# Doane - Stats - Chap 007 - Test answers

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Chapter 07Continuous Probability Distributions

True / False Questions1.A continuous uniform distribution is always symmetric.TrueFalse

2.The height and width of a continuous uniform distribution's PDF are the same.TrueFalse

3.A continuous uniform distribution U(0,800) will have = 400 and = 230.94.TrueFalse

4.A continuous uniform distribution U(100,200) will have the same standard deviation as a continuous uniform distribution U(200,300).TrueFalse

5.For a continuous uniform distribution U(200,400), the parameters are = 300 and = 100.TrueFalse

6.The exponential distribution describes the number of arrivals per unit of time.TrueFalse

7.The exponential distribution is always skewed right.TrueFalse

8.If arrivals follow a Poisson distribution, waiting times follow the exponential distribution.TrueFalse

9.The triangular distribution is used in "what-if" analysis for business planning.TrueFalse

10.The triangular distribution is symmetric.TrueFalse

11.The triangular distribution T(0,10,20) is skewed left.TrueFalse

12.A triangular distribution can be skewed either left or right.TrueFalse

13.For a continuous random variable, the total area beneath the PDF will be greater than zero but less than one.TrueFalse

14.The exponential distribution is continuous and the Poisson distribution is discrete, yet the two distributions are closely related.TrueFalse

15.The mean, median, and mode of a normal distribution will always be the same.TrueFalse

16.There is a simple formula for normal areas, but we prefer a table for greater accuracy.TrueFalse

17.Normal distributions differ only in their means and variances.TrueFalse

18.Any normal distribution has a mean of 0 and a standard deviation of 1.TrueFalse

19.We would use a normal distribution to model the waiting time until the next Florida hurricane strike.TrueFalse

20.Experience suggests that 4 percent of all college students have had a tonsillectomy. In a sample of 300 college students, we need to find the probability that at least 10 have had a tonsillectomy. It is acceptable to use the normal distribution to estimate this probability.TrueFalse

21.The normal is a good approximation to the binomial when n is greater than or equal to 10.TrueFalse

22.The true proportion of accounts receivable with some kind of error is 4 percent for Venal Enterprises. If an auditor randomly samples 50 accounts receivable, it is acceptable to use the normal approximation to estimate the probability that fewer than two will contain errors.TrueFalse

23.The normal distribution is a good approximation to the binomial if both 10 and n 10.TrueFalse

24.The normal distribution is a good approximation to the binomial if n = 200 and = .03.TrueFalse

25.The normal distribution is a good approximation to the binomial if n = 25 and = .50.TrueFalse

26.The exponential distribution can be either right-skewed or left-skewed, depending on .TrueFalse

27.The number of lightning strikes in a day in Miami is a continuous random variable.TrueFalse

28.The area under a normal curve is 1 only if the distribution is standardized N(0,1).TrueFalse

29.The area under an exponential curve can exceed 1 because the distribution is right-skewed.TrueFalse

Multiple Choice Questions30.A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, the average amount of water dispensed by the machine is:

A.12 ounces.

B.13 ounces.

C.14 ounces.

D.16 ounces.

31.A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, the standard deviation of the amount of water dispensed is about:

A.1.73 ounces.

B.3.00 ounces.

C.0.57 ounce.

D.3.51 ounces.

32.A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, what is the probability that 13 or more ounces will be dispensed in a given glass?

A..1666

B..3333

C..5000

D..6666

33.A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. The mean of this distribution is:

A.30.5.

B.31.5.

C.32.5.

D.33.5.

34.A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. The standard deviation of this distribution is approximately:

A.52.1.

B.32.5.

C.6.85.

D.7.22.

35.A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. What is P(30 X 40)?

A..20

B..40

C..60

D..80

36.The Excel function =800*RAND() would generate random numbers with standard deviation approximately equal to:

A.200.

B.188.

C.231.

D.400.

37.The Excel function =40*RAND() would generate random numbers with standard deviation approximately equal to

A.13.33.

B.20.00.

C.11.55.

D.19.27.

38.If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of waiting more than 0.5 hour for the next arrival is:

A..2407.

B..1653.

C..1222.

D..5000.

39.If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of waiting less than 0.5 hour for the next arrival is:

A..7122.

B..8105.

C..8347.

D..7809.

40.If arrivals occur at a mean rate of 2.6 events per minute, the exponential probability of waiting more than 1.5 minutes for the next arrival is:

A..0202.

B..0122.

C..0535.

D..2564.

41.If arrivals occur at a mean rate of 1.6 events per minute, the exponential probability of waiting less than 1 minute for the next arrival is:

A..2019.

B..7104.

C..8812.

D..7981.

42.Bob's z-score for the last exam was 1.52 in Prof. Axolotl's class BIO 417, "Life Cycle of the Ornithorhynchus." Bob said, "Oh, good, my score is in the top 10 percent." Assuming a normal distribution of scores, is Bob right?

A.Yes.

B.No.

C.Must have n to answer.

43.The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What proportion of brook trout caught will be between 12 and 18 inches in length?

A..6563

B..6826

C..2486

D..4082

44.The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. The first quartile for the lengths of brook trout would be:

A.16.01 inches.

B.11.00 inches.

C.11.98 inches.

D.10.65 inches.

45.The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What lower limit should the State Game Commission set on length if it is desired that 80 percent of the catch may be kept by fishers?

A.12.80 inches

B.11.48 inches

C.12.00 inches

D.9.22 inches

46.In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. What percent of customers require less than 32 minutes for a simple haircut?

A.95.99 percent

B.99.45 percent

C.97.72 percent

D.45.99 percent

47.In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. The slowest quartile of customers will require longer than how many minutes for a simple haircut?

A.3(n + 1)/4 minutes

B.26 minutes

C.25.7 minutes

D.27.7 minutes

48.In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. For a simple haircut, the middle 90 percent of the customers will require:

A.between 18.4 and 31.6 minutes.

B.between 19.9 and 30.1 minutes.

C.between 20.0 and 30.0 minutes.

D.between 17.2 and 32.8 minutes.

