Answer to Question #345816 in Statistics and Probability for gre

Question

Scores on a scholarship aptitude exam are normally distributed with a mean of 72 and a standard deviation of 8. What is the lowest score that will place an applicant at the top 10% of the distribution?

Answer 1

Let the minimum score required to be in the upper 10% of the group be c. Then

$P(X \ge c)=0.10$

$P(X<c)=1-0.1=0.90$

$P(Z<\frac{c-72}{8})=0.9$

Then we found z-value from p from z-table:

So,

$\frac{c-72}{8}=1.28$

c-72=10.24

c=82.24