Statistics 12th Edition by James T. McClave-Terry T Sincich - Test Bank

Statistics 12th Edition by James T. McClave-Terry T Sincich – Test Bank

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Ch. 5 Continuous Random Variables
5.1 Continuous Probability Distributions
1 Understand Probability Distributions for Continuous Random Variable
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Answer the question True or False.
1) The number of children in a family can be modelled using a continuous random variable.
A) True B) False
2) For any continuous probability distribution, P(x = c) = 0 for all values of c.
A) True B) False
3) The total area under a probability distribution equals 1.
A) True B) False
4) For a continuous probability distribution, the probability that x is between a and b is the same regardless of
whether or not you include the endpoints, a and b, of the interval.
A) True B) False
5.2 The Uniform Distribution
1 Find Probability
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find the probability.
1) Suppose x is a random variable best described by a uniform probability distribution with c = 30 and d = 90.
Find P(30 ≤ x ≤ 45).
A) 0.25 B) 0.15 C) 0.35 D) 0.025
2) Suppose x is a random variable best described by a uniform probability distribution with c = 10 and d = 50.
Find P(40 ≤ x ≤ 50).
A) 0.25 B) 0.1 C) 0.35 D) 0.025
3) Suppose x is a random variable best described by a uniform probability distribution with c = 10 and d = 90.
Find P(x < 42).
A) 0.4 B) 0.32 C) 0.5 D) 0.04
4) Suppose x is a random variable best described by a uniform probability distribution with c = 20 and d = 60.
Find P(x > 52).
A) 0.2 B) 0.08 C) 0.8 D) 0.02
5) Suppose x is a random variable best described by a uniform probability distribution with c = 20 and d = 60.
Find P(x > 60).
A) 0 B) 1 C) 0.5 D) 0.4
6) Suppose x is a random variable best described by a uniform probability distribution with c = 60 and d = 20.
Find P(x ≥ 60).
A) 1 B) 0 C) 0.5 D) 0.4
7) Suppose x is a uniform random variable with c = 20 and d = 60. Find P(x > 44).
A) 0.4 B) 0.6 C) 0.1 D) 0.9
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8) Suppose x is a uniform random variable with c = 20 and d = 50. Find P(23 < x < 45).
Round to the nearest hundredth when necessary.
A) 0.73 B) 0.27 C) 0.5 D) 1
Solve the problem.
9) Suppose x is a random variable best described by a uniform probability distribution with c = 3 and d = 5. Find
the value of a that makes the following probability statement true: P(x ≤ a) = 0.75.
A) 4.5 B) 3.5 C) 4.7 D) 1.5
10) Suppose x is a random variable best described by a uniform probability distribution with c = 6 and d = 14. Find
the value of a that makes the following probability statement true: P(x ≥ a) = 0.6.
A) 9.2 B) 10.8 C) 9.4 D) 4.8
11) Suppose x is a random variable best described by a uniform probability distribution with c = 3 and d = 9. Find
the value of a that makes the following probability statement true: P(3.5 ≤ x ≤ a) = 0.5.
A) 6.5 B) 6 C) 4 D) 1.2
12) Suppose x is a random variable best described by a uniform probability distribution with c = 2 and d = 6. Find
the value of a that makes the following probability statement true: P(x ≤ a) = 1.
A) a ≥ 6 B) a ≥ 2 C) a ≤ 6 D) a ≤ 2
13) High temperatures in a certain city for the month of August follow a uniform distribution over the interval 68°F
to 90°F. What is the probability that the high temperature on a day in August exceeds 73°F?
A) 0.7727 B) 0.2273 C) 0.462 D) 0.0455
14) High temperatures in a certain city for the month of August follow a uniform distribution over the interval 69°F
to 99°F. Find the temperature which is exceeded by the high temperatures on 90% of the days in August.
A) 72°F B) 96°F C) 79°F D) 99°F
15) The diameters of ball bearings produced in a manufacturing process can be described 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 4.6 millimeters?
A) 0.45 B) 0.8364 C) 2 D) 0.5111
16) The diameters of ball bearings produced in a manufacturing process can be described using a uniform
distribution over the interval 2.5 to 4.5 millimeters. Any ball bearing with a diameter of over 4.25 millimeters or
under 2.75 millimeters is considered defective. What is the probability that a randomly selected ball bearing is
defective?
A) .25 B) .50 C) .75 D) 0
17) A machine is set to pump cleanser into a process at the rate of 7 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 6.5 to 9.5 gallons per minute. What is the probability that at the time the machine is checked it is
pumping more than 8.0 gallons per minute?
A) .50 B) .7692 C) .667 D) .25
18) A machine is set to pump cleanser into a process at the rate of 10 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 10.0 to 13.0 gallons per minute. Would you expect the machine to pump more than 12.85 gallons per
minute?
A) No, since .05 is a low probability. B) Yes, since .95 is a high probability.
C) Yes, since .05 is a high probability. D) No, since .95 is a high probability.
