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 is 2.3 and variance of X is ${\mathrm{\sigma }}^2,$ then 100${\mathrm{\sigma }}^2$ is equal to:

Given Probability distribution of X is

Since $\sum \mathrm{P}(\mathrm{X}=\mathrm{x})=1$ then

$$\begin{gathered} \Rightarrow \frac{1}{5}+\mathrm{a}+\frac{1}{3}+\frac{1}{5}+\mathrm{b}=1 \\ \Rightarrow \frac{2}{5}+\frac{1}{3}+\mathrm{a}+\mathrm{b}=1 \\ \Rightarrow \mathrm{a}+\mathrm{b}=1-\frac{2}{5}-\frac{1}{3} \\ \Rightarrow \mathrm{a}+\mathrm{b}=\frac{15-6-5}{15} \\ \Rightarrow \mathrm{a}+\mathrm{b}=\frac{4}{15} -\left(1\right) \end{gathered}$$

Given mean $=2.3$

$$\begin{gathered} \Rightarrow \sum {\mathrm{x}}_{\mathrm{i}} {\mathrm{P}}_{\mathrm{i}}=2.3 \\ \Rightarrow \frac{1}{5}(-2)-\mathrm{a}+1+\frac{4}{5}+6\mathrm{b}=\frac{23}{10} \\ \Rightarrow -\mathrm{a}+6\mathrm{b}=\frac{9}{10} -\left(2\right) \\ \left(1\right)+\left(2\right) \\ \Rightarrow 7\mathrm{b}=\frac{4}{15}+\frac{9}{10} \\ \Rightarrow 7\mathrm{b}=\frac{8+27}{30} \\ \Rightarrow 7\mathrm{b}=\frac{35}{30} \\ \Rightarrow \mathrm{b}=\frac{1}{6} \end{gathered}$$

from (1)   $\mathrm{a}+\frac{1}{6}=\frac{4}{15}$

$\begin{array}{rcl} & \Rightarrow & \mathrm{a}=\frac{4}{15}-\frac{1}{6}\\ & =& \frac{8-5}{30}\\ & =& \frac{3}{30}\\ & =& \frac{1}{10}\end{array}$

Variance $\sum {\mathrm{P}}_{\mathrm{i}} {{\mathrm{x}}_{\mathrm{i}}}^2-{\left(\mathrm{mean}\right)}^2$

$\begin{array}{rcl}{\mathrm{\sigma }}^2& =& \left(\frac{1}{5}\left(4\right)+\mathrm{a}+3+\frac{6}{5}+36\mathrm{b}\right)-{(2.3)}^2\\ & =& \left[\frac{4}{5}+12+\frac{16}{5}+\frac{1}{10}+6\right]-{\left[\frac{23}{10}\right]}^2\end{array}$

$\begin{array}{rcl} & =& \frac{131}{10}-\frac{529}{100}\\ & =& \frac{1310-529}{100}\\ {\mathrm{\sigma }}^2& =& \frac{781}{100}\\ & \Rightarrow & 100{\mathrm{\sigma }}^2=781\end{array}$