking to identify the highest 1.5% of the fill amounts poured by this

machine. For what fill amount are they searching? Round to the nearest thousandth.

A) 12.317 oz B) 11.913 oz C) 12.143 oz D) 12.087 oz


distribution with a mean of 4.5 minutes and a standard deviation of 1 minute. Find the cut-off time which

75.8% of the college students exceed when trying to find a parking spot in the library parking lot.

A) 5.2 min B) 5.0 min C) 4.8 min D) 5.3 min

57) The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a normal distribution with

a standard deviation of 0.12 ounce. Every can that has more than 12.30 ounces of soda poured into it causes a

spill and the can needs to go through a special cleaning process before it can be sold. What is the mean amount

of soda the machine should dispense if the company wants to limit the percentage that need to be cleaned

because of spillage to 3%?

A) 12.0744 oz B) 12.5256 oz C) 12.0396 oz D) 12.5604 oz


by employees. Before the new system was installed, the amount being spent on personal calls follows a normal

distribution with an average of $400 per month and a standard deviation of $50 per month. Refer to such

expenses as PCEʹs (personal call expenses). Find the point in the distribution below which 2.5% of the PCEʹs

fell.

A) $302.00 B) $498.00 C) $10.00 D) $390.00

59) A brewery has a beer dispensing machine that dispenses beer into the companyʹs 12 ounce bottles. The

distribution for the amount of beer dispensed by the machine follows a normal distribution with a standard

deviation of 0.19 ounce. The company can control the mean amount of beer dispensed by the machine. What

value of the mean should the company use if it wants to guarantee that 98.5% of the bottles contain at least 12

ounces (the amount on the label)? Round to the nearest thousandth.

A) 12.412 oz B) 12.462 oz C) 12.003 oz D) 12.001 oz


mean of 60,000 miles and a standard deviation of 2400 miles. What warranty should the company use if they

want 96% of the tires to outlast the warranty?

A) 55,800 mi B) 64,200 mi C) 57,600 mi D) 62,400 mi


61) The board of examiners that administers the real estate brokerʹs examination in a certain state found that the

mean score on the test was 494 and the standard deviation was 72. If the board wants to set the passing score

so that only the best 10% of all applicants pass, what is the passing score? Assume that the scores are normally

distributed.

62) The board of examiners that administers the real estate brokerʹs examination in a certain state found that the

mean score on the test was 527 and the standard deviation was 72. If the board wants to set the passing score

so that only the best 80% of all applicants pass, what is the passing score? Assume that the scores are normally

distributed.

63) Farmers often sell fruits and vegetables at roadside stands during the summer. One such roadside stand has a

daily demand for tomatoes that is approximately normally distributed with a mean equal to 125 tomatoes per

day and a standard deviation equal to 30 tomatoes per day. How many tomatoes must be available on any

given day so that there is only a 1.5% chance that all tomatoes will be sold?

7.3 Assessing Normality

1 Use normal probability plots to assess normality.


Use a normal probability plot to asses whether the sample data could have come from a population that is normally

distributed.

1) An industrial psychologist conducted an experiment in which 40 employees that were identified as ʺchronically

tardyʺ by their managers were divided into two groups of size 20. Group 1 participated in the new ʺItʹs Great to

be Awake!ʺ program, while Group 2 had their pay pay docked. The following data represent the number of

minutes that employees in Group 1 were late for work after participating in the program.

118 126 141 145 141

157 142 177 132 135

147 123 132 147 152

153 153 154 125 147

A) normally distributed B) not normally distributed

2) The following data represent a random sample of the number of shares of a pharmaceutical companyʹs stock

traded for 20 days in 2000.

4.28 9.2 10.7 7.82 14.14

5.59 9.99 9.24 12.4 13.55

4.54 8.84 11.89 7.05 27.84

8.01 10.44 4.58 7.54 11.18

A) not normally distributed B) normally distributed


3) A normal probability plot is a graph that plots _____________ versus _____________.

A) observed data, normal scores B) normal score, observed data

C) normal data, observed scores D) observed scores, normal data

4) If sample data are taken from a population that is normally distributed, a normal probability plot of the

observed data values versus the expected z scores will

A) be approximately linear. B) have no discernable pattern.

C) look exponential in nature. D) have a correlation coefficient near 0.

7.4 The Normal Approximation to the Binomial Probability Distribution

1 Approximate binomial probabilities using the normal distribution.


Compute P(x) using the binomial probability formula. Then determine whether the normal distribution can be used as

an approximation for the binomial distribution. If so, approximate P(x) and compare the result to the exact probability.

1) n = 80, p = 0.3, x = 20 (Round the standard deviation to three decimal places to work the problem.)

A) Exact: 0.0626; Approximate: 0.0620 B) Exact: 0.0626; Approximate: 0.0628

C) Exact: 0.0634; Approximate: 0.0628 D) Exact: 0.0626; Approximate: 0.0612


2) A student answers all 48 questions on a multiple-choice test by guessing. Each question has four possible

answers, only one of which is correct. Find the probability that the student gets exactly 15 correct answers. Use

the normal distribution to approximate the binomial distribution.