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19) A machine is set to pump cleanser into a process at the rate of 9 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 8.5 to 11.5 gallons per minute. Find the probability that between 9.0 gallons and 10.0 gallons are
pumped during a randomly selected minute.
A) 0.33 B) 0.67 C) 0 D) 1
20) Suppose a uniform random variable can be used to describe the outcome of an experiment with outcomes
ranging from 40 to 80. What is the probability that this experiment results in an outcome less than 50? Round to
the nearest hundredth when necessary.
A) 0.25 B) 0.08 C) 0.31 D) 1
21) The age of customers at a local hardware store follows a uniform distribution over the interval from 18 to 60
years old. Find the probability that the next customer who walks through the door exceeds 50 years old. Round
to the nearest ten-thousandth.
A) 0.8333 B) 0.3600 C) 0.2381 D) 0.7619
22) After a particular heavy snowstorm, the depth of snow reported in a mountain village followed a uniform
distribution over the interval from 15 to 22 inches of snow. Find the probability that a randomly selected
location in this village had between 17 and 18 inches of snow. Round to the nearest ten-thousandth.
A) 0.2857 B) 0.1429 C) 0.4286 D) 0.5714
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
23) High temperatures in a certain city for the month of August follow a uniform distribution over the interval
75°F to 95°F. What is the probability that a random day in August has a high temperature that exceeds 80°F?
24) The diameters of ball bearings produced in a manufacturing process can be described using a uniform
distribution over the interval 2.5 to 8.5 millimeters. What is the probability of a randomly selected ball bearing
having a diameter less than 4.5 millimeters?
25) A machine is set to pump cleanser into a process at the rate of 9 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 8.5 to 9.5 gallons per minute. Find the probability that the machine pumps less than 8.75 gallons during
a randomly selected minute.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
26) Which geometric shape is used to represent areas for a uniform distribution?
A) Rectangle B) Circle C) Bell curve D) Triangle
2 Find Mean, Variance, Standard Deviation
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
27) The diameters of ball bearings produced in a manufacturing process can be described using a uniform
distribution over the interval 6.5 to 8.5 millimeters. What is the mean diameter of ball bearings produced in this
manufacturing process?
A) 7.5 millimeters B) 7.0 millimeters C) 8.0 millimeters D) 8.5 millimeters
28) Suppose x is a uniform random variable with c = 40 and d = 70. Find the standard deviation of x.
A) σ = 8.66 B) σ = 31.75 C) σ = 1.58 D) σ = 3.03
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29) 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 variance of the distribution.
A) 0.75 B) 0.33 C) 2.08 D) 12.00
30) Suppose a uniform random variable can be used to describe the outcome of an experiment with outcomes
ranging from 30 to 80. What is the mean outcome of this experiment?
A) 55 B) 30 C) 60 D) 80
31) The age of customers at a local hardware store follows a uniform distribution over the interval from 18 to 60
years old. Find the average age of customers to this hardware store.
A) 50 years old B) 39 years old C) 45 years old D) 60 years old
32) After a particular heavy snowstorm, the depth of snow reported in a mountain village followed a uniform
distribution over the interval from 15 to 22 inches of snow. Find the standard deviation of the snowfall
amounts.
A) 18.5 inches B) 4.08 inches C) 2.02 inches D) 1.42 inches
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
33) Suppose x is a uniform random variable with c = 10 and d = 80. Find the mean of the random variable x.
5.3 The Normal Distribution
1 Use Standard Normal Distribution
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
1) Use the standard normal distribution to find P(0 < z < 2.25).
A) .4878 B) .5122 C) .8817 D) .7888
2) Use the standard normal distribution to find P(-2.25 < z < 0).
A) .4878 B) .5122 C) .6831 D) .0122
3) Use the standard normal distribution to find P(-2.25 < z < 1.25).
A) .8821 B) .0122 C) .4878 D) .8944
4) Use the standard normal distribution to find P(-2.50 < z < 1.50).
A) .9270 B) .8822 C) .6167 D) .5496
5) Use the standard normal distribution to find P(z < -2.33 or z > 2.33).
A) .0198 B) .9809 C) .7888 D) .0606
6) Find a value of the standard normal random variable z, called z0, such that P(-z0 ≤ z ≤ z0) = 0.98.
A) 2.33 B) 1.645 C) 1.96 D) .99
7) Find a value of the standard normal random variable z, called z0, such that P(z ≥ z0) = 0.70.
A) -.53 B) -.98 C) -.81 D) -.47
8) Find a value of the standard normal random variable z, called z0, such that P(z ≤ z0) = 0.70.
A) .53 B) .98 C) .81 D) .47
9) Which shape is used to represent areas for a normal distribution?
A) Bell curve B) Circle C) Rectangle D) Triangle
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10) For a standard normal random variable, find the probability that z exceeds the value -1.65.
A) 0.4505 B) 0.0495 C) 0.9505 D) 0.5495
11) For a standard normal random variable, find the point in the distribution in which 11.9% of the z -values fall
below.
A) -1.18 B) 1.18 C) -0.30 D) -1.45
Answer the question True or False.
12) The mean of the standard normal distribution is 1 and the standard deviation is 0.
