Data Analysis And Decision Making 4th Edition By S. Christian Albright – Test Bank
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NARRBEGIN: SA_56_61 A popular retail store knows that the distribution of purchase amounts by its customers is approximately normal with a mean of $30 and a standard deviation of $9. Below you will find normal probability and percentile calculations related to the customer purchase amounts.
P(Sales < $ 15.00) = 0.048, P(Sales < $ 20.00) = 0.133, P(Sales < $ 25.00) = 0.289, P(Sales < $ 35.00) = 0.711
1st Percentile = $9.06, 5th Percentile = $15.20, 95th Percentile = $44.80, 99th Percentile = $50.94 NARREND
What is the probability that a randomly selected customer will spend less than $15? ANS:
0.048
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that a randomly selected customer will spend $20 or more? ANS:
1.0 – 0.133 = 0.867
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that a randomly selected customer will spend $30 or more? ANS:
0.50
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that a randomly selected customer will spend between $20 and $35? ANS:
0.711 – 0.133 = 0.578
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What two dollar amounts, equidistant from the mean of $30, such that 90% of all customer purchases are between these values? ANS:
$15.20 to $44.80
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What two dollar amounts, equidistant from the mean of $30, such that 98% of all customer purchases are between these values? ANS:
$9.06 to $50.94
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_62_63 The weekly demand for General Motors (GM) car sales follows a normal distribution with a mean of 40,000 cars and a standard deviation of 12,000 cars. NARREND
There is a 5% chance that GM will sell more than what number of cars during the next year? ANS:
Let X represent the number of cars that will be sold during the next year. Assuming that demands in different weeks are independent, then the yearly distribution is normal with mean and standard deviation given by:
Therefore the 95th percentile (using Excel) is 2.222 million cars.
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that GM will sell between 2.0 and 2.3 million cars during the next year? ANS:
== 0.8184
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_64_65 The height of a typical American male adult is normally distributed with a mean of 68 inches and a standard deviation of 5 inches. We observe the heights of 12 American male adults. NARREND
What is the probability that exactly half the male adults will be less than 62 inches tall? ANS:
Let X be the height of a male adult. Then P(X<62) = P(Z<-1.2) = 0.1151 Number of male adults who are less than 62 inches tall follows a binomial distribution with n =12 and p = 0.1151. Then P(exactly 6 under 62 inches) = 0.00103
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Let Y be the number of the 12 male adults who are less than 62 inches tall. Determine the mean and standard deviation of Y. ANS:
Y follows a binomial distribution with n =12 and p = 0.1151. Then,
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_66_71 The weekly demand for a particular automobile manufacturer follows a normal distribution with a mean of 40,000 cars and a standard deviation of 10,000. Below you will find probability and percentile calculations related to the customer purchase amounts. Use this information to answer the following questions.
P(Sales < 2,000,000) = 0.134, P(Sales < 2,050,000) = 0.339 P(Sales < 2,100,000) = 0.609, P(Sales < 2,150,000) = 0.834
1st percentile = 1,912,245, 5th percentile = 1,961,388 95th percentile = 2,198,612, 99th percentile = 2,247,755 NARREND
Calculate the mean, variance, and standard deviation for the entire year (assume 52 weeks in the year). ANS:
Mean = 2,080,000, Variance = 5,200,000,000, Std. Dev. = 72,111
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions There is a 1% chance that this company will sell more than what number of cars during the next year? ANS:
2,247,755
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that this company will sell more than 2 million cars next year? ANS:
1.0 – 0.134 = 0.866
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that this company will sell between 2.0 and 2.15 million cars next year? ANS:
0.834 – 0.134 = 0.700
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What number of cars, equidistant from the mean, such that 90% of car sales are between these values? ANS:
1,961,388 to 2,198,612
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What number of cars, equidistant from the mean, such that 98% of car sales are between these values? ANS:
1,912,245 to 2,247,755
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_71_75 Wendy’s fast-food restaurant sells hamburgers and chicken sandwiches. On a typical weekday, the demand for hamburgers is normally distributed with a mean of 450 and standard deviation of 80 and the demand for chicken sandwiches is normally distributed with a mean of 120 and standard deviation of 30. Use this information to answer the following questions. NARREND
How many hamburgers must the restaurant stock to be 99% sure of not running out on a given day? ANS:
The 99th percentile for hamburger sandwiches = 636.1 (To be on the safe side, round up to 637).
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions How many chicken sandwiches must the restaurant stock to be 99% sure of not running out on a given day? ANS:
The 99th percentile for chicken sandwiches = 189.79 (To be on the safe side, round up to 190).
