Download Sst4e Tif 07 Last modified January 9, 2017 at 1:21 pm

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 64°F
to 92°F. What is the probability that a randomly selected August day has a high temperature that exceeded
69°F?
A) 0.8214
B) 0.1786
C) 0.4423
D) 0.0357
2) High temperatures in a certain city for the month of August follow a uniform distribution over the interval 73°F
to 103°F. Find the high temperature which 90% of the August days exceed.
A) 76°F
B) 100°F
C) 83°F
D) 103°F
3) The diameter of ball bearings produced in a manufacturing process can be explained using a uniform
distribution over the interval 7.5 to 9.5 millimeters. What is the probability that a randomly selected ball
bearing has a diameter greater than 8.6 millimeters?
A) 0.45
B) 0.9053
C) 4
D) 0.5059
4) Suppose x is a uniform random variable with c = 10 and d = 60. Find the probability that a randomly selected
observation exceeds 30.
A) 0.6
B) 0.4
C) 0.1
D) 0.9
5) Suppose x is a uniform random variable with c = 30 and d = 80. Find the probability that a randomly selected
observation is between 33 and 75.
A) 0.84
B) 0.16
C) 0.5
D) 0.8
6) A machine is set to pump cleanser into a process at the rate of 5 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 4.5 to 7.5 gallons per minute. Find the probability that between 5.0 gallons and 6.0 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 70. What is the probability that this experiment results in an outcome less than 40?
A) 0.25
B) 0.1
C) 0.31
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
Page 216
2 Graph a normal curve.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Provide an appropriate response.
1) 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.
2) 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.
3) Draw a normal curve with μ = 50 and σ = 16. Label the mean and the inflection points.
A)
34 50
66
B)
18
50
82
Page 217
C)
42 50
58
D)
34
50
66
3 State the properties of the normal curve.
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Provide an appropriate response.
1) 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.
Page 218
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
2) The graph of a normal curve is given. Use the graph to identify the value of μ and σ.
3
6
A) μ = 12, σ = 3
9 12
15
18 B) μ = 3, σ = 12
21
C) μ = 12, σ = 9
D) μ = 9, σ = 12
3) 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
4) 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
5) Approximately ____% of the area under the normal curve is between μ - 3σ and μ + 3σ.
A) 99.7
B) 68
C) 95
D) 50
Determine whether the graph can represent a normal curve. If it cannot, explain why.
6)
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.
Page 219
7)
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.
8)
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.
D) The graph can represent a normal density function.
Page 220
9)
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.
10)
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.
Page 221
11)
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.
D) The graph can represent a normal density function.
12)
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.
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Provide an appropriate response.
1) 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.
Page 222
2) The weight of 2-year old hyraxes is known to be normally distributed with mean
μ = 2200 grams and standard deviation σ = 365 grams
(a) Draw a normal curve with the parameters labeled.
(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.
3) 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. (a) Draw a normal curve with the parameters labeled.
(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.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
4) 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
5) 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
6) 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.
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 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
Page 223
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 = 0.3 and z = 1.4
A) 0.3013
B) 0.6179
C) 0.9192
D) 0.3821
13) z = -0.7 and z = 0.7
A) 0.516
B) 0.242
C) 0.758
D) 0.5
14) z = -2 and z = -0.1
A) 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 = 60 and standard deviation σ = 16.
Compute the probability P(X < 80).
A) 0.8944
B) 0.8849
C) 0.1056
D) 0.9015
16) Assume that the random variable X is normally distributed, with mean μ = 100 and standard deviation σ = 15.
Compute the probability P(X > 112).
A) 0.2119
B) 0.1977
C) 0.7881
D) 0.2420
17) Assume that the random variable X is normally distributed, with mean μ = 50 and standard deviation σ = 16.
Compute the probability P(6 < X < 70).
A) 0.8914
B) 0.8819
C) 0.8944
D) 0.7888
Provide an appropriate response.