A) True B) False
13) P(-1 < x < 0) = P(0 < x < 1) for any random variable x that is normally distributed.
A) True B) False
14) Nearly 100% of the observed occurrences of a random variable x that is normally distributed will fall within
three standard deviations of the mean of the distribution of x.
A) True B) False
2 Use Normal Distribution
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
15) 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. Find the probability that a randomly selected boy in secondary school can
run the mile in less than 325 seconds.
A) .0107 B) .4893 C) .9893 D) .5107
16) 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 40 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) 408.8 seconds B) 511.2 seconds C) 525.8 seconds D) 394.2 seconds
17) 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 50 seconds. Between what times do we expect approximately 95% of the boys to run the
mile?
A) between 362 and 558 seconds B) between 377.75 and 542.28 seconds
C) between 365 and 555 seconds D) between 0 and 542.28 seconds
18) The weight 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
proportion of the 24-ounce bags contain more than the advertised 24 ounces of chips?
A) .9938 B) .4938 C) .0062 D) .5062
19) The volume of soda a dispensing machine pours into a 12 -ounce can of soda follows a normal distribution
with a mean of 12.54 ounces and a standard deviation of 0.36 ounce. The company receives complaints from
consumers who actually measure the amount of soda in the cans and claim that the volume is less than the
advertised 12 ounces. What proportion of the soda cans contain less than the advertised 12 ounces of soda?
A) .0668 B) .5668 C) .9332 D) .4332
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20) The amount of soda a dispensing machine pours into a 12 -ounce can of soda follows a normal distribution
with a mean of 12.54 ounces and a standard deviation of 0.36 ounce. Each can holds a maximum of 12.90
ounces of soda. Every can that has more than 12.90 ounces of soda poured into it causes a spill and the can
must go through a special cleaning process before it can be sold. What is the probability that a randomly
selected can will need to go through this process?
A) .1587 B) .3413 C) .8413 D) .6587
21) The amount of soda a dispensing machine pours into a 12 -ounce can of soda follows a normal distribution
with a standard deviation of 0.16 ounce. Every can that has more than 12.40 ounces of soda poured into it
causes a spill and the can must 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 must be cleaned
because of spillage to 3%?
A) 12.0992 ounces B) 12.7008 ounces C) 12.0528 ounces D) 12.7472 ounces
22) Before a new phone system was installed, the amount a company spent on personal calls followed a normal
distribution with an average of $700 per month and a standard deviation of $50 per month. Refer to such
expenses as PCE’s (personal call expenses). Using the distribution above, what is the probability that during a
randomly selected month PCE’s were between $575.00 and $790.00?
A) .9579 B) .0421 C) .9999 D) .0001
23) Before a new phone system was installed, the amount a company 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 PCE’s (personal call expenses). Find the point in the distribution below which 2.5% of the PCE’s
fell.
A) $402.00 B) $598.00 C) $12.50 D) $487.50
24) Before a new phone system was installed, the amount a company spent on personal calls followed a normal
distribution with an average of $700 per month and a standard deviation of $50 per month. Refer to such
expenses as PCE’s (personal call expenses). Find the probability that a randomly selected month had PCE’s
below $550.
A) 0.0013 B) 0.7857 C) 0.2143 D) 0.9987
25) The preventable monthly loss at a company has a normal distribution with a mean of $4700 and a standard
deviation of $40. A new policy was put into place, and the preventable loss the next month was $4460. What
inference can you make about the new policy?
A) Because the probability that the monthly loss would be as low as $4460 is small, the new policy is
working.
B) Because the probability that the monthly loss would be as low as $4460 is not very small, the new policy is
not working.
C) While the probability that the monthly loss would be as low as $4460 is small, it is not unexpected.
D) The new policy is probably less effective than the one it replaced.
26) 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 2300 miles. What is the probability a particular tire of this
brand will last longer than 57,700 miles?
A) .8413 B) .1587 C) .2266 D) .7266
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 1700 miles. What is the probability a certain tire of this brand
will last between 56,430 miles and 56,940 miles?
A) .0180 B) .9813 C) .4920 D) .4649
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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 warranty should the company use if they
want 96% of the tires to outlast the warranty?
A) 54,925 miles B) 65,075 miles C) 57,100 miles D) 62,900 miles
29) The price of a gallon of milk follows a normal distribution with a mean of $3.20 and a standard deviation of
$0.10. Find the price for which 12.3% of milk vendors exceeded.
A) $3.316 B) $3.084 C) $3.215 D) $3.238
30) The price of a gallon of milk follows a normal distribution with a mean of $3.20 and a standard deviation of
$0.10. What proportion of the milk vendors had prices that were less than $3.075 per gallon?
A) 0.8944 B) 0.3944 C) 0.2112 D) 0.1056
31) A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed
into a can is known to have a normal distribution with a mean of 5 milliliters (ml) and a standard deviation of
0.4 ml. Answer the following questions based on this information. What proportion of the paint cans contain
less than 5.54 ml of the dye?