49.The area under the normal curve between z = 0 and z = 1 is ______________ the area under the normal curve between z = 1 and z = 2.

A.less than

B.greater than

C.equal to

50.The price-earnings ratio for firms in a given industry follows the normal distribution. In this industry, a firm whose price-earnings ratio has a standardized value of z = 1.00 is approximately in the highest ______ percent of firms in the industry.

A.16 percent

B.34 percent

C.68 percent

D.75 percent

51.A student's grade on an examination was transformed to a z value of 0.67. Assuming a normal distribution, we know that she scored approximately in the top:

A.15 percent.

B.50 percent.

C.40 percent.

D.25 percent.

52.The MPG (miles per gallon) for a certain compact car is normally distributed with a mean of 31 and a standard deviation of 0.8. What is the probability that the MPG for a randomly selected compact car would be less than 32?

A.0.3944

B.0.8944

C.0.1056

D.0.5596

53.The probability is .80 that a standard normal random variable is between -z and +z. The value of z is approximately:

A.1.28.

B.1.35.

C.1.96.

D.1.45.

54.The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What proportion of the citizens will require less than one hour?

A.0.4772

B.0.9772

C.0.9974

D.0.9997

55.The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. The slowest 10 percent of the citizens would need at least how many minutes to complete the form?

A.27.2

B.35.8

C.52.8

D.59.6

56.The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What is the third quartile (in minutes) for the time required to complete the form?

A.44.75

B.46.75

C.47.50

D.52.50

57.Exam scores were normal in BIO 200. Jason's exam score was one standard deviation above the mean. What percentile is he in?

A.68th

B.75th

C.78th

D.84th

58.Compared to the area between z = 1.00 and z = 1.25, the area between z = 2.00 and z = 2.25 in the standard normal distribution will be:

A.smaller.

B.larger.

C.the same.

D.impossible to compare without knowing and .

59.A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of applicants would you expect to have scores of 600 or above?

A.0.0401

B.0.4599

C.0.5401

D.0.0852

60.A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of the applicants would you expect to have a score of 400 or above?

A.0.2734

B.0.7734

C.0.7266

D.0.7500

61.A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. The top 2.5 percent of the applicants would have a score of at least (choose the nearest integer):

A.606.

B.617.

C.600.

D.646.

62.If the random variable Z has a standard normal distribution, then P(1.25 Z 2.17) is:

A.0.0906.

B.0.9200.

C.0.4700.

D.0.3944.

63.If the random variable Z has a standard normal distribution, then P(Z -1.37) is:

A.0.9147.

B.0.4147.

C.0.5016.

D.0.0853.

64.Assume that X is normally distributed with a mean = $64. Given that P(X $75) = 0.2981, we can calculate that the standard deviation of X is approximately:

A.$20.76.

B.$13.17.

C.$5.83.

D.$7.05.

65.The standard deviation of a normal random variable X is $20. Given that P(X $10) = 0.1841. From this we can determine that the mean of the distribution is equal to:

A.$13.

B.$26.

C.$20.

D.$28.

66.The random variable X is normally distributed with mean of 80 and variance of 36. The 67th percentile of the distribution is:

A.72.00.

B.95.84.

C.90.00.

D.82.64.

67.The area under the normal curve between the 20th and 70th percentiles is equal to:

A.0.7000.

B.0.5000.

C.0.9193.

68.The variable in a normal distribution can assume any value between

A.-3 and +3

B.-4 and +4

C.-1 and +1

D.- and +

69.What are the mean and standard deviation for the standard normal distribution?

A. = 0, = 0

B. = 1, = 1

C. = 1, = 0

D. = 0, = 1

70.Any two normal curves are the same except for their:

A.standard deviations.

B.means.

C.standard deviations and means.

D.standard deviations, means, skewness, and kurtosis.

71.Light bulbs are normally distributed with an average lifetime of 1000 hours and a standard deviation of 250 hours. The probability that a light bulb picked at random will last less than 1500 hours is about:

A.97.72 percent.

B.95.44 percent.

C.75.00 percent.

D.68.00 percent.

72.To convert a normally distributed variable X into a standard Z score we would:

A.subtract the mean from the original observation and divide the result by the variance.

B.subtract the mean from the original observation and divide the result by the standard deviation.

C.add the mean and the original observation, then divide by the variance.

D.subtract the mean from the standard deviation and divide by the variance.

73.Regarding continuous probability distributions, which statement is incorrect?

A.The triangular distribution may be skewed left or right.

B.The uniform distribution is never skewed.

C.The normal distribution is sometimes skewed.

D.The exponential distribution is always skewed right.

74.Which model best describes your waiting time until you get the next nonworking web URL ("This page cannot be displayed") as you click on various websites for Florida condo rentals?

A.Triangular

B.Uniform

C.Normal

D.Exponential

75.On average, a major earthquake (Richter scale 6.0 or above) occurs 3 times a decade in a certain California county. What is the probability that less than six months will pass before the next earthquake?

A..1393

B..8607

C..0952

D..9048

76.If the mean time between in-flight aircraft engine shutdowns is 12,500 operating hours, the 90th percentile of waiting times to the next shutdown will be:

A.20,180 hours.

B.28,782 hours.

C.23,733 hours.

D.18,724 hours.

77.On average, 15 minutes elapse between discoveries of fraudulent corporate tax returns in a certain IRS office. What is the probability that less than 30 minutes will elapse before the next fraudulent corporate tax return is discovered?

A..1353

B..6044

C..7389

D..8647

78.If the mean time between unscheduled maintenance of LCD displays in a hospital's CT scan facility is 4,000 operating hours, what is the probability of unscheduled maintenance in the next 5,000 hours?

A..8000

B..7135

C..2865

D..5000

79.A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. What is the probability that less than 6 minutes will elapse before the next breakdown?

A..8187

B..0488

C..1813

D..2224

80.A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. The median time between breakdowns is:

A.30.0 minutes.

B.35.7 minutes.

C.25.4 minutes.

D.20.8 minutes.

81.Which probability model is most appropriate to describe the waiting time (working days) until an office photocopier breaks down (i.e., requires unscheduled maintenance)?

