Percentile

[vicki]

th of time for for students to complete a test is normally distributed with mean 75 minutes and standard deviation 15 minutes. How much time should be allowed so that 90% of students will finish the exam?

[Adrian]

We first find the 90th percentile of the standard normal random variable. From the normal table, P(Z < 1.28)=0.90

Let x be the 90th percentile for the length of time for for students to complete a test.
ie 90% of students would finish the exam before x.

(x-mean)/sd = 1.28
(x-75)/15 = 1.28
x = 75 + 1.28*15
x = 94.2 minutes