mean and standard deviation

Question

A particular fruit's weights are normally distributed, with a mean of 595 grams and a standard deviation of 29 grams.

If you pick 16 fruits at random, then 3% of the time, their mean weight will be greater than how many grams?

Answer 1

z = (xbar - mu)/(SD/sqrt(n))

z for 3rd percentile = -2.78

mu = 595

SD = 29

n = 16

so

-2.78 = (xbar - 595)/(29/sqrt(16))

solve for xbar