Q:

Find the Variance of a Normal Random Variable

I'm working on the following problem and could use some help. Unless I'm not making a connection to material covered in my course, we haven't covered this and I'm not sure how to work this problem out.

Suppose that X is a normal RV with mean 5.  If $P(X>9)=0.2$, approximately what is $\text{Var}(x)$?

This is what I have so far:
Let $z=\frac{x-\mu}{\sigma}$. Then, 
$$ P(X>9)\Rightarrow P(z\ge \frac{9-\mu}{\sigma})=0.2 \, \, \text{and} \, \, P(z\le \frac{9-\mu}{\sigma})=0.8 $$
That's about all I can decern from the problem. I know that the variance is $\sigma$ and the mean is $\mu$. So, it makes me believe that I have to set up the problem somehow to solve for $\sigma$, which would be the approximate value of $\text{Var}(x)$. I'm just not sure where to begin with that, so any help would be appreciated. 
Thank you!

A:

$\sigma$ is the standard deviation while $\sigma^2$ is the variance. 
We have
$$\begin{align*}
P(X\gt9)=0.2
&\iff 1-P(X<9)=0.2\\\\
&\iff1-\Phi\left(\frac{9-5}{\sigma}\right)=0.2\\\\
&\iff\Phi\left(\frac{9-5}{\sigma}\right)=0.8
\end{align*}$$
By a z-table, $\Phi(z)=0.8$ when $z\approx 0.84$ so 
$$\frac{9-5}{\sigma}=0.84\Rightarrow\sigma^2 \approx 22.68$$
More accurately, R statistical software gives $z\approx 0.8416$
> qnorm(.8)
[1] 0.8416212

and $\sigma^2\approx22.59$ 
> (4/qnorm(.8))^2
[1] 22.58846

