Question

Suppose that scores on an exam are normally distributed with a mean of 80 and variance of 81. What is the first quartile?

A.) 97.6

B.) 63.3

C.) 86.1

D.) 73.9

E.) 87.1

Answer 1

Solution:-

Given that,

mean = = 80

standard deviation = = `81=9

Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 9+80

x = 73.9

correct anser is D