# Suppose that X is a normal random variable with mean 5. If P{X>9}=.2, approximat

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

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Ched1950

Given: X is a normal random variable with mean $$\displaystyle\mu={5}$$ and $$\displaystyle{P}{\left[{X}{>}{9}\right]}={0.2}$$
To find: variance of X
Solution: Let $$\displaystyle\sigma^{{2}}$$ be the variance of X. Now,
$$\displaystyle{P}{\left[{X}{>}{9}\right]}={0.2}$$
$$\displaystyle\Rightarrow{P}{\left[{X}\leq{9}\right]}={1}-{0.2}={0.8}$$
$$\displaystyle\Rightarrow{P}{\left[{\frac{{{x}-\mu}}{{\sigma}}}\leq{\frac{{{9}-{4}}}{{\sigma}}}\right]}={0.8}$$
$$\displaystyle\Rightarrow{P}{\left[{Z}\leq{\frac{{{4}}}{{\sigma}}}\right]}={0.8}$$
$$\displaystyle\Rightarrow\phi{\left({\frac{{{4}}}{{\sigma}}}\right)}={0.8}$$
$$\displaystyle{\Rightarrow}{\frac{{{4}}}{{\sigma}}}={0.845}$$
$$\displaystyle\Rightarrow\sigma={\frac{{{4}}}{{{0.845}}}}={4.73}$$
Thus variance of X is $$\displaystyle{v}{a}{r}{\left({X}\right)}=\sigma^{{2}}={\left({4.73}\right)}^{{2}}={22.37}={22}$$