What must a student score to be in the 90th percentile?

Question

Scores on a national exam were normally with a mean of 400 and standard deviation of 75.

if 16 students take the exam what is the probability that the mean score is at most 390 ??

Answer 1

z = (390 - 400) / ( 75 / sqrt(16)) =

(-10)/ ( 75 / 4) =

-10/1 * 4/ 75 =

-40/75

= -8/15 = -0.53

Per the normal table,

the probability is 0.296901 or just under 29.7%