A) 0.0885 B) 0.9885 C) 0.5885 D) 0.9115
32) A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed
into a can is known to have a normal distribution with a mean of 5 milliliters (ml) and a standard deviation of
0.4 ml. Answer the following questions based on this information. Find the dye amount that represents the 9th
percentile of the distribution.
A) 4.464 ml B) 5.536 ml C) 4.964 ml D) 4.836 ml E) 4.936 ml
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
33) The rate of return for an investment can be described by a normal distribution with mean 27% and standard
deviation 3%. What is the probability that the rate of return for the investment will be at least 22.5%?
34) The rate of return for an investment can be described by a normal distribution with mean 47% and standard
deviation 3%. What is the probability that the rate of return for the investment exceeds 53%?
35) The board of examiners that administers the real estate broker’s examination in a certain state found that the
mean score on the test was 435 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.
36) The board of examiners that administers the real estate broker’s examination in a certain state found that the
mean score on the test was 419 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.
37) 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 2600 miles. If the manufacturer guarantees the tread life of the
tires for the first 56,880 miles, what proportion of the tires will need to be replaced under warranty?
38) 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 of 405 tomatoes and a
standard deviation of 30 tomatoes. If there are 363 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?
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39) 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 of 124 tomatoes and a
standard deviation of 30 tomatoes. 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?
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
40) Suppose a random variable x is best described by a normal distribution with μ = 60 and σ = 5. Find the z-score
that corresponds to the value x = 65.
A) 1 B) 5 C) 65 D) 12
41) Suppose a random variable x is best described by a normal distribution with μ = 60 and σ = 13. Find the z-score
that corresponds to the value x = 112.
A) 4 B) 52 C) 15
13 D) 13
42) Suppose a random variable x is best described by a normal distribution with μ = 60 and σ = 12. Find the z-score
that corresponds to the value x = 60.
A) 0 B) 1 C) 12 D) 5
43) Suppose a random variable x is best described by a normal distribution with μ = 60 and σ = 8. Find the z-score
that corresponds to the value x = 30.
A) -3.75 B) 3.75 C) -8 D) 8
44) IQ test scores are normally distributed with a mean of 96 and a standard deviation of 12. An individual’s IQ
score is found to be 110. Find the z-score corresponding to this value.
A) 1.17 B) -1.17 C) 0.86 D) -0.86
5.4 Descriptive Methods for Assessing Normality
1 Determine if Data is Normally Distributed
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Solve the problem.
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 pounds and a standard deviation of 4 pounds. You randomly sample
30 preschool children and find their weights (in pounds) 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
Draw a histogram to display the data. Is it reasonable to assume that the weights are normally distributed?
Why?
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2) The printout below contains summary statistics of the heights of a sample of 200 adult men in the United
States.
Descriptive Statistics: HT
Variable N Mean StDev Minimum Q1 Median Q2 Maximum
HT 200 70.187 2.716 62.375 67.875 69.625 71.500 91.125
Use the information in the printout to determine whether the distribution of heights is approximately normal.
Explain your reasoning.
3) The following data represent the scores of a sample of 50 students on a statistics exam. The mean score is
x = 80.3, and the standard deviation is s = 11.37.
49 51 59 63 66 68 68 69 70 71
71 71 73 74 76 76 76 77 78 79
79 79 79 80 80 82 83 83 83 85
85 86 86 88 88 88 88 89 89 89
90 91 92 92 93 95 96 97 97 98
What percentage of the scores fall in each of the intervals x ± s, x ± 2s, and x ± 3s? Based on these percentages,
do you believe that the distribution of scores is approximately normal? Explain.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
4) Which of the following statements is not a property of the normal curve?
A) P(μ – σ < x < μ + σ) ≈ .95 B) symmetric about μ
C) mound-shaped (or bell shaped) D) P(μ – 3σ < x < μ + 3σ) ≈ .997
5) Which of the following is not a method used for determining whether data are from an approximately normal
distribution?
A) Construct a histogram or stem-and-leaf display. The shape of the graph or display should be uniform
(evenly distributed).
B) Compute the intervals x ± s, x ± 2s, and x ± 3s. The percentages of measurements falling in each should be
approximately 68%, 95%, and 100% respectively.
C) Find the interquartile range, IQR, and standard deviation, s, for the sample. Then IQR
s ≈ 1.3.
D) Construct a normal probability plot. The points should fall approximately on a straight line.
6) Which one of the following suggests that the data set is approximately normal?
A) A data set with Q1 = 14, Q3 = 68, and s = 41.
B) A data set with Q1 = 2.2, Q3 = 7.3, and s = 2.1.
C) A data set with Q1 = 105, Q3 = 270, and s = 33.
D) A data set with Q1 = 1330, Q3 = 2940, and s = 2440.
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7) Which one of the following suggests that the data set is not approximately normal?
A) A data set with 68% of the measurements within x ± 2s.
B)
C)
Stem Leaves
3 0 3 9
4 2 4 7 7
5 1 3 4 8 8 9 9 9
6 0 0 5 6 6 7 8
7 1 1 5
8 2 7
D) A data set with IQR = 752 and s = 574.
