Question

An index that is a standardized measure used in observing infants over time is approximately normal with a mean of 93 and a standard deviation of 5.

Find the index score that is the 72nd percentile.

Answer 1

solution

Given that,

mean = $\mu$ = 93

standard deviation = $\sigma$ =5

Using standard normal table,

P(Z < z) = 72%

= P(Z < z) = 0.72  

= P(Z < 0.58) = 0.72

z = 0.58 Using standard normal z table,

Using z-score formula  

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

x= 0.58 *5+93

x= 95.9