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

3) If the probability of a newborn kitten being female is 0.5, find the probability that in 100 births, 55 or more will

be female. Use the normal distribution to approximate the binomial distribution.

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

4) A local concert center reports that it has been experiencing a 15% rate of no-shows on advanced reservations.

Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows. Round the

standard deviation to three decimal places to work the problem.

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

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

A) 0.1190 B) 0.8810 C) 0.0871 D) 0.3875

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

A) 0.9147 B) 0.1190 C) 0.8810 D) 0.0853

7) A salesperson found that there was a 1% chance of a sale from her phone solicitations. Find the probability of

getting 5 or more sales for 1000 telephone calls. Round the standard deviation to three decimal places to work

the problem.

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


8) The author of an economics book has trouble deciding whether to use the words ʺheʺ or ʺsheʺ in the bookʹs

examples. To solve the problem, the author flips a coin each time the problem arises. If a head shows, the

author uses ʺheʺ and if a tail shows, the author uses ʺsheʺ. If this problem occurs 100 times in the book, what is

the probability that ʺsheʺ will be used 58 times?

9) In a recent survey, 86% of the community favored building more parks in their neighborhood. You randomly

select 22 citizens and ask each if he or she thinks the community needs more parks. Decide whether you can

use the normal distribution to approximate the binomial distribution. If so, find the mean and standard

deviation. If not, explain why.

10) A recent survey found that 72% of all adults over 50 own cell phones. You randomly select 31 adults over 50,

and ask if he or she owns a cell phone. Decide whether you can use the normal distribution to approximate the

binomial distribution. If so, find the mean and standard deviation, If not, explain why. Round to the nearest

hundredth when necessary.

11) According to government data, the probability than an adult never had the flu is 14%. You randomly select 66

adults and ask if he or she ever had the flu. Decide whether you can use the normal distribution to

approximate the binomial distribution, If so, find the mean and standard deviation, If not, explain why. Round

to the nearest hundredth when necessary.

12) For the following conditions, determine if it is appropriate to use the normal distribution to approximate a

binomial distribution with n = 7 and p = 0.1.

13) For the following conditions, determine if it is appropriate to use the normal distribution to approximate a

binomial distribution with n = 35 and p = 0.5.

14) The failure rate in a German class is 30%. In a class of 50 students, find the probability that exactly five students

will fail. Use the normal distribution to approximate the binomial distribution. Round the standard deviation to

three decimal places to work the problem.

15) A local rental car agency has 50 cars. The rental rate for the winter months is 60%. Find the probability that in a

given winter month at least 35 cars will be rented. Use the normal distribution to approximate the binomial

distribution. Round the standard deviation to three decimal places to work the problem.

16) A local rental car agency has 200 cars. The rental rate for the winter months is 60%. Find the probability that in

a given winter month fewer than 140 cars will be rented. Use the normal distribution to approximate the

binomial distribution.


17) True or False: In order to use a normal approximation to the binomial probability distribution, np(1 - p) ≥ 10.

A) True B) False

18) Assuming that all conditions are met to approximate a binomial probability distribution with the standard

normal distribution, then to compute P(x ≥ 19) from the binomial distribution we must compute

as the normal approximation.

A) P(x ≥ 18.5) B) P(x ≤ 18.5) C) P(x ≥ 19.1) D) P(x ≤ 18.9)


normal distribution, then to compute P(x ≤ 23) from the binomial distribution we must compute


A) P(x ≤ 23.5) B) P(x ≥ 23.5) C) P(x ≤ 22.9) D) P(x ≤ 23.1)


normal distribution, then to compute P(12 ≤ x ≤ 15) from the binomial distribution we must compute


A) P(11.5 < x < 15.5) B) P(12.5 < x < 14.5)

C) P(x > 15.5) and P(x < 11.5) D) P(x > 12.5) and P(x < 14.5)

Ch. 7 The Normal Probability DistributionAnswer Key

7.1 Properties of the Normal Distribution1 Use the uniform probability distribution.

1) A

2) A

3) A

4) A

5) A

6) A

7) A

8) A

9) A

2 Graph a normal curve.

10) A

11) A

12) A

3 State the properties of the normal curve.

13) (a)

It is not reasonable to assume that the heights are normally distributed since the histogram is skewed.

(b) μ = 31, σ = 3.86

(c) Yes. The mean and standard deviation are close.

14) A

15) A

16) A

17) A

18) A

19) A

20) A

21) A

22) A

23) A

24) D

4 Explain the role of area in the normal density function.

25) (a), (b)

(c) The two interpretations are: (1) the proportion of test takers who scored less than 725 is 0.8413 and (2) the

probability that a randomly selected test taker has a score less than 725 is 0.8413.

