For the probability distribution given by following

\(\operatorname{Var}(\mathrm{X})=\)

We have \(0.07+0.2+0.3+\mathrm{k}+0.07+0.04+0.02=1\) \(\therefore \mathrm{k}=0.3\)
\(\begin{array}{|l|l|l|l|}
\hline \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}{ }^2 \\
\hline 5 & 0.07 & 0.35 & 1.75 \\
\hline 6 & 0.2 & 1.2 & 7.2 \\
\hline 7 & 0.3 & 2.1 & 14.7 \\
\hline 8 & 0.3 & 2.4 & 19.2 \\
\hline 9 & 0.07 & 0.63 & 5.67 \\
\hline 10 & 0.04 & 0.4 & 4 \\
\hline 11 & 0.02 & 0.22 & 2.42 \\
\hline \text{Total } & 7.3 & 54.94 \\
\hline
\end{array}\)
Variance \((\mathrm{x})=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=(54.94)-(7.3)^2=1.65\)