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 450 seconds and a
standard deviation of 50 seconds. Find the probability that a randomly selected boy in secondary school can
run the mile in less than 335 seconds.
A) 0.0107
B) 0.4893
C) 0.9893
D) 0.5107
19) 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 60 seconds. Find the probability that a randomly selected boy in secondary school will
take longer than 302 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 12.13 ounces and a standard
deviation of 0.04 ounce. Find the probability that the bottle contains fewer than 12.03 ounces of beer.
A) 0.0062
B) 0.4938
C) 0.9938
D) 0.5062
Page 224
21) 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 13.14 onces and a standard
deviation of 0.04 ounce. Find the probability that the bottle contains more than 13.14 ounces of beer.
A) 0.5
B) 1
C) 0
D) 0.4
22) 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 12.29 ounces and a standard
deviation of 0.04 ounce. Find the probability that the bottle contains between 12.19 and 12.25 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 3.0 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 2.5 minutes.
A) 0.3085
B) 0.1915
C) 0.3551
D) 0.2674
24) 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 take between 5.0 and 7.5 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 12.48 ounces and a standard deviation of 0.32 ounce. The cans only hold 12.80 ounces of soda. Every
can that has more than 12.80 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
27) 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 follows a normal
distribution with an average of $900 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 $750.
A) 0.0013
B) 0.8333
C) 0.1667
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 2900 miles. What is the probability a particular tire of this
brand will last longer than 57,100 miles?
A) 0.8413
B) 0.1587
C) 0.2266
D) 0.7266
29) 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 2900 miles. What is the probability a certain tire of this brand
will last between 53,910 miles and 54,780 miles?
A) 0.0180
B) 0.9813
C) 0.4920
D) 0.4649
Page 225
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
30) A firm believes the internal rate of return for its proposed investment can best be described by a normal
distribution with mean 26% and standard deviation 3%. What is the probability that the internal rate of return
for the investment will be at least 21.5%?
31) A firm believes the internal rate of return for its proposed investment can best be described by a normal
distribution with mean 36% and standard deviation 3%. What is the probability that the internal rate of return
for the investment exceeds 42%?
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 413 tomatoes per
day and a standard deviation equal to 30 tomatoes per day. If there are 371 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?
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
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.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find the indicated z-score.
1) 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
2) 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
3) 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
4) Find the z-score for which the area under the standard normal curve to its left is 0.04.
A) -1.75
B) -1.89
C) -1.48
D) -1.63
5) Find the z-score for which the area under the standard normal curve to its left is 0.70.
A) 0.53
B) 0.98
C) 0.81
D) 0.47
6) 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
7) Find the z-score for which the area under the standard normal curve to its right is 0.70.
A) -0.53
B) -0.98
C) -0.81
D) -0.47
8) Find the z-score for which the area under the standard normal curve to its right is 0.09.
A) 1.34
B) 1.39
C) 1.26
D) 1.45
9) Find the z-score having area 0.86 to its right under the standard normal curve; that is, find z
A) -1.08
B) 1.08
C) 0.8051
Page 226
.
0.86
D) 0.5557
10) 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
11) For a standard normal curve, find the z-score that separates the bottom 30% from the top 70%.
A) -0.53
B) -0.98
C) -0.47
D) -0.12
12) For a standard normal curve, find the z-score that separates the bottom 70% from the top 30%.
A) 0.53
B) 0.98
C) 0.47
D) 0.12
13) 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.
B) -1.39, 1.39
C) -1.46, 1.46
D) -1.45, 1.45
A) -1.53, 1.53
14) 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
15) 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)
16) 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 z α.
17) z 0.05
A) 1.645
B) -1.645
C) 0.52
D) 1.75
Find the indicated percentile.
18) Assume that the random variable X is normally distributed with mean μ = 50 and standard deviation σ = 12.
Find the 24th percentile for X.
