4.3.5 Percentiles of an Arbitrary Normal Distribution

The $\left(100p\right)$th percentile of a normal distribution with mean $\mu$ and standard deviation $\sigma$ is easily related to the $\left(100p\right)$th percentile of the standard normal distribution.

Proposition

$$\begin{array}{rl}& \begin{array}{c}\left(100p\right)\text{ th percentile }\\ \text{ for normal }\left(\mu ,\sigma \right)\end{array}\\ & =\mu +\left[\begin{array}{c}\left(100p\right)\text{ th for }\\ \text{ standard normal }\end{array}\right]\cdot \sigma \end{array}$$

Another way of saying this is that

if $z$ is the desired percentile for the standard normal distribution, then the desired percentile for the normal $\left(\mu ,\sigma \right)$ distribution is $z$ standard deviations from $\mu$ .

EX 4.18