A.Normal

B.Uniform

C.Exponential

D.Poisson

82.Bob's z-score for the last exam was -1.15 in FIN 417, "Capital Budgeting Strategies." Bob said, "Yipe! My score is within the bottom quartile." Assuming a normal distribution, is Bob right?

A.Yes

B.No

C.Must know the class size to answer

83.Exam scores were normal in MIS 200. Jason's exam score was 1.41 standard deviations above the mean. What percentile is he in?

A.68th

B.75th

C.84th

D.92nd

84.Compared to the area between z = 0.50 and z = 0.75, the area between z = 1.50 and z = 1.75 in the standard normal distribution will be:

A.smaller

B.larger

C.the same

85.If GMAT scores for applicants at Oxnard Graduate School of Business are N(500, 50) then the top 5 percent of the applicants would have a score of at least (choose the nearest integer):

A.575.

B.582.

C.601.

D.608.

86.If the random variable Z has a standard normal distribution, then P(1.17 Z 2.26) is:

A.0.1091.

B.0.1203.

C.0.2118.

D.0.3944.

87.If the random variable Z has a standard normal distribution, then P(Z -1.72) is:

A.0.9573.

B.0.0446.

C.0.5016.

D.0.0427.

88.Excel's =100*RAND() function produces continuous random numbers that are uniformly distributed between 0 and 100. The standard deviation of this distribution is approximately:

A.50.00.

B.28.87.

C.33.33.

D.25.00.

89.Excel's =RAND() function produces random numbers that are uniformly distributed between 0 and 1. The mean of this distribution is approximately

A..5000

B..2500

C..3333

D..2887

90.Excel's =RAND() function produces random numbers that are uniformly distributed from 0 to 1. What is the probability that the random number exceeds .75?

A.75 percent

B.50 percent

C.25 percent

91.Which is the correct Excel formula for the 80th percentile of a distribution that is N(475, 33)?

A.=NORM.DIST(80, 475, 33,1)

B.=NORM.INV(0.80, 475, 33)

C.=NORM.S.INV((80 - 475)/33)

92.If arrivals follow a Poisson distribution with mean 1.2 arrivals per minute, find the 75th percentile of waiting times until the next arrival (i.e., 75 percent below).

A.1.155 minutes (69.3 seconds)

B.0.240 minute (14.4 seconds)

C.1.919 minutes (115.1 seconds)

93.A software developer makes 175 phone calls to its current customers. There is an 8 percent chance of reaching a given customer (instead of a busy signal, no answer, or answering machine). The normal approximation of the probability of reaching at least 20 customers is:

A..022.

B..007.

C..063.

D..937.

94.For Gardyloo Manufacturing, the true proportion of accounts receivable with some kind of error is .20. If an auditor randomly samples 225 accounts receivable, what is the approximate normal probability that 39 or fewer will contain errors?

A..1797

B..2097

C..1587

D..0544

95.A letter is mailed to a sample of 500 homeowners. Based on past experience, the probability of an undeliverable letter is 0.06. The normal approximation to the binomial probability of 40 or more undeliverable letters is:

A.0.9632

B.0.0368

C.0.2305

D.0.7695

96.In a T-F exam with 100 questions, passing requires a score of at least 60. What is the approximate normal probability that a "guesser" will score at least 60 points?

A..0287

B..4713

C..0251

D..0377

97.A multiple choice exam has 100 questions. Each question has five choices. What would be the approximate probability that a "guesser" could achieve a score of 30 or more?

A.0.0088

B.0.0062

C.0.0015

D.0.4913

98.For which binomial distribution would a normal approximation be most acceptable?

A.n = 50, = 0.05

B.n = 100, = 0.04

C.n = 40, = 0.25

D.n = 400, = 0.02

99.Historically, the default rate on a certain type of commercial loan is 20 percent. If a bank makes 100 of these loans, what is the approximate probability that at least 26 will result in default?

A.0.2000

B.0.0668

C.0.0846

D.0.0336

100.A company employs 300 employees. Each year, there is a 30 percent turnover rate for employees. We want to do a normal approximation to the binomial distribution of the number of employees who leave each year. For this normal approximation, the mean is ______ and the standard deviation is _____.

A.90, 63

B.90, 7.937

C.90, 30

D.90, 15

101.The probability that a rental car will be stolen is 0.001. If 25,000 cars are rented from Hertz, what is the normal approximation to the probability that fewer than 20 will be stolen?

A..2577

B..1335

C..1128

D..8335

102.If adult male heights are normally distributed with a mean of 180 cm and a standard deviation of 7 cm, how high should an aircraft lavatory door be to ensure that 99.9 percent of adult males will not have to stoop as they enter?

A.195.7 cm

B.201.6 cm

C.207.3 cm

D.201.4 cm

103.TotCo is developing a new deluxe baby bassinet. If the length of a newborn baby is normally distributed with a mean of 50 cm and a standard deviation of 5 cm, what should be the interior length of the bassinet to ensure that 99 percent of newborn babies will fit, with a safety margin of 15 cm on each end of the bassinet?

A.95.45 cm

B.85.22 cm

C.91.63 cm

D.98.92 cm

104.The triangular distribution T(4, 12, 26) has a mean of:

A.14.

B.18.

C.12.

D.13.

105.The triangular distribution T(0, 10, 20) has a standard deviation of:

A.4.082.

B.3.775.

C.3.024.

D.2.994.

106.The triangular distribution T(5, 23, 62) has a mean of:

A.23.

B.30.

C.33.

D.35.

107.The triangular distribution T(10, 20, 50) has a standard deviation of:

A.9.498.

B.9.225.

C.8.498.

D.7.710.

108.Which statement is incorrect?

A.The triangular distribution always has a single mode.

B.The mean of the triangular distribution is (a + b + c)/3.

C.The triangular distribution is right-skewed.

109.Bob used a triangular distribution of T(20, 30, 61) to represent his daily commute time (minutes). Which statement is incorrect?

A.The distribution is right-skewed.

B.The mode of the distribution exceeds the mean.

