Consider our earlier continuous example 5.4.4 in which we found \(\mu = \frac{5}{4}\) and \(\sigma = \sqrt{\frac{51}{80}}\text{.}\) Then,
\begin{equation*}
P(0 < X < 1) = P \left ( \frac{0-\frac{5}{4}}{\sqrt{\frac{51}{80}}} < \frac{X - \frac{5}{4}}{\sqrt{\frac{51}{80}}} < \frac{1-\frac{5}{4}}{\sqrt{\frac{51}{80}}} \right )
\end{equation*}
gives the middle term is Z and the other endpoints are now in standard units that indicate the number of standard deviations from the mean rather than actual problem units.