A) 41.48
B) 58.52
C) 42.92
D) 38.6
Provide an appropriate response.
19) 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 50 seconds. The fitness association wants to recognize the fastest 10% of the boys 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) 376 sec
B) 504 sec
C) 522.25 sec
D) 357.75 sec
20) 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 460 seconds and a
standard deviation of 60 seconds. Between what times do we expect most (approximately 95%) of the boys to
run the mile?
B) between 361.3 and 558.736 sec
A) between 342.4 and 577.6 sec
C) between 365 and 555 sec
D) between 0 and 558.736 sec
21) The amount of corn chips dispensed into a 16-ounce bag by the dispensing machine has been identified as
possessing a normal distribution with a mean of 16.5 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) 16.59 oz
B) 16.09 oz
C) 16.63 oz
D) 16.13 oz
Page 227
22) 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 12.47 ounces and a standard
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) 12.557 oz
B) 11.913 oz
C) 12.383 oz
D) 12.087 oz
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 5.5 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) 6.2 min
B) 6.0 min
C) 5.8 min
D) 6.3 min
24) The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a normal distribution with
a standard deviation of 0.06 ounce. Every can that has more than 12.15 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) 12.0372 oz
B) 12.2628 oz
C) 12.0198 oz
D) 12.2802 oz
25) 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 follows a normal
distribution with an average of $800 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) $702.00
B) $898.00
C) $20.00
D) $780.00
26) 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 0.10 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) 12.217 oz
B) 12.243 oz
C) 12.002 oz
D) 12.001 oz
27) 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 3000 miles. What warranty should the company use if they
want 96% of the tires to outlast the warranty?
A) 54,750 mi
B) 65,250 mi
C) 57,000 mi
D) 63,000 mi
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
28) The board of examiners that administers the real estate brokerʹs examination in a certain state found that the
mean score on the test was 493 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.
29) The board of examiners that administers the real estate brokerʹs examination in a certain state found that the
mean score on the test was 426 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.
30) 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?
Page 228
7.3 Assessing Normality
1 Use normal probability plots to assess normality.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Use a normal probability plot to asses whether the sample data could have come from a population that is normally
distributed.
1) Determine whether the following normal probability plot indicates that the sample data could have come from
a population that is normally distributed.
A) normally distributed
B) not normally distributed
2) Determine whether the following normal probability plot indicates that the sample data could have come from
a population that is normally distributed.
A) normally distributed
B) not normally distributed
Page 229
3) Determine whether the following normal probability plot indicates that the sample data could have come from
a population that is normally distributed.
A) not normally distributed
B) normally distributed
4) Determine whether the following normal probability plot indicates that the sample data could have come from
a population that is normally distributed.
A) not normally distributed
B) normally distributed
5) 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.
124 132 147 151 147
163 148 183 138 141
153 129 138 153 158
159 159 160 131 153
A) normally distributed
B) not normally distributed
6) The following data represent a random sample of the number of shares of a pharmaceutical companyʹs stock
traded for 20 days in 2000.
4.71 9.63 11.13 8.25 14.57
6.02 10.42 9.67 12.83 13.98
4.97 9.27 12.32 7.48 28.27
8.44 10.87 5.01 7.97 11.61
A) not normally distributed
B) normally distributed
Page 230
Provide an appropriate response.
7) 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
8) 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
B) have no discernable pattern.
A) be approximately linear.
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.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
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 = 0.7, x = 62 (Round the standard deviation to three decimal places to work the problem.)
A) Exact: 0.0343; Approximate: 0.0342
B) Exact: 0.0343; Approximate: 0.0351
C) Exact: 0.0352; Approximate: 0.0351
D) Exact: 0.0343; Approximate: 0.0333
Provide an appropriate response.
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
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
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?
Page 231
9) In a recent survey, 84% of the community favored building more parks in their neighborhood. You randomly
select 19 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 73% of all adults over 50 own cell phones. You randomly select 42 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 15%. You randomly select 75
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 = 6 and p = 0.2.