C.The mean of the distribution is 37.

D.The midrange of the distribution is 40.5.

110.Phyllis used a triangular distribution of T(10, 15, 20) to represent her daily commute time (minutes). Which statement is incorrect?

A.The distribution is right-skewed.

B.The mode of the distribution is at the mean.

C.The mean of the distribution is 15.

D.The midrange of the distribution is 15.

111.In a continuous distribution:

A.P(X < 5) is the same as P(X 5).

B.P(X < 5) is less than P(X 5).

C.P(X < 5) is more than P(X 5).

112.In a continuous distribution the

A.PDF is usually higher than the CDF.

B.CDF is used to find left-tail probabilities.

C.PDF shows the area under the curve.

D.CDF is usually the same as the PDF.

113.If the mean waiting time for the next arrival is 12 minutes, what is the median waiting time?

A.7.2 minutes

B.8.3 minutes

C.9.1 minutes

D.12 minutes

114.If the mean waiting time for the next arrival is 18 minutes, what is the first quartile (25th percentile) for waiting times?

A.13 minutes

B.7.9 minutes

C.5.2 minutes

D.3.1 minutes

115.Could this function be a PDF?

A.Yes.

B.No.

C.It depends on x.

116.Could this function be a PDF?

A.Yes.

B.No.

C.It depends on x.

117.The ages of job applicants for a security guard position are uniformly distributed between 25 and 65. Could a 25-year-old job applicant be two standard deviations below the mean (or more than two standard deviations)?

A.Yes.

B.No.

C.Impossible to determine from given information.

Chapter 07 Continuous Probability Distributions Answer Key

True / False Questions1.A continuous uniform distribution is always symmetric.TRUEThe PDF is the same height for all X values.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-02 Calculate uniform probabilities.Topic: Uniform Continuous Distribution

2.The height and width of a continuous uniform distribution's PDF are the same.FALSEThe PDF height must be 1/(b - a) so that the total area is unity.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-02 Calculate uniform probabilities.Topic: Uniform Continuous Distribution

3.A continuous uniform distribution U(0,800) will have = 400 and = 230.94.TRUEApply the formulas for the mean and standard deviation.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-02 Calculate uniform probabilities.Topic: Uniform Continuous Distribution

4.A continuous uniform distribution U(100,200) will have the same standard deviation as a continuous uniform distribution U(200,300).TRUEIn the standard deviation formula, (b - a)2 is the same for both these examples.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-02 Calculate uniform probabilities.Topic: Uniform Continuous Distribution

5.For a continuous uniform distribution U(200,400), the parameters are = 300 and = 100.FALSEThe standard deviation is [(400 - 200)2/12]1/2 = 57.7.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-02 Calculate uniform probabilities.Topic: Uniform Continuous Distribution

6.The exponential distribution describes the number of arrivals per unit of time.FALSEArrivals per unit of time would be Poisson (but waiting time is exponential).

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

7.The exponential distribution is always skewed right.TRUEThe PDF clearly shows extreme right-skewness.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

8.If arrivals follow a Poisson distribution, waiting times follow the exponential distribution.TRUEReview the definition of an exponential distribution.

AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

9.The triangular distribution is used in "what-if" analysis for business planning.TRUESimplicity in visualizing planning scenarios is an attraction of the triangular distribution.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional).Topic: Triangular Distribution (Optional)

10.The triangular distribution is symmetric.FALSETriangular distribution is symmetric only if the mode is at the axis midpoint.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional).Topic: Triangular Distribution (Optional)

11.The triangular distribution T(0,10,20) is skewed left.FALSEOnly left-skewed if the mode is right of the axis midpoint.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional).Topic: Triangular Distribution (Optional)

12.A triangular distribution can be skewed either left or right.TRUELeft-skewed if the mode is right of the axis midpoint, and vice versa.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional).Topic: Triangular Distribution (Optional)

13.For a continuous random variable, the total area beneath the PDF will be greater than zero but less than one.FALSEThe total area must be 1 if it is a PDF.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-01 Define a continuous random variable.Topic: Describing a Continuous Distribution

14.The exponential distribution is continuous and the Poisson distribution is discrete, yet the two distributions are closely related.TRUEPoisson arrivals (discrete) imply exponential waiting times (continuous).

AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

15.The mean, median, and mode of a normal distribution will always be the same.TRUEA normal distribution is perfectly symmetric.

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 07-03 Know the form and parameters of the normal distribution.Topic: Normal Distribution

16.There is a simple formula for normal areas, but we prefer a table for greater accuracy.FALSEWe have a formula for the PDF, but there is no exact formula for areas under the curve.

AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning Objective: 07-03 Know the form and parameters of the normal distribution.Topic: Normal Distribution

17.Normal distributions differ only in their means and variances.TRUEAll normal distributions look the same except for scaling.

AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning Objective: 07-03 Know the form and parameters of the normal distribution.Topic: Normal Distribution

18.Any normal distribution has a mean of 0 and a standard deviation of 1.FALSEOnly the standardized normal is N(0,1).

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-03 Know the form and parameters of the normal distribution.Topic: Normal Distribution

19.We would use a normal distribution to model the waiting time until the next Florida hurricane strike.FALSEHurricane arrivals might be regarded as Poisson events, so waiting times are exponential.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

20.Experience suggests that 4 percent of all college students have had a tonsillectomy. In a sample of 300 college students, we need to find the probability that at least 10 have had a tonsillectomy. It is acceptable to use the normal distribution to estimate this probability.TRUEThe quick rule is n 10 and n(1 - ) 10, which is the case in this example.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.Topic: Normal Approximations

21.The normal is a good approximation to the binomial when n is greater than or equal to 10.FALSEThe quick rule is n 10 and n(1 - ) 10.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.Topic: Normal Approximations

22.The true proportion of accounts receivable with some kind of error is 4 percent for Venal Enterprises. If an auditor randomly samples 50 accounts receivable, it is acceptable to use the normal approximation to estimate the probability that fewer than two will contain errors.FALSEThe quick rule is n 10 and n(1 - ) 10, which is not fulfilled in this case.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.Topic: Normal Approximations

23.The normal distribution is a good approximation to the binomial if both 10 and n 10.FALSEThe quick rule is n 10 and n(1 - ) 10.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.Topic: Normal Approximations

24.The normal distribution is a good approximation to the binomial if n = 200 and = .03.FALSEThe quick rule is n 10 and n(1 - ) 10, which is not fulfilled in this case.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.Topic: Normal Approximations

25.The normal distribution is a good approximation to the binomial if n = 25 and = .50.TRUEThe quick rule is n 10 and n(1 - ) 10, which is fulfilled in this case.