26) (a), (b)

(c) The two interpretations are: (1) the proportion of hyraxes who weighed more than 2930 is 0.0228 and (2) the

probability that a randomly selected hyrax weighs more than 2930 is 0.0228.

27) (a), (b)

(c) The two interpretations are: (1) the proportion of mpg between 29.3 and 23.7 is 0.6419 and (2) the probability that a

randomly selected mpg is between 29.3 and 23.7 is 0.6419.

28) A

29) A

30) A

7.2 Applications of the Normal Distribution1 Find and interpret the area under a normal curve.

1) A

2) A

3) A

4) A

5) A

6) A

7) A

8) A

9) A

10) A

11) A

12) A

13) A

14) A

15) A

16) A

17) A

18) A

19) A

20) A

21) A

22) A

23) A

24) A

25) A

26) A

27) A

28) A

29) A

30) Let x be the internal rate of return. Then x is a normal random variable with μ = 47% and σ = 3%. To determine the

probability that x is at least 42.5%, we need to find the z-value for x = 42.5%.

z = x - μ

σ =

42.5 - 47

3 = -1.5

P(x ≥ 42.5%) = P(z ≥ -1.5) = 1 - P(z ≤ -1.5) = 1 - 0.0668 = 0.9332

31) Let x be the internal rate of return. Then x is a normal random variable with μ = 25% and σ = 3%. To determine the

probability that x exceeds 31%, we need to find the z-value for x = 31%.

z = x - μ

σ =

31 - 25

3 = 2

P(x > 31%) = P(z ≥ 2) = 1 - P(z ≤ 2) = 1 - 0.9772 = 0.0228

32) Let x be the number of tomatoes sold per day. Then x is a normal random variable with μ = 571 and σ = 30.

To determine if all 83 tomatoes will be sold, we need to find the z-value for x = 529.

z = x - μ

σ =

529 - 571

30 = -1.4

P(x ≥ 529) = P(z ≥ -1.4) = 1 - P(z ≤ -1.4) = 1 - 0.0808 = 0.9192

33) A

2 Find the value of a normal random variable.

34) A

35) A

36) A

37) A

38) A

39) A

40) A

41) A

42) A

43) A

44) A

45) A

46) A

47) A

48) A

49) A

50) A

51) A

52) A

53) A

54) A

55) A

56) A

57) A

58) A

59) A

60) A

61) Let x be a score on this exam. Then x is a normally distributed random variable with μ = 494 and σ = 72. We want to

find the value of x0, such that P(x > x0) = 0.10. The z-score for the value x = x0 is

z = x0 - μ

σ =

x0 - 494

72.

P(x > x0) = P z > x0 - 494

72 = 0.10

We find x0 - 494

72 ≈ 1.28.

x0 - 494 = 1.28(72) ⇒ x0 = 494 + 1.28(72) = 586.16

62) Let x be a score on this exam. Then x is a normally distributed random variable with μ = 527 and σ = 72. We want to

find the value of x0, such that P(x > x0) = 0.80. The z-score for the value x = x0 is

z = x0 - μ

σ =

x0 - 527

72.

P(x > x0) = P z > x2 - 527

72 = 0.80

We find x0 - 527

72 ≈ -0.84.

x0 - 527 = -0.84(72) ⇒ x0 = 527 - 0.84(72) = 466.52

63) Let x be the number of tomatoes sold per day. Then x is a normal random variable with μ = 125 and σ = 30.

We want to find the value x0, such that P(x > x0) = .015. The z-value for the point x = x0 is

z = x - μ

σ =

x0 - 125

30.

P(x > x0) = P(z > x0 - 125

30)= 0.015

We find x0 - 125

30 = 2.17

x0 - 125 = 2.17(30) ⇒ x0 = 125 + 2.17(30) = 190

7.3 Assessing Normality1 Use normal probability plots to assess normality.

1) A

2) A

3) A

4) A

7.4 The Normal Approximation to the Binomial Probability Distribution1 Approximate binomial probabilities using the normal distribution.

1) A

2) A

3) A

4) A

5) A

6) A

7) A

8) P(57.5 < x < 58.5) = P(1.50 < z < 1.70) = 0.9554 - 0.9332 = 0.0222

9) cannot use normal distribution, nq = (22)(0.14) = 3.08 < 5

10) use normal distribution, μ = 22.32 and σ = 2.5.

11) use normal distribution, μ = 9.24 and σ = 2.82.

12) cannot use normal distribution

13) can use normal distribution

14) P(4.5 < X < 5.5) = P(-3.24 < z < -2.93) = 0.0017 - 0.0006 = 0.0011

15) P(x ≥ 34.5) = 0.0968

16) P(x ≤ 139.5) = 0.9975

17) A

18) A

19) A

20) A

Ch. 7 The Normal Probability Distributionsite.iugaza.edu.ps/mriffi/files/2018/02/Statistics... ·...

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