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions If the restaurant stocks 600 hamburgers and 150 chicken sandwiches for a given day, what is the probability that it will run out of hamburgers or chicken sandwiches (or both) that day? Assume that the demands for hamburgers and chicken sandwiches are probabilistically independent. ANS:
P( stock out of at least one) = 1-P (no stock out of either =1 – 0.8158 = 0.1842
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Why is the independence assumption in Question 74 probably not realistic? Using a more realistic assumption, do you think the probability in Question 74 would increase or decrease? ANS:
It might not be realistic to assume independence. If hamburgers and chicken sandwiches are “substitute” products; that is, a customer will order one if there is none of the other, then the two demands would be positively correlated. Still, it is not clear that this would have a large effect on the probability in Question 132.
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_76_78 A set of final exam scores in an organic chemistry course was found to be normally distributed, with a mean of 73 and a standard deviation of 8. NARREND
What percentage of students scored between 81 and 89 on this exam? ANS:
P(81 X 89) = P(1.00 Z 2.00) = 0.9772 – 0.8413 = 0.1359 Therefore, 13.59% of students scored between 81 and 89 on this exam.
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability of getting a score higher than 85 on this exam? ANS:
P(X > 85) = P(Z > 1.5) = 0.0668
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Only 5% of the students taking the test scored higher than what value? ANS:
P(X > A) = 0.05. Then, P(Z ) = 0.95 A = 73 + 1.645(8) = 86.16 Therefore, only 5% of the students taking the test scored higher than about 86.2.
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_79_90 The service manager for a new appliances store reviewed sales records of the past 20 sales of new microwaves to determine the number of warranty repairs he will be called on to perform in the next 90 days. Corporate reports indicate that the probability any one of their new microwaves needs a warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are independent of one another and is interested in predicting the number of warranty repairs he will be called on to perform in the next 90 days for this batch of 20 new microwaves sold. NARREND
What type of probability distribution will most likely be used to analyze warranty repair needs on new microwaves in this situation? ANS:
Binomial distribution
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that none of the 20 new microwaves sold will require a warranty repair in the first 90 days? ANS:
P(X= 0) = 0.3585
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that exactly two of the 20 new microwaves sold will require a warranty repair in the first 90 days? ANS:
P(X = 2) = 0.1887
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that less than two of the 20 new microwaves sold will require a warranty repair in the first 90 days? ANS:
P(X < 2) = 0.7358
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that at most two of the 20 new microwaves sold will require a warranty repair in the first 90 days? ANS:
P(X 2) = 0.9245
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that only one of the 20 new microwaves sold will require a warranty repair in the first 90 days? ANS:
P(X = 1) = 0.3774
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that more than one of the 20 new microwaves sold will require a warranty repair in the first 90 days? ANS:
P(X > 1) = 0.2642
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that at least one of the 20 new microwaves sold will require a warranty repair in the first 90 days? ANS:
P(X 1) = 0.6415
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that between two and four (inclusive) of the 20 new microwaves sold will require a warranty repair in the first 90 days? ANS:
P(2 X 4) = 0.2616
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that between three and six (exclusive) of the 20 new microwaves sold will require a warranty repair in the first 90 days? ANS:
P(3 X 6) = 0.0156
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the expected number of the new microwaves sold that will require a warranty repair in the first 90 days? ANS:
E(X) = np = 1.0
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the standard deviation of the number of the new microwaves sold that will require a warranty repair in the first 90 days? ANS:
= 0.9747
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_91_94 Consider a binomial random variable X with n = 5 and p = 0.40. NARREND
Find the probability distribution of X. ANS:
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find P(X < 3). ANS:
0.6826
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find P(2X4). ANS:
0.6528
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find the mean and the variance of X. ANS:
E(X) = 2, and Var(X) = 1.2
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_95_100 A recent survey in Michigan revealed that 60% of the vehicles traveling on highways, where speed limits are posted at 70 miles per hour, were exceeding the limit. Suppose you randomly record the speeds of ten vehicles traveling on US 131 where the speed limit is 70 miles per hour. Let X denote the number of vehicles that were exceeding the limit. NARREND
Describe the probability distribution of X. ANS:
Binomial with n = 10 and p = 0.60.
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find P(X = 10). ANS:
0.006
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find P(4 < X < 9). ANS:
0.788
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find P(X = 2). ANS:
0.01
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find P(3X6). ANS:
0.607
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Suppose that an highway patrol officer can obtain radar readings on 500 vehicles during a typical shift. How many traffic violations would be found in a shift? ANS:
300
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_101_103 Past experience indicates that 20% of all freshman college students taking an intermediate algebra course withdraw from the class. NARREND
(A) Using the binomial distribution, find the probability that 6 or more of the 30 students taking this course in a given semester will withdraw from the class. (B) Using the normal approximation to the binomial, find the probability that 6 or more of the 30 students taking this course in a given semester will withdraw from the class.