26.The exponential distribution can be either right-skewed or left-skewed, depending on .FALSEThe PDF of the exponential shows that it is always right-skewed.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

27.The number of lightning strikes in a day in Miami is a continuous random variable.FALSEThe "number of . . ." anything is discrete.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-01 Define a continuous random variable.Topic: Describing a Continuous Distribution

28.The area under a normal curve is 1 only if the distribution is standardized N(0,1).FALSEAny normal distribution has a total area of one under the PDF.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-03 Know the form and parameters of the normal distribution.Topic: Normal Distribution

29.The area under an exponential curve can exceed 1 because the distribution is right-skewed.FALSEIf it's a PDF, the total area under the PDF is one.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

Multiple Choice Questions30.A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, the average amount of water dispensed by the machine is:

A.12 ounces.

B.13 ounces.

C.14 ounces.

D.16 ounces.

The mean is halfway between the end points of the distribution.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-02 Calculate uniform probabilities.Topic: Uniform Continuous Distribution

31.A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, the standard deviation of the amount of water dispensed is about:

A.1.73 ounces.

B.3.00 ounces.

C.0.57 ounce.

D.3.51 ounces.

The standard deviation is [(16 - 10)2/12]1/2 = 1.73.

32.A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, what is the probability that 13 or more ounces will be dispensed in a given glass?

A..1666

B..3333

C..5000

D..6666

Half the area is above 13.

33.A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. The mean of this distribution is:

A.30.5.

B.31.5.

C.32.5.

D.33.5.

The mean is halfway between the end points of the distribution.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-02 Calculate uniform probabilities.Topic: Uniform Continuous Distribution

34.A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. The standard deviation of this distribution is approximately:

A.52.1.

B.32.5.

C.6.85.

D.7.22.

The standard deviation is [(45 - 20)2/12]1/2 = 7.22.

35.A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. What is P(30 X 40)?

A..20

B..40

C..60

D..80

The desired area is 10/25 = .40.

36.The Excel function =800*RAND() would generate random numbers with standard deviation approximately equal to:

A.200.

B.188.

C.231.

D.400.

The standard deviation is [(800 - 0)2/12]1/2 = 230.94.

37.The Excel function =40*RAND() would generate random numbers with standard deviation approximately equal to

A.13.33.

B.20.00.

C.11.55.

D.19.27.

The standard deviation is [(40 - 0)2/12]1/2 = 11.55.

38.If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of waiting more than 0.5 hour for the next arrival is:

A..2407.

B..1653.

C..1222.

D..5000.

P(X > .50) = exp(-3.6 0.50) = .1653.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

39.If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of waiting less than 0.5 hour for the next arrival is:

A..7122.

B..8105.

C..8347.

D..7809.

P(X < .50) = 1 - exp(-3.6 0.50) = 1 - .1653 = .8347.

40.If arrivals occur at a mean rate of 2.6 events per minute, the exponential probability of waiting more than 1.5 minutes for the next arrival is:

A..0202.

B..0122.

C..0535.

D..2564.

P(X > 1.5) = exp(-2.6 1.50) = .0202.

41.If arrivals occur at a mean rate of 1.6 events per minute, the exponential probability of waiting less than 1 minute for the next arrival is:

A..2019.

B..7104.

C..8812.

D..7981.

(X < 1) = 1 - exp(-1.6 1) = 1 - .2019 = .7981.

42.Bob's z-score for the last exam was 1.52 in Prof. Axolotl's class BIO 417, "Life Cycle of the Ornithorhynchus." Bob said, "Oh, good, my score is in the top 10 percent." Assuming a normal distribution of scores, is Bob right?

A.Yes.

B.No.

C.Must have n to answer.

P(Z < 1.52) = .9357.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.Topic: Standard Normal Distribution

43.The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What proportion of brook trout caught will be between 12 and 18 inches in length?

A..6563

B..6826

C..2486

D..4082

P(12 < X < 18) = P(-.67 < Z < 1.33) = .6568 (from Appendix C) or .6563 using Excel.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.Topic: Standard Normal Distribution

44.The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. The first quartile for the lengths of brook trout would be:

A.16.01 inches.

B.11.00 inches.

C.11.98 inches.

D.10.65 inches.

Using Excel =NORM.INV(.25,14,3) = 11.98, or Q1 = 14 - 0.675(3) = 11.975 using Appendix C.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.Topic: Standard Normal Distribution

45.The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What lower limit should the State Game Commission set on length if it is desired that 80 percent of the catch may be kept by fishers?

A.12.80 inches

B.11.48 inches

C.12.00 inches

D.9.22 inches

Using Excel =NORM.INV(.20,14,3) = 11.475, or X = 14 - 0.84(3) = 11.48 using Appendix C.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.Topic: Standard Normal Distribution

46.In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. What percent of customers require less than 32 minutes for a simple haircut?

A.95.99 percent

B.99.45 percent

C.97.72 percent

D.45.99 percent

Using Excel =NORMDIST(32,25,4,1) = 0.9599, or use z = (32 - 25)/4 = 1.75 with Appendix C.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.Topic: Standard Normal Distribution

47.In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. The slowest quartile of customers will require longer than how many minutes for a simple haircut?

A.3(n + 1)/4 minutes

B.26 minutes

C.25.7 minutes

D.27.7 minutes

Using Excel =NORM.INV(.75,25,4) = 27.698, or Q3 = 25 + 0.675(4) = 27.7 using Appendix C.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.Topic: Standard Normal Distribution

48.In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. For a simple haircut, the middle 90 percent of the customers will require:

A.between 18.4 and 31.6 minutes.