8) If a data set is normally distributed, what is the proportion of measurements you would expect to fall within
μ ± σ?
A) 68% B) 50% C) 95% D) 100%
9) A statistician received some data to analyze. The sender of the data suggested that the data was normally
distributed. Which of the following methods can be used to determine if the data is, in fact, normally
distributed?
I. Construct a histogram and/or stem-and-leaf display of the data and check the shape.
II. Compute the intervals x ± s, x ± 2s, and x ± 3s, and determine the percentage of measurements falling in
each. Compare these percentages to 68%, 95%, and 100%.
III. Calculate a value of IQR
s . If this value is approximately 1.3, then the data is normal.
IV. Construct a normal probability plot of the data. If the points fall on a straight line, then the data is
normal.
A) I only
B) II only
C) III only
D) IV only
E) I, II, III, and IV
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10) Data has been collected and a normal probability plot for one of the variables is shown below. Based on your
knowledge of normal probability plots, do you believe the variable in question is normally distributed? The
data are represented by the”o” symbols in the plot.
A) Yes. The plot reveals a straight line and this indicates the variable is normally distributed.
B) No. The plot does not reveal a straight line and this indicates the variable is not normally distributed.
C) Yes. The plot reveals a curve and this indicates the variable is normally distributed.
Answer the question True or False.
11) When the points on a normal probability plot lie approximately on a straight line, the data are approximately
normally distributed.
A) True B) False
5.5 Approximating a Binomial Distribution with a Normal Distribution (Optional)
1 Determine if Normal Distribution Can be Used to Approximate Binomial
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Solve the problem.
1) Determine if it is appropriate to use the normal distribution to approximate a binomial distribution when n = 7
and p = 0.3.
2) Determine if it is appropriate to use the normal distribution to approximate a binomial distribution when n = 36
and p = 0.3.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
3) A study of college students stated that 25% of all college students have at least one tattoo. In a random sample
of 80 college students, let x be the number of the students that have at least one tattoo. Can the normal
approximation be used to estimate the binomial distribution in this problem?
A) Yes B) No
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Answer the question True or False.
4) The continuity correction factor is the name given to the .5 adjustment necessary when estimating the binomial
with the normal distribution.
A) True B) False
2 Find Probability
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find the probability.
5) Assume that x is a binomial random variable with n = 400 and p = 0.30. Use a normal approximation to find
P(x > 100).
A) 0.9834 B) 0.9875 C) 0.5854 D) 0.5910
6) Assume that x is a binomial random variable with n = 400 and p = 0.30. Use a normal approximation to find
P(x ≥ 110).
A) 0.8749 B) 0.8508 C) 0.3749 D) 0.5517
7) Assume that x is a binomial random variable with n = 100 and p = 0.60. Use a normal approximation to find
P(x ≤ 65).
A) 0.8686 B) 0.8212 C) 0.5910 D) 0.1314
8) Assume that x is a binomial random variable with n = 100 and p = 0.60. Use a normal approximation to find
P(x < 65).
A) 0.8212 B) 0.8686 C) 0.5753 D) 0.1788
9) Assume that x is a binomial random variable with n = 1000 and p = 0.80. Use a normal approximation to find
P(800 < x ≤ 820).
A) 0.4314 B) 0.4222 C) 0.4474 D) 0.0517
Solve the problem.
10) Transportation officials tell us that 80% of drivers wear seat belts while driving. Find the probability that more
than 576 drivers in a sample of 700 drivers wear seat belts.
A) 0.0594 B) 0.9406 C) 0.8 D) 0.2
11) Transportation officials tell us that 80% of drivers wear seat belts while driving. What is the probability of
observing 633 or fewer drivers wearing seat belts in a sample of 850 drivers?
A) approximately 0 B) 0.8 C) 0.2 D) approximately 1
12) Transportation officials tell us that 80% of drivers wear seat belts while driving. What is the probability that
between 538 and 546 drivers in a sample of 700 drivers wear seat belts?
A) 0.0837 B) 0.0166 C) 0.1003 D) 0.8997
13) A certain baseball player hits a home run in 6% of his at-bats. Consider his at-bats as independent events. Find
the probability that this baseball player hits more than 32 home runs in 750 at-bats?
A) 0.9726 B) 0.0274 C) 0.06 D) 0.94
14) A certain baseball player hits a home run in 4% of his at-bats. Consider his at-bats as independent events. Find
the probability that this baseball player hits at most 16 home runs in 650 at-bats?
A) 0.0287 B) 0.9713 C) 0.04 D) 0.96
15) A study of college students stated that 25% of all college students have at least one tattoo. In a random sample
of 80 college students, let x be the number of the students that have at least one tattoo. Find the approximate
probability that more than 30 of the sampled students had at least one tattoo.
A) 0.4929 B) 0.9929 C) 0.0071 D) 0.0034
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16) A study of college students stated that 25% of all college students have at least one tattoo. In a random sample
of 80 college students, let x be the number of the students that have at least one tattoo. Find the approximate
probability that more than 17 and less than 26 of the sampled students had at least one tattoo.
