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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!

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  • $\begingroup$ You also know $\mu=5$. $\endgroup$ Mar 13, 2018 at 5:17

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$\sigma$ 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
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  • $\begingroup$ So, I don't understand how to interpret this. I'm not sure how $\Phi (X) =0.8$ when $z\approx 0.84$. What is the mathematical backing behind this result? As I mentioned above, I don't believe this was covered in my class, so I'm not sure about the background needed for this problem. $\endgroup$
    – NoVa
    Mar 13, 2018 at 3:51
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    $\begingroup$ Do you have a z-table in front of you so I can walk you through it? $\endgroup$
    – Remy
    Mar 13, 2018 at 3:52
  • $\begingroup$ Yes, is this one okay? stat.ufl.edu/~athienit/Tables/Ztable.pdf $\endgroup$
    – NoVa
    Mar 13, 2018 at 3:53
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    $\begingroup$ $\Phi(z)$ is the probability that the random variable $X$ takes on a value less than $z$. The $z$ table gives the $z$-scores with their associated probabilities. We must look for the probability closest to $0.8$ and find the associated $z$-score. The closest probability you can find is $0.7995$ which happens when $z=0.84$ $\endgroup$
    – Remy
    Mar 13, 2018 at 3:54
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    $\begingroup$ Better to say $\Phi(z)$ is the probability that a standard normal random variable, let us call it $Z$, takes some value less than $z$. Don't confuse it with the already used $X$, which is not standard normal distributed. $\endgroup$ Mar 13, 2018 at 5:23

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