The probability function of a random variable X is given by $\mathrm{P}(\mathrm{X}=\mathrm{k})={\mathrm{ck}}^2,$ where c is a constant and $\mathrm{k}\in \left\{\right),1,2,3,4\} .$ If ${\mathrm{\sigma }}^2$ is the variance of X and $\mathrm{\mu }$ is the mean of X, then ${\mathrm{\sigma }}^2+{\mathrm{\mu }}^2=$

Since $\sum _{\mathrm{k}=0}^4 \mathrm{P}(\mathrm{X}=\mathrm{k})=1$

$$\begin{gathered} \Rightarrow \sum _{\mathrm{k}=0}^4 {\mathrm{Ck}}^2=1 \\ \Rightarrow \mathrm{C}=\frac{1}{30} \end{gathered}$$

Mean $\left(\mathrm{\mu }\right)=\sum {\mathrm{x}}_{\mathrm{i}} \mathrm{P}(\mathrm{x}={\mathrm{x}}_{\mathrm{i}})$

                 $\begin{array}{rcl} & =& 0+\frac{1}{30}+\frac{8}{30}+\frac{27}{30}+\frac{64}{30}\\ & =& \frac{100}{30}\\ \mathrm{\mu }& =& \frac{10}{3}\end{array}$

 $\begin{array}{rcl}variance\left({\sigma }^2\right)& =& \sum {x_i}^{2}P(X=x_i)-{\left(\mu \right)}^2\\ \Rightarrow {\sigma }^2& +& {\mu }^2=0+\frac{1}{30}+\frac{16}{30}+\frac{81}{30}+\frac{256}{30} \\ =\frac{354}{30}& & \\ =11.8& & \\ & & \\ & & \\ & & \\ & & \end{array}$