B.between 19.9 and 30.1 minutes.

C.between 20.0 and 30.0 minutes.

D.between 17.2 and 32.8 minutes.

The 90 percent range is 1.645.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.Topic: Standard Normal Distribution

49.The area under the normal curve between z = 0 and z = 1 is ______________ the area under the normal curve between z = 1 and z = 2.

A.less than

B.greater than

C.equal to

The standard normal PDF grows closer to the axis as z increases to the right of zero.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.Topic: Standard Normal Distribution

50.The price-earnings ratio for firms in a given industry follows the normal distribution. In this industry, a firm whose price-earnings ratio has a standardized value of z = 1.00 is approximately in the highest ______ percent of firms in the industry.

A.16 percent

B.34 percent

C.68 percent

D.75 percent

About 15.86 percent of the area is above one standard deviation.

51.A student's grade on an examination was transformed to a z value of 0.67. Assuming a normal distribution, we know that she scored approximately in the top:

A.15 percent.

B.50 percent.

C.40 percent.

D.25 percent.

P(Z > 0.67) = 1 - P(Z < 0.67) = 1 - .2514 = .7486.

52.The MPG (miles per gallon) for a certain compact car is normally distributed with a mean of 31 and a standard deviation of 0.8. What is the probability that the MPG for a randomly selected compact car would be less than 32?

A.0.3944

B.0.8944

C.0.1056

D.0.5596

P(X < 32) = P(Z < 1.25) = .8944.

53.The probability is .80 that a standard normal random variable is between -z and +z. The value of z is approximately:

A.1.28.

B.1.35.

C.1.96.

D.1.45.

For tail areas of .1000 we would use z = 1.282.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.Topic: Standard Normal Distribution

54.The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What proportion of the citizens will require less than one hour?

A.0.4772

B.0.9772

C.0.9974

D.0.9997

P(X < 60) = P(Z < 2.00) = .9772.

55.The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. The slowest 10 percent of the citizens would need at least how many minutes to complete the form?

A.27.2

B.35.8

C.52.8

D.59.6

Using Excel =NORM.INV(.90,40,10) = 52.82, or 40 + 1.282(10) = 52.82 using Appendix C.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.Topic: Standard Normal Distribution

56.The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What is the third quartile (in minutes) for the time required to complete the form?

A.44.75

B.46.75

C.47.50

D.52.50

Using Excel =NORM.INV(.75,40,10) = 46.75, or Q3 = 40 + 0.675(10) = 46.75 using Appendix C

57.Exam scores were normal in BIO 200. Jason's exam score was one standard deviation above the mean. What percentile is he in?

A.68th

B.75th

C.78th

D.84th

About 15.87 percent of the area lies above one standard deviation.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.Topic: Standard Normal Distribution

58.Compared to the area between z = 1.00 and z = 1.25, the area between z = 2.00 and z = 2.25 in the standard normal distribution will be:

A.smaller.

B.larger.

C.the same.

D.impossible to compare without knowing and .

The normal PDF approaches the axis as z increases beyond zero.

59.A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of applicants would you expect to have scores of 600 or above?

A.0.0401

B.0.4599

C.0.5401

D.0.0852

P(X > 600) = P(Z > 1.75) = .0401.

60.A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of the applicants would you expect to have a score of 400 or above?

A.0.2734

B.0.7734

C.0.7266

D.0.7500

P(X > 400) = P(Z > -0.75) = .7734.

61.A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. The top 2.5 percent of the applicants would have a score of at least (choose the nearest integer):

A.606.

B.617.

C.600.

D.646.

Using z = 1.96, we get X = 460 + 1.96 80 = 616.8.

62.If the random variable Z has a standard normal distribution, then P(1.25 Z 2.17) is:

A.0.0906.

B.0.9200.

C.0.4700.

D.0.3944.

P(Z 2.17) - P(Z 1.25) = .0906.

63.If the random variable Z has a standard normal distribution, then P(Z -1.37) is:

A.0.9147.

B.0.4147.

C.0.5016.

D.0.0853.

From Appendix C we get the left tail area of .0853.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.Topic: Standard Normal Distribution

64.Assume that X is normally distributed with a mean = $64. Given that P(X $75) = 0.2981, we can calculate that the standard deviation of X is approximately:

A.$20.76.

B.$13.17.

C.$5.83.

D.$7.05.

For a right-tail area of .2981 we need z = -.53, so with x = 75 we set z = (x - )/ and solve for .

65.The standard deviation of a normal random variable X is $20. Given that P(X $10) = 0.1841. From this we can determine that the mean of the distribution is equal to:

A.$13.

B.$26.

C.$20.

D.$28.

For a left-tail area of .1841 we need z = -.90, so with x = 10 we set z = (x - )/ and solve for .

66.The random variable X is normally distributed with mean of 80 and variance of 36. The 67th percentile of the distribution is:

A.72.00.

B.95.84.

C.90.00.

D.82.64.

Since P(Z < 0.44) = .6700, (from Appendix C) we get 80 + 0.44(6) = 82.64.

67.The area under the normal curve between the 20th and 70th percentiles is equal to:

A.0.7000.

B.0.5000.

C.0.9193.

Logically, this must be .70 - .20 = .50, as you can verify from Appendix C.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.Topic: Normal Distribution

68.The variable in a normal distribution can assume any value between

A.-3 and +3

B.-4 and +4

C.-1 and +1

D.- and +

Almost all the area is within -3 and +3, but the curve never quite touches the z-axis.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-03 Know the form and parameters of the normal distribution.Topic: Normal Distribution

69.What are the mean and standard deviation for the standard normal distribution?

A. = 0, = 0

B. = 1, = 1

C. = 1, = 0

D. = 0, = 1

We have standardized so the mean must be zero.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 07-03 Know the form and parameters of the normal distribution.Topic: Standard Normal Distribution

70.Any two normal curves are the same except for their:

A.standard deviations.

B.means.

C.standard deviations and means.