A) 0.1800 B) 0.6644 C) 0.4222 D) 0.2422
3 Find Mean, Standard Deviation, z-Score
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
17) A certain baseball player hits a home run in 3% of his at-bats. Consider his at-bats as independent events.
How many home runs do we expect the baseball player to hit in 600 at-bats?
A) 18 B) 603 C) 3 D) 17.46
18) A study of college students stated that 25% of all college students have at least one tattoo. In a random sample
of 80 college students, let x be the number of the students that have at least one tattoo. Find the mean and
standard deviation for this binomial distribution.
A) Mean = 20, Standard Deviation = 3.87 B) Mean = 80, Standard Deviation = 15
C) Mean = 20, Standard Deviation = 15 D) Mean = 80, Standard Deviation = 3.87
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
19) It is against the law to discriminate against job applicants because of race, religion, sex, or age. Of the
individuals who apply for an accountant’s position in a large corporation, 39% are over 45 years old. If the
company decides to choose 96 of a very large number of applicants for closer credential screening, claiming
that the selection will be random and not age-biased, what is the z-value associated with fewer than 46 of
those chosen being over 45 years old? (Assume that the applicant pool is large enough so that x, the number in
the sample over 45 years old, has a binomial probability distribution.)
20) A loan officer has 66 loan applications to screen during the next week. If past record indicates that she turns
down 17% of the applicants, what is the z-value associated with 61 or more of the 66 applications being
rejected?
21) Suppose that 88% of the stocks listed on a particular exchange increased in value yesterday. Let x be the
number of stocks that increased in value yesterday in a random of 72 stocks listed on the exchange. Find the
mean and standard deviation of x.
22) Suppose that 67% of the employees of a company participate in the company’s medical savings program. Let x
be the number of employees who participate in the program in a random sample of 50 employees. Find the
mean and standard deviation of x.
5.6 The Exponential Distribution (Optional)
1 Identify the Characteristics of an Exponential Distribution
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Answer the question True or False.
1) The exponential distribution is sometimes called the waiting-time distribution, because it is used to describe
the length of time between occurrences of random events.
A) True B) False
2) The probability density function for an exponential random variable x has a graph called a bell curve.
A) True B) False
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3) The exponential distribution has the property that its mean equals its standard deviation.
A) True B) False
4) The exponential distribution is governed by two quantities, μ and σ, that determine its shape and location
A) True B) False
2 Find Probability
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
5) Determine the value of e-a/θ for θ =2 and a = 3.
A) 0.223130 B) 0.513417 C) -4.481689 D) 4.481689
6) Suppose that x has an exponential distribution with θ =1.5. Find P(x > 1).
A) 0.513417 B) 0.223130 C) 0.486583 D) 0.776870
7) Suppose that x has an exponential distribution with θ = 2.5. Find P(x ≥ 4).
A) 0.201897 B) 0.535261 C) 0.464739 D) 0.798103
8) Suppose that x has an exponential distribution with θ = 2. Find P(x < 1.5).
A) 0.527633 B) 0.472367 C) 0.736403 D) 0.263597
9) Suppose that x has an exponential distribution with θ = 5. Find P(x ≤ 10).
A) 0.864665 B) 0.393469 C) 0.606531 D) 0.135335
10) The time between customer arrivals at a furniture store has an approximate exponential distribution with mean
θ = 8.5 minutes. If a customer just arrived, find the probability that the next customer will arrive in the next 5
minutes.
A) 0.444694 B) 0.817316 C) 0.555306 D) 0.182684
11) The time between customer arrivals at a furniture store has an approximate exponential distribution with mean
θ = 8.5 minutes. If a customer just arrived, find the probability that the next customer will not arrive for at least
20 minutes.
A) 0.095089 B) 0.653770 C) 0.904911 D) 0.346230
12) The time (in years) until the first critical-part failure for a certain car is exponentially distributed with a mean
of 3.4 years. Find the probability that the time until the first critical-part failure is 5 years or more.
A) 0.229790 B) 0.506617 C) 0.770210 D) 0.493383
13) The time (in years) until the first critical-part failure for a certain car is exponentially distributed with a mean
of 3.4 years. Find the probability that the time until the first critical-part failure is less than 1 year.
A) 0.254811 B) 0.745189 C) 0.033373 D) 0.966627
14) The waiting time (in minutes) between ordering and receiving your meal at a certain restaurant is
exponentially distributed with a mean of 10 minutes. The restaurant has a policy that your meal is free if you
have to wait more than 25 minutes after ordering. What is the probability of receiving a free meal?
A) 0.082085 B) 0.670320 C) 0.917915 D) 0.329680
15) The time between arrivals at an ATM machine follows an exponential distribution with θ = 10 minutes. Find
the probability that more than 25 minutes will pass between arrivals.
A) 0.917915 B) 0.670320 C) 0.329680 D) 0.082085
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16) The time between arrivals at an ATM machine follows an exponential distribution with θ = 10 minutes. Find
the probability that less than 25 minutes will pass between arrivals.
