If the average (xbar) on an assignment in a class was 82% with a standard deviation (s) of 4.8, what is the z-score for someone scoring 75% (round to the one hundredths place or two decimal places)?

Answer 1

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

Answer 2

82-75= 7/4.8=1.45