13) For the following conditions, determine if it is appropriate to use the normal distribution to approximate a
binomial distribution with n = 38 and p = 0.6.
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 100 cars. The rental rate for the winter months is 60%. Find the probability that in
a given winter month at least 70 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.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
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)
19) Assuming that all conditions are met to approximate a binomial probability distribution with the standard
normal distribution, then to compute P(x ≤ 23) from the binomial distribution we must compute
as the normal approximation.
A) P(x ≤ 23.5)
B) P(x ≥ 23.5)
C) P(x ≤ 22.9)
D) P(x ≤ 23.1)
20) Assuming that all conditions are met to approximate a binomial probability distribution with the standard
normal distribution, then to compute P(12 ≤ x ≤ 15) from the binomial distribution we must compute
as the normal approximation.
A) P(11.5 < x < 15.5)
C) P(x > 15.5) and P(x < 11.5)
B) P(12.5 < x < 14.5)
D) P(x > 12.5) and P(x < 14.5)
Page 232
Ch. 7 The Normal Probability Distribution
Answer Key
7.1 Properties of the Normal Distribution
1 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.
1) A
2) A
3) A
3 State the properties of the normal curve.
1) (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.
2) A
3) A
4) A
5) A
6) A
7) A
8) A
9) A
10) A
11) A
12) D
Page 233
4 Explain the role of area in the normal density function.
1) (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.
2) (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.
3) (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.
4) A
5) A
6) A
7.2 Applications of the Normal Distribution
1 Find and interpret the area under a normal curve.
1) A
2) A
3) A
Page 234
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 = 26% and σ = 3%. To determine the
probability that x is at least 21.5%, we need to find the z-value for x = 21.5%.
x - μ 21.5 - 26
= = -1.5
z = 3
σ
P(x ≥ 21.5%) = 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 = 36% and σ = 3%. To determine the
probability that x exceeds 42%, we need to find the z-value for x = 42%.
z = x - μ 42 - 36
= = 2
3
σ
P(x > 42%) = 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 = 413 and σ = 30.
To determine if all 83 tomatoes will be sold, we need to find the z -value for x = 371.
z = x - μ 371 - 413
= = -1.4
30
σ
P(x ≥ 371) = 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.
1) A
2) A
3) A
4) A
5) A
6) A
Page 235
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) Let x be a score on this exam. Then x is a normally distributed random variable with μ = 493 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 - 493
72
P(x > x0 ) = P z > We find x0 - 493
72
.
x0 - 493
72
= 0.10
≈ 1.28.
x0 - 493 = 1.28(72) ⇒ x0 = 493 + 1.28(72) = 585.16
29) Let x be a score on this exam. Then x is a normally distributed random variable with μ = 426 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 - 426
72
P(x > x0 ) = P z > We find x0 - 426
72
.
x2 - 426
72
= 0.80
≈ -0.84.
x0 - 426 = -0.84(72) ⇒ x0 = 426 - 0.84(72) = 365.52
30) 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 x0 - 125
= .
σ
30
P(x > x0 ) = P(z > We find x0 - 125
30
x0 - 125
30
)= 0.015
= 2.17
x0 - 125 = 2.17(30) ⇒ x0 = 125 + 2.17(30) = 190
Page 236
7.3 Assessing Normality
1 Use normal probability plots to assess normality.
1) A
2) A
3) A
4) A
5) A
6) A
7) A
8) A
7.4 The Normal Approximation to the Binomial Probability Distribution
1 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 = (19)(0.16) = 3.04 < 5
10) use normal distribution, μ = 30.66 and σ = 2.88.
11) use normal distribution, μ = 11.25 and σ = 3.09.
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 ≥ 69.5) = 0.0262
16) P(x ≤ 139.5) = 0.9975
17) A
18) A
19) A
20) A
Page 237