A) 0.917915 B) 0.670320 C) 0.329680 D) 0.082085
17) The time between arrivals at an ATM machine follows an exponential distribution with θ = 10 minutes. Find
the probability that between 15 and 25 minutes will pass between arrivals.
A) 0.223130 B) 0.082085 C) 0.141045 D) 0.305215
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
18) Suppose that x has an exponential distribution with θ = 1.75. Find each probability.
a. P(x ≥ 2)
b. P(x < 2)
19) The length of time (in months) that a cashier works for a certain fast food restaurant is exponentially
distributed with a mean of 7 months.
a. Find the probability that a cashier works for the restaurant for at least 2 years.
b. Find the probability that a cashier works for the restaurant for less than 1 month.
20) The time between equipment failures (in days) at a particular factory is exponentially distributed with a mean
of 4.5 days. A machine just failed and was repaired today.
a. Find the probability that another machine will fail within the next day.
b. Find the probability that there will be no more equipment failures in the next week.
21) An online retailer reimburses a customer’s shipping charges if the customer does not receive his order within
one week. Delivery time (in days) is exponentially distributed with a mean of 3.2 days. What percentage of
customers have their shipping charges reimbursed?
3 Find Mean, Standard Deviation
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
22) Suppose that the random variable x has an exponential distribution with θ = 1.5. Find the mean and standard
deviation of x.
A) μ = 1.5; σ = 1.5 B) μ = 0; σ = 1 C) μ = 1.5; σ = 1 D) μ = 0; σ = 1.5
23) Suppose that the random variable x has an exponential distribution with θ = 1.5. Find the probability that x
will assume a value within the interval μ ± 2σ.
A) .950213 B) .049787 C) .716531 D) .864665
24) The time between arrivals at an ATM machine follows an exponential distribution with θ = 10 minutes. Find
the mean and standard deviation of this distribution.
A) Mean = 10, Standard Deviation = 3.16 B) Mean = 10, Standard Deviation = 10
C) Mean = 10, Standard Deviation = 100 D) Mean = 3.16, Standard Deviation = 3.16
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SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
25) Suppose that the random variable x has an exponential distribution with θ = 3.
a. Find the probability that x assumes a value more than three standard deviations from μ.
b. Find the probability that x assumes a value less than one standard deviation from μ.
c. Find the probability that x assumes a value within a half standard deviation of μ.
Page 147Copyright © 2013 Pearson Education, Inc.
Ch. 5 Continuous Random Variables
Answer Key
5.1 Continuous Probability Distributions
1 Understand Probability Distributions for Continuous Random Variable
1) B
2) A
3) A
4) A
5.2 The Uniform Distribution
1 Find Probability
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) C
22) B
23) Let x = high temperature in August. Then x is a uniform random variable with c = 75°F and
d = 95°F.
P(x > a) = d – a
d – c
P(x > 80) = 95 – 80
95 – 75 = 15
20 = .75
24) Let x = ball bearing diameter. Then x is a uniform random variable with c = 2.5 and d = 8.5.
P(x < a) = a – c
d – c
P(x < 4.5) = 4.5 – 2.5
8.5 – 2.5 = 2
6 = 0.33
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25) Let x = gallons pumped per minute. Then x is a uniform random variable with c = 8.5 and d = 9.5.
P(x < a) = a – c
d – c
P(x < 8.75) = 8.75 – 8.5
9.5 – 8.5 = .25
1 = .25
26) A
2 Find Mean, Variance, Standard Deviation
27) A
28) A
29) A
30) A
31) B
32) C
33) μ = c + d
2 = 10 + 80
2 = 45
5.3 The Normal Distribution
1 Use Standard Normal Distribution
1) A
2) A
3) A
4) A
5) A
6) A
7) A
8) A
9) A
10) C
11) A
12) B
13) B
14) A
2 Use Normal Distribution
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) D
31) D
32) A
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33) Let x be the rate of return. Then x is a normal random variable with μ = 27% and σ = 3%. To determine the probability
that x is at least 22.5%, we need to find the z-value for x = 22.5%.
z = x – μ
σ = 22.5 – 27
3 = -1.5
P(x ≥ 22.5%) = P(-1.5 ≤ z) = .5 + P(-1.5 ≤ z ≤ 0) = .5 + .4332 = .9332
34) Let x be the rate of return. Then x is a normal random variable with μ = 47% and σ = 3%. To determine the probability
that x exceeds 53%, we need to find the z-value for x = 53%.
z = x – μ
σ = 53 – 47
3 = 2
P(x > 53%) = P(z ≥ 2) = .5 – P(0 ≤ z ≤ 2) = .5 – .4772 = .0228
35) Let x be a score on this exam. Then x is a normally distributed random variable with μ = 435 and σ = 72. We want to
find the value of x0, such that P(x > x0) = .10. The z-score for the value x = x0 is
z = x0 – μ
σ = x0 – 435
72 .
