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 \(\displaystyle{v}{a}{r}{\left({X}\right)}=\sigma^{{2}}={\left({4.73}\right)}^{{2}}={22.37}={22}\)