D.standard deviations, means, skewness, and kurtosis.

We write N(,) so show the similarity of all normals.

71.Light bulbs are normally distributed with an average lifetime of 1000 hours and a standard deviation of 250 hours. The probability that a light bulb picked at random will last less than 1500 hours is about:

A.97.72 percent.

B.95.44 percent.

C.75.00 percent.

D.68.00 percent.

P(Z < 1500) = P(Z < 2.00) = .9772 from Appendix C (or from Excel).

72.To convert a normally distributed variable X into a standard Z score we would:

A.subtract the mean from the original observation and divide the result by the variance.

B.subtract the mean from the original observation and divide the result by the standard deviation.

C.add the mean and the original observation, then divide by the variance.

D.subtract the mean from the standard deviation and divide by the variance.

Review the z-score transformation.

AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning Objective: 07-03 Know the form and parameters of the normal distribution.Topic: Standard Normal Distribution

73.Regarding continuous probability distributions, which statement is incorrect?

A.The triangular distribution may be skewed left or right.

B.The uniform distribution is never skewed.

C.The normal distribution is sometimes skewed.

D.The exponential distribution is always skewed right.

Review the characteristics of these four distributions.

AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning Objective: 07-03 Know the form and parameters of the normal distribution.Topic: Normal Distribution

74.Which model best describes your waiting time until you get the next nonworking web URL ("This page cannot be displayed") as you click on various websites for Florida condo rentals?

A.Triangular

B.Uniform

C.Normal

D.Exponential

Waiting time until the next event resembles an exponential distribution.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

75.On average, a major earthquake (Richter scale 6.0 or above) occurs 3 times a decade in a certain California county. What is the probability that less than six months will pass before the next earthquake?

A..1393

B..8607

C..0952

D..9048

Set = 3/120 = 0.025 earthquake/month so P(X < 6) = 1 - exp(-0.025 6) = 1 - .8607 = .1393.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

76.If the mean time between in-flight aircraft engine shutdowns is 12,500 operating hours, the 90th percentile of waiting times to the next shutdown will be:

A.20,180 hours.

B.28,782 hours.

C.23,733 hours.

D.18,724 hours.

Set = 1/12500 and solve for x in left-tail area of 1 - exp(-x) = .90 by taking logs of both sides.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-08 Solve for x for a given exponential probability.Topic: Exponential Distribution

77.On average, 15 minutes elapse between discoveries of fraudulent corporate tax returns in a certain IRS office. What is the probability that less than 30 minutes will elapse before the next fraudulent corporate tax return is discovered?

A..1353

B..6044

C..7389

D..8647

P(X < 30) = 1 - exp(-x) = 1 - exp(-(1/15) 30) = 1 - .1353 = .8647.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

78.If the mean time between unscheduled maintenance of LCD displays in a hospital's CT scan facility is 4,000 operating hours, what is the probability of unscheduled maintenance in the next 5,000 hours?

A..8000

B..7135

C..2865

D..5000

P(X < 5000) = 1 - exp(-x) = 1 - exp(-(1/4000) 5000) = 1 - .2865 = .7135.

79.A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. What is the probability that less than 6 minutes will elapse before the next breakdown?

A..8187

B..0488

C..1813

D..2224

(X < 6) = 1 - exp(-x) = 1 - exp(-(1/30) 6) = 1 - .8187 = .1813.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-07 Find the exponential probability for a given x.Topic: Exponential Distribution

80.A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. The median time between breakdowns is:

A.30.0 minutes.

B.35.7 minutes.

C.25.4 minutes.

D.20.8 minutes.

Set = 1/30 and solve for x in right-tail area of 1 - exp(-x) = .50 by taking logs of both sides.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-08 Solve for x for a given exponential probability.Topic: Exponential Distribution

81.Which probability model is most appropriate to describe the waiting time (working days) until an office photocopier breaks down (i.e., requires unscheduled maintenance)?

A.Normal

B.Uniform

C.Exponential

D.Poisson

Poisson breakdowns suggest exponential waiting time.

82.Bob's z-score for the last exam was -1.15 in FIN 417, "Capital Budgeting Strategies." Bob said, "Yipe! My score is within the bottom quartile." Assuming a normal distribution, is Bob right?

A.Yes

B.No

C.Must know the class size to answer

The bottom quartile would be below z = -.675 so Bob is indeed below that point.

83.Exam scores were normal in MIS 200. Jason's exam score was 1.41 standard deviations above the mean. What percentile is he in?

A.68th

B.75th

C.84th

D.92nd

P(Z < 1.41) = .9207.

84.Compared to the area between z = 0.50 and z = 0.75, the area between z = 1.50 and z = 1.75 in the standard normal distribution will be:

A.smaller

B.larger

C.the same

The normal PDF approaches the axis as z increases beyond zero so areas get smaller.

85.If GMAT scores for applicants at Oxnard Graduate School of Business are N(500, 50) then the top 5 percent of the applicants would have a score of at least (choose the nearest integer):

A.575.

B.582.

C.601.

D.608.

The top 5 percent would require z = 1.645 so x = 500 + 1.645(50) = 582.25.

86.If the random variable Z has a standard normal distribution, then P(1.17 Z 2.26) is:

A.0.1091.

B.0.1203.

C.0.2118.

D.0.3944.

Subtract P(Z 2.26) - P(Z 1.17).

87.If the random variable Z has a standard normal distribution, then P(Z -1.72) is:

A.0.9573.

B.0.0446.

C.0.5016.

D.0.0427.

Use Appendix C or Excel =NORM.S.DIST(-1.72,1).

88.Excel's =100*RAND() function produces continuous random numbers that are uniformly distributed between 0 and 100. The standard deviation of this distribution is approximately:

A.50.00.

B.28.87.

C.33.33.

D.25.00.