P(x > x0) = P z > x0 – 435
72 = .10
We find x0 – 435
72 ≈ 1.28.
x0 – 435 = 1.28(72) ⇒ x0 = 435 + 1.28(72) = 527.16
36) Let x be a score on this exam. Then x is a normally distributed random variable with μ = 419 and σ = 72. We want to
find the value of x0, such that P(x > x0) = .80. The z-score for the value x = x0 is
z = x0 – μ
σ = x0 – 419
72 .
P(x > x0) = P z > x2 – 419
72 = .80
We find x0 – 419
72 ≈ -.84.
x0 – 419 = -.84(72) ⇒ x0 = 419 – .84(72) = 358.52
37) Let x be the tread life of this brand of tire. Then x is a normal random variable with μ = 60,000 and σ = 2600. To
determine what proportion of tires fail before reaching 56,880 miles, we need to find the z-value for x = 56,880.
z = x – μ
σ = 56,880 – 60,000
2600 = -1.20
P(x ≤ 56,880) = P(z ≤ -1.20) = .5 – P(-1.20 ≤ z ≤ 0) = .5 – .3849 = .1151
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38) Let x be the number of tomatoes sold per day. Then x is a normal random variable with μ = 405 and σ = 30.
To determine the probability that all 363 tomatoes will be sold, we need to find the z-value for x = 363.
z = x – μ
σ = 363 – 405
30 = -1.4
P(x ≥ 363) = P(z ≥ -1.4) = .5 + P(-1.4 ≤ z ≤ 0) = .5 + .4192 = .9192
39) Let x be the number of tomatoes sold per day. Then x is a normal random variable with μ = 124 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 – 124
30 .
P(x > x0) = P(z > x0 – 124
30 )= .015
We find x0 – 124
30 = 2.17
x0 – 124 = 2.17(30) ⇒ x0 = 124 + 2.17(30) = 189
40) A
41) A
42) A
43) A
44) A
5.4 Descriptive Methods for Assessing Normality
1 Determine if Data is Normally Distributed
1)
It is not reasonable to assume that the heights are normally distributed since the histogram is not mound -shaped and
symmetric about the mean of 31 pounds.
2) IQR
s = 71.500 – 67.875
2.176 ≈ 1.33; Since this number is reasonably close to 1.3, the distribution of heights is approximately
normal.
3) The percentages are 70%, 96%, and 100%, respectively. Since these percentages are reasonably close to 68%, 95%, and
100%, we conclude that the distribution of scores is approximately normal.
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4) A
5) A
6) A
7) A
8) A
9) E
10) B
11) A
5.5 Approximating a Binomial Distribution with a Normal Distribution (Optional)
1 Determine if Normal Distribution Can be Used to Approximate Binomial
1) cannot use normal distribution
2) can use normal distribution
3) A
4) A
2 Find Probability
5) A
6) A
7) A
8) A
9) A
10) A
11) A
12) A
13) A
14) A
15) D
16) B
3 Find Mean, Standard Deviation, z-Score
17) A
18) A
19) x is a binomial random variable with n = 96 and p = 0.39.
z = (x + .5) – np
np(1 – p) = (46 + .5) – 96(0.39)
96(0.39)(1 – 0.39) = 1.90
20) Let x be the number of the 66 applications rejected. Then x is a binomial random variable with n = 66 and p = 0.17.
z = (x – .5) – np
np(1 – p) = (61 – .5) – 66(0.17)
(66)(0.17)(1 – 0.17) = 16.15
21) Mean = μ = .88(72) = 63.36; standard deviation = σ = 72(.88)(.12) ≈ 2.76
22) Mean = μ = .67(72) = 33.5; standard deviation = σ = 50(.67)(.33) ≈ 3.32
5.6 The Exponential Distribution (Optional)
1 Identify the Characteristics of an Exponential Distribution
1) A
2) B
3) A
4) B
2 Find Probability
5) A
6) A
7) A
8) A
9) A
10) A
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11) A
12) A
13) A
14) A
15) D
16) A
17) C
18) a. P(x ≥ 2) = e-2/1.75 ≈ .318907
b. P(x < 2) = 1 – P(x ≥ 2) = 1 – e-2/1.75 ≈ 1 – .318907 = .681093
19) a. P(x ≥ 24) = e-24/7 ≈ .032433
b. P(x < 1) = 1 – P(x ≥ 1) = 1 – e-1/7 ≈ 1 – .866878 = .133122
20) a. P(x < 1) = 1 – P(x ≥ 1) = 1 – e-1/4.5 ≈ 1 – .800737 = .199263
b. P(x > 7) = e-7/4.5 ≈ .211072
21) a. P(x ≥ 7) = e-7/3.2 ≈ .112197; about 11.2 percent
3 Find Mean, Standard Deviation
22) A
23) A
24) B
25) a. P(x < -6) + P(x > 12) = 0 + P(x > 12) = e-12/3 = e-4 ≈ .018316
b. P(0 < x < 6) = P(x < 6) = 1 – P(x ≥ 6) = 1 – e-6/3 = 1 – e-2 ≈ .864665
c. P(1.5 < x < 4.5) = P(x ≥ 1.5) – P(x ≥ 4.5)= e-1.5/3 – e-4.5/3 = e-.5 – e-1.5 ≈ .383400
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