Let X be a random variable with distribution. x - 2 - 1 3 4 6 P ( X = x ) 1 5 a 1 3 1 5 b If the mean of X…

Given mean is $\mu =\sum p_i{x}_i=2.3$

$\Rightarrow \frac{-2}{5}-a+1+\frac{4}{5}+6\mathrm{b}=2.3$

$\Rightarrow 6b-a=0.9 ...\left(1\right)$

Also $\sum p_i=1$

$\Rightarrow \frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$

$\Rightarrow a+b=\frac{4}{15} ...\left(2\right)$

Solving, we get $a=\frac{1}{10}; b=\frac{1}{6}$

Variance is ${\sigma }^2=\sum p_i{x}_i^2-{\mu }^2$

$\Rightarrow {\sigma }^2=\frac{1}{5}\left(4\right)+\frac{1}{10}\left(1\right)+\frac{1}{3}\left(9\right)+\frac{1}{5}\left(16\right)+\frac{1}{6}\left(36\right)-5.29$

$\Rightarrow {\sigma }^2=7.81$

$\Rightarrow 100{\sigma }^2=781$