Question

In a survey of men in a certain country? (ages 20minus??29), the mean height was 63.5 inches with a standard deviation of 2.6 inches. ?(a) What height represents the 99th ?percentile? ?(b) What height represents the first? quartile?

Answer 1

(a). It is given that

$\mu = 63.5$

$\sigma = 2.6$

The z-score with a left tail of 99% is 2.33

$z= \frac{X-\mu}{\sigma}$

$2.33= \frac{X-63.5}{2.6}$

(b). For first quartile, that is, z-score for a left tail of 25%,

$z = \frac{X-\mu}{\sigma}$

$-0.6745 = \frac{X - 63.5}{2.6}$

inches