The standard deviation is [(b - a)2/12]1/2 = [(100 - 0)2/12]1/2.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-02 Calculate uniform probabilities.Topic: Normal Distribution

89.Excel's =RAND() function produces random numbers that are uniformly distributed between 0 and 1. The mean of this distribution is approximately

A..5000

B..2500

C..3333

D..2887

The mean is halfway between the end points.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 07-02 Calculate uniform probabilities.Topic: Uniform Continuous Distribution

90.Excel's =RAND() function produces random numbers that are uniformly distributed from 0 to 1. What is the probability that the random number exceeds .75?

A.75 percent

B.50 percent

C.25 percent

This is the upper 25 percent.

91.Which is the correct Excel formula for the 80th percentile of a distribution that is N(475, 33)?

A.=NORM.DIST(80, 475, 33,1)

B.=NORM.INV(0.80, 475, 33)

C.=NORM.S.INV((80 - 475)/33)

Review Excel functions in Appendix J.

AACSB: TechnologyBlooms: RememberDifficulty: 2 MediumLearning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.Topic: Standard Normal Distribution

92.If arrivals follow a Poisson distribution with mean 1.2 arrivals per minute, find the 75th percentile of waiting times until the next arrival (i.e., 75 percent below).

A.1.155 minutes (69.3 seconds)

B.0.240 minute (14.4 seconds)

C.1.919 minutes (115.1 seconds)

Set = 1.2 and solve for x in right-tail area of exp(-x) = .25 by taking logs of both sides.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-08 Solve for x for a given exponential probability.Topic: Exponential Distribution

93.A software developer makes 175 phone calls to its current customers. There is an 8 percent chance of reaching a given customer (instead of a busy signal, no answer, or answering machine). The normal approximation of the probability of reaching at least 20 customers is:

A..022.

B..007.

C..063.

D..937.

Set n = 175 and = .08. Calculate = n = (175)(.08) = 14 and = [n(1 - )]1/2 = [175(.08)(1 - .08)]1/2 = 3.588872. Use x = 19.5 (with the continuity correction) and calculate the binomial P(X 20) P(z 1.532515) using z = (x - )/ = 1.532515.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.Topic: Normal Approximations

94.For Gardyloo Manufacturing, the true proportion of accounts receivable with some kind of error is .20. If an auditor randomly samples 225 accounts receivable, what is the approximate normal probability that 39 or fewer will contain errors?

A..1797

B..2097

C..1587

D..0544

Set = n and = [n(1 - )]1/2 and convert x = 39.5 (using the continuity correction) to a z score with z = (x - )/. Set n = 225 and = .20. Calculate = n = (225)(.20) = 45 and = [n(1 - )]1/2 = [225(.20)(1 - .20)]1/2 = 6.000. Use x = 39.5 (with the continuity correction) and calculate the binomial P(X 39) P(z -.916667) using z = (x - )/ = -.916667.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.Topic: Normal Approximations

95.A letter is mailed to a sample of 500 homeowners. Based on past experience, the probability of an undeliverable letter is 0.06. The normal approximation to the binomial probability of 40 or more undeliverable letters is:

A.0.9632

B.0.0368

C.0.2305

D.0.7695

Set n = 500 and = .06. Calculate = n = (500)(.06) = 30 and = [n(1 - )]1/2 = [500(.06)(1 - .06)]1/2 = 5.31037. Use x = 39.5 (with the continuity correction) and calculate the binomial P(X 40) P(z 1.78895) using z = (x - )/ = 1.78895.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.Topic: Normal Approximations

96.In a T-F exam with 100 questions, passing requires a score of at least 60. What is the approximate normal probability that a "guesser" will score at least 60 points?

A..0287

B..4713

C..0251

D..0377

A guesser would have a 50 percent chance of a correct answer, so we set = .50. There are n = 100 questions, so we calculate = n = (100)(.50) = 50 and = [n(1 - )]1/2 = [100(.50)(1 - .50)]1/2 = 5. Use x = 59.5 (with the continuity correction) and calculate the binomial P(X 60) P(z 1.90) using z = (x - )/ = 1.90.

97.A multiple choice exam has 100 questions. Each question has five choices. What would be the approximate probability that a "guesser" could achieve a score of 30 or more?

A.0.0088

B.0.0062

C.0.0015

D.0.4913

A guesser would have a 20 percent chance of a correct answer (1 out of 5) so we set = .20. There are n = 100 questions, so we calculate = n = (100)(.20) = 20 and = [n(1 - )]1/2 = [100(.20)(1 - .20)]1/2 = 4. Use x = 29.5 (with the continuity correction) and calculate the binomial P(X 30) P(z 2.375) using z = (x - )/ = 2.375.

98.For which binomial distribution would a normal approximation be most acceptable?

A.n = 50, = 0.05

B.n = 100, = 0.04

C.n = 40, = 0.25

D.n = 400, = 0.02

We want n 10 and n(1 - ) 10.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.Topic: Normal Approximations

99.Historically, the default rate on a certain type of commercial loan is 20 percent. If a bank makes 100 of these loans, what is the approximate probability that at least 26 will result in default?

A.0.2000

B.0.0668

C.0.0846

D.0.0336

Set n = 100 and = .20. Calculate = n = (100)(.20) = 20 and = [n(1 - )]1/2 = [100(.20)(1 - .20)]1/2 = 4. Use x = 25.5 (with the continuity correction) and calculate the binomial P(X 26) P(z 1.375) using z = (x - )/ = 1.375.

100.A company employs 300 employees. Each year, there is a 30 percent turnover rate for employees. We want to do a normal approximation to the binomial distribution of the number of employees who leave each year. For this normal approximation, the mean is ______ and the standard deviation is _____.

A.90, 63

B.90, 7.937

C.90, 30

D.90, 15

Use n = 300 and = .30, and then calculate = n and = [n(1 - )]1/2.

101.The probability that a rental car will be stolen is 0.001. If 25,000 cars are rented from Hertz, what is the normal approximation to the probability that fewer than 20 will be stolen?

A..2577

B..1335

C..1128

D..8335

Set n = 25,000 and = .001. Calculate = n = (25000)(.001) = 25 and = [n(1 - )]1/2 = [25000(.001)(1 - .001)]1/2 = 4.9975 Use x = 19.5 (with the continuity correction) and calculate the binomial P(X < 20) P(z