Question

Assume that adults have IQ scores that are normally distributed with a mean of 95.9 and a standard deviation 18.2. Find the first quartile Q1 which is the IQ score separating the bottom​ 25% from the top​ 75%.

Answer 1

Solution:-

Given that,

mean = $\mu$ = 95.9

standard deviation = $\sigma$ = 18.2

Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.67 ) = 0.25

z = -0.67

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.67 * 18.2+95.9

x = 83.706

First quartile =Q1 = 84