Question

The director of a MPA program finds the incoming 200 students in the department with a mean GPA of 3.04 and a standard deviation of .57. Assuming that students' GAPs are normally distributed, what is the range of GPA of 95% students?

Answer 1

P ( a < Z < b ) = 0.95

0.95 / 2 = 0.475

Sine in the Normal curve 0.5 area above and below the mean

a = 0.5 - 0.475 = 0.025

b= 0.5 + 0.475 = 0.975

Looking for the probability 0.025 & 0.975 in standard normal table to find the critical value Z = -1.96 & Z = 1.96

-1.96 = ( X - 3.04 ) / 0.57

X = 1.92 i.e a = 1.92

1.96 = ( X - 3.04 ) / 0.57

X = 4.16 i.e   b = 4.16

P ( 1.92 < Z < 4.16 ) = 0.95