In a standardized test for lawyers, scores are normally distributed and the standard deviation is known to be 28.5 points. How large of a sample must to find the estimate the mean score of all lawyers taking this test if you want the margin of error to be less than 10 points. Use a 95% level of confidence

To find the required sample size for the mean score, we can use the formula:

n = (zc sigma / E)^2

where

n is the sample size

zc is the critical value associated with 95% confidence, in this example zc = 1.96

sigma is standard deviation

E is margin of error.

Thus we have:

n = (zc sigma / E)^2

n = (1.96 * 28.5 / 10)^2

n = 31.2

We always round samples sizes in these questions up to the next higher number so the sample size here would be 32