(C) Compare the results obtained in (A) and (B). Under what conditions will the normal approximation to this binomial probability become even more accurate? ANS:
(A) Parameters for binomial are n = 30 and p = 0.20. Then, (using Excel)
(B) Parameters for normal approximation to the binomial are mean = and standard deviation. Then (using Excel). (C) The results are not identical, but very close. We have used the continuity correction (making the interval longer by 0.5) for the normal approximation in (B). Generally, the normal approximation to the binomial gets better as n increases and p is not too close to 0 or 1.
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_104_106 A large retailer has purchased 10,000 DVDs. The retailer is assured by the supplier that the shipment contains no more than 1% defective DVDs (according to agreed specifications). To check the supplier’s claim, the retailer randomly selects 100 DVDs and finds six of the 100 to be defective. NARREND
(A) Assuming the supplier’s claim is true, compute the mean and the standard deviation of the number of defective DVDs in the sample. (B) Based on your answer to (A), is it likely that as many as six DVDs would be found to be defective, if the claim is correct?
(C) Suppose that six DVDs are indeed found to be defective. Based on your answer to (A), what might be a reasonable inference about the manufacturer’s claim for this shipment of 10,000 DVDs? ANS:
(A) Mean = np =1, and standard deviation = 0.995
(B) No. If you were 3 standard deviations to the right of the mean, the value would be 3.985. It is unlikely you would observe 6 defects out of 100. (C) You would have to infer that the manufacturer’s claim is incorrect. Based on this observation, the supplier appears to have a higher defect rate than 1%.
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_107_111 Suppose that the number of customers arriving each hour at the only checkout counter at a local convenience store is approximately Poisson distributed with an expected arrival rate of 30 customers per hour. Let X represent the number of customers arriving per hour. The probabilities associated with X are shown below.
P(X < 20) = 0.0219, P(X < 25) = 0.1572, P(X < 30) = 0.4757 P(X = 30) = 0.0726, P(X = 31) = 0.0703, P(X = 32) = 0.0659, P(X = 33) = 0.0599, P(X = 34) = 0.0529, P(X = 35) = 0.0453 NARREND
What is the probability that at least 25 customers arrive at this checkout counter in a given hour? ANS:
0.843
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that at least 20 customers, but fewer than 30 customers arrive at this checkout counter in a given hour? ANS:
0.4538
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that fewer than 33 customers arrive at this checkout counter in a given hour? ANS:
0.6718
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that the number of customers who arrive at this checkout counter in a given hour will be between 30 and 35 (inclusive)? ANS:
0.3669
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that the number of customers who arrive at this checkout counter in a given hour will be greater than 35? ANS:
0.1574
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_112_114 The number of arrivals at a local gas station between 3:00 and 5:00 P.M. has a Poisson distribution with a mean of 12. NARREND
Find the probability that the number of arrivals between 3:00 and 5:00 P.M. is at least 10. ANS:
=12; P(X 10) = 0.758
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find the probability that the number of arrivals between 3:30 and 4:00 P.M. is at least 10. ANS:
= 3; P(X 10) = 0.001
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find the probability that the number of arrivals between 4:00 and 5:00 P.M. is exactly two. ANS:
= 6; P(X = 2) = 0.045
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_115_116 A used car salesman in a small town states that, on the average, it takes him 5 days to sell a car. Assume that the probability distribution of the length of time between sales is exponentially distributed. NARREND
What is the probability that he will have to wait at least 8 days before making another sale? ANS:
0.2019
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that he will have to wait between 6 and 10 days before making another sale? ANS:
0.1659
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_117_120 The time it takes a technician to fix a computer problem is exponentially distributed with a mean of 15 minutes. NARREND
What is the probability density function for the time it takes a technician to fix a computer problem? ANS:
f(x) = (15), x 0
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that it will take a technician less than 10 minutes to fix a computer problem? ANS:
0.4866
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the variance of the time it takes a technician to fix a computer problem? ANS:
Var (X) = 225
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that it will take a technician between 10 to 15 minutes to fix a computer problem? ANS:
0.1455
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions NARRBEGIN: SA_121_124 A continuous random variable X has the probability density function: f(x) = 2, 0 NARREND
What is the distribution of X and what are the parameters? ANS:
Exponential with parameter = 2
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Find the mean and standard deviation of X. ANS:
E(X) = Stdev (X) = 1/ = 0.5
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that X is between 1 and 3? ANS:
0.1329
PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions What is the probability that X is at most 2? ANS:
0.9817 PTS: 1 MSC: AACSB: Analytic | AACSB: Probability Distributions Download 55.59 Kb. Share with your friends:
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