The monthly earnings of a group of business students are are normally distributed with a standard

deviation of 589 dollars. A researcher wants to estimate the mean monthly earnings of all business

students. Find the sample size needed to have a confidence level of 95% and a margin of error of

132 dollars.

Margin of error = + / - Z*SD/sqrt (n)

Where, Z = 1.96 at 95% confidence interval, SD = standard deviation = 589 dollars, n = sample size (to be calculated)

For the values given;

132 = 1.96*589/Sqrt (n) = > Sqrt (n) = 1.96*589/132 = 8.7458 = > n = 8.7458^2 = 76.49

Rounding up to near a person, n (sample size) = 77 students