The life of shoes is normally distributed with mean $8$ months and standard deviation $2$ months. If $5000$ pairs are issued, how many pairs would be expected to need replacement within $12$ months.
Given: Mean $= \mu = 8$; $\;$ Standard deviation $= \sigma = 2$
Let $X$ be a normal variate.
To find: $P \left(X < 12\right)$
Standard normal variate $Z = \dfrac{X - \mu}{\sigma}$
When $X = 12$, $\;$ $Z = \dfrac{12 - 8}{2} = 2$
$\begin{aligned}
\therefore \; P \left(X < 12\right) = P \left(Z < 2\right) \\\\
& = \left(\text{area between } Z = - \infty \text{ to } Z = 0\right) \\\\
& \hspace{1.5cm} + \left(\text{area between } Z = 0 \text{ to } Z = 2\right) \\\\
& = 0.5000 + 0.4772 \;\;\; \left[\text{from Normal Distribution Table}\right] \\\\
& = 0.9772
\end{aligned}$
$\therefore \;$ Number of pairs of shoes that would need replacement within $12$ months $= 5000 \times 0.9772 = 4886$