Question

According to U.S. News & World Report's publication America's Best Colleges, the average cost to attend the University of Southern California (USC) after deducting grants based on need is $27,775 . Assume the population standard deviation is $8,200. Suppose that a random sample of 60 USC students will be taken from this population. Use z-table.

a. What is the value of the standard error of the mean? (to nearest whole number)

b. What is the probability that the sample mean will be more than 27,775? (to 2 decimals)

c. What is the probability that the sample mean will be within $1,500 of the population mean? (to 4 decimals)

d. How would the probability in part (c) change if the sample size were increased to 110? (to 4 decimals)

Answer 1

Solution: Given that mean = 27775, standard deviation = 8200, n = 60

a. the standard error = s/sqrt(n) = 8200/sqrt(60) = 1058.615 = 1059

b. P(X > 27775) = P((X-mean)/sd > (27775-27775)/1059)
= P(Z > 0)
= 0.5

c. Z = 1500/(8200/sqrt(60)) = +/- 1.4169

P(-1.4169 < Z < +1.4169) = 0.8444

d. if the sample size increases,then the sample mean will estimate the population mean better and thus the probability that the sample mean is withij 1500 of the population mean would then increase