MBA535 (Analytical Tools for Decision-Making) Marist College Week 4 Assessment (20 points each question; 160 points total) Question 1. a) For some positive value of Z, the probability that a standard normal variable is between 0 and Z is 0.3770. What is the value of Z ? (Explain/show how you obtain your answer.) (7 pts.) b) For some value of Z, the probability that a standard normal variable is below Z is 0.2090. What is the value of Z ? (Explain/show how you obtain your answer.) (7 pts.) c) For some positive value of Z, the probability that a standard normal variable is between 0 and Z is 0.3340. What is the value of Z ? (Explain/show how you obtain your answer.) (6 pts.) Question 2. Suppose we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute. a) What is the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes? (Explain/show how you obtain your answer.) (7 pts.) b) What is the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot? (Explain/show how you obtain your answer.) (7 pts.) c) What is the point in the distribution in which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot? (Explain/show how you obtain your answer.) (6 pts.) Question 3. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. a) What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order? (Explain/show how you obtain your answer.) (7 pts.) b) What proportion of customers having to hold more than 1.5 minutes will hang up before placing an order? (Explain/show how you obtain your answer.) (7 pts.) c) Find the waiting time at which only 10% of the customers will continue to hold. (Explain/show how you obtain your answer.) (6 pts.) Question 4. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. a) What proportion of the plan recipients would receive payments beyond age 75? (Explain/show how you obtain your answer.) (7 pts.) b) What proportion of the plan recipients die before they reach the standard retirement age of 65? (Explain/show how you obtain your answer.) (7 pts.) c) Find the age at which payments have ceased for approximately 86% of the plan participants. (Explain/show how you obtain your answer.) (6 pts.) Question 5. Major league baseball salaries averaged $3.26 million with a standard deviation of $1.2 million in 2009. Suppose a sample of 100 major league players was taken. a) What was the standard error for the sample mean salary? (Explain/show how you obtain your answer.) (4 pts.) b) Find the approximate probability that the mean salary of the 100 players exceeded $3.5 million. (Explain/show how you obtain your answer.) (4 pts.) c) Find the approximate probability that the mean salary of the 100 players exceeded $4.0 million. (Explain/show how you obtain your answer.) (4 pts.) d) Find the approximate probability that the mean salary of the 100 players was no more than $3.0 million. (Explain/show how you obtain your answer.) (4 pts.) e) Find the approximate probability that the mean salary of the 100 players was less than $2.5 million. (Explain/show how you obtain your answer.) (4 pts.) Question 6. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. a) What is the standard error for the sample mean? (Explain/show how you obtain your answer.) (5 pts.) b) What is the probability that the sample mean will be between 0.99 and 1.01 centimeters? (Explain/show how you obtain your answer.) (5 pts.) c) What is the probability that the sample mean will be below 0.95 centimeters? (Explain/show how you obtain your answer.) (5 pts.) d) Above what value do 2.5% of the sample means fall? (Explain/show how you obtain your answer.) (5 pts.) Question 7. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. a) If a sample of 16 fish is taken, what would the standard error of the mean weight equal? (Explain/show how you obtain your answer.) (5 pts.) b) If a sample of 25 fish yields a mean of 3.6 pounds, what is the Z-score for this observation? (Explain/show how you obtain your answer.) (5 pts.) c) If a sample of 64 fish yields a mean of 3.4 pounds, what is probability of obtaining a sample mean this large or larger? (Explain/show how you obtain your answer.) (5 pts.) d) What percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds? (Explain/show how you obtain your answer.) (5 pts.) Question 8. Online customer service is a key element to successful online retailing. According to a marketing survey, 37.5% of online customers take advantage of the online customer service. Random samples of 200 customers are selected. a) What is the population mean of all possible sample proportions? (Explain/show how you obtain your answer.) (4 pts.) b) What is the standard error of all possible sample proportions? (Explain/show how you obtain your answer.) (4 pts.) c) What percent of the samples are likely to have between 35% and 40% who take advantage of online customer service? (Explain/show how you obtain your answer.) (4 pts.) d) What percent of the samples are likely to have less than 37.5% who take advantage of online customer service? (Explain/show how you obtain your answer.) (4 pts.) e) Please fill-in the blanks: 95% of the samples proportions symmetrically around the population proportion will have between ________% and ________% of the customers who take advantage of online customer service. (Explain/show how you obtain your answer.) (4 pts.)