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.)