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So I'm trying to get an edge on Discrete Math for next fall, and I'm working on this problem:

Suppose that X is a normal random variable with mean 3. If P{X > 4} = 0.1, what is the variance of X? What should be c such that P{X < c} = 0.2? Plot the PDF of this r.v., then plot its CDF.

would the Variance just be:

$Var(X) = (4 - 3)^2*.1 = (4-3)*.1 = 1 * .1 = .1$?

Also, how do I compute the p.d.f. for a normal random variable? Is it just:

P(X) = $(1 / (\sqrt(2 * (\pi) * (\sigma ^ 2)))$ * e ^$-((x - (\mu))^2$/$2*(\sigma ^2))$

I think I'm doing something very wrong, but I can't figure out what.

Thank you.

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up vote 0 down vote accepted

$$ 0.1=\Pr(X>4) = \Pr\left(\frac{X-\text{mean}}{\text{standard deviation}} > \frac{4-\text{mean}}{\text{standard deviation}}\right) $$ $$ = \Pr\left(\frac{X-3}{\sigma} >\frac{4-3}{\sigma}\right) = \Pr\left(Z>\frac 1 \sigma\right). $$

So go to the table (or the software that you use) to find the value of $z_0$ such that $0.1=\Pr(Z>z_0)$. Set $1/\sigma = \text{that value}$. That gives you the standard deviation $\sigma$.

The pdf is $$ \frac{1}{\sqrt{2\pi}\,\sigma} e^{-((x-\mu)/\sigma)^2/2}. $$

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