finite math

Question

Suppose the test scores in a math class are normally distributed with a mean of 75 and a standard deviation of 6.25. What score would a student need in order to be in the 90th percentile for that test?

Answer 1

find  area of .9 in the table for normal distributions.

 

the z score corresponding to this area is 1.29

 

now solve the z score formula for x

 

z= x -mu/sigma

1.29 = x -75/6.25

1.29(6.25) +75 = x

 

x = 83.06 

 

a student who scored 83.06 or 83 would be in the 90th percentile