Question

Suppose that X is a normal random variable with mean $5$. If $P\left(X>9\right)=.2$, approximately what is $Var\left(X\right)$?

Short answer

The required variance is 22.66.

Step-by-step solution

Step 1. Given Information.

$X$is a random variable with mean $5$.

Also,$P\left(X>9\right)=0.2.$

Step 2. Find VarX.

Here,

$$\begin{gathered} P\left(X>9\right)=0.2 \\ P\left(\frac{X-\mu }{\sigma }>\frac{9-\mu }{\sigma }\right)=0.2 \\ 1-\phi \left(\frac{9-\mu }{\sigma }\right)=0.2 \\ \phi \left(\frac{9-\mu }{\sigma }\right)=0.8 \\ \frac{9-\mu }{\sigma }=0.84 \\ \sigma =\frac{9-\mu }{0.84} \\ \sigma =\frac{9-5}{0.84} \\ \sigma =4.76 \\ {\sigma }^2={\left(4.76\right)}^2 \\ {\sigma }^2=22.66 \end{gathered}$$

Hence, Var$\left(X\right)$=22.66.