Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution

${x_i}$	0	1	2	3	4	5
${f_i}$	$k + 2$	$2k$	${k^2} - 1$	${k^2} - 1$	${k^2} + 1$	$k - 3$

where $\sum f_{i}=62$. If $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^{2}+\sigma^{2}\right]$ is equal to :

We have, $\Sigma f_i=62$ <br/><br/>$$ \begin{aligned} & \left.(K+2)+2 K+\left(K^2-1\right)\right)+\left(K^2-1\right)+\left(K^2+1\right)+(K-3)=62 \\\\ & \Rightarrow 3 K^2+4 K-64=0 \\\\ & \Rightarrow (3 K+16)(K-4)=0 \\\\ & \Rightarrow K=4 \quad \end{aligned} $$ <br/><br/>$\left(\because k=\frac{-16}{3} \text { is not possible }\right)$ <br/><br/>$$ \begin{array}{|r|c|c|c|} \hline x_i & f_i & f_i x_i & f_i x_i^2 \\ \hline 0 & 6 & 0 & 0 \\ 1 & 8 & 8 & 8 \\ 2 & 15 & 30 & 60 \\ 3 & 15 & 45 & 135 \\ 4 & 17 & 68 & 272 \\ 5 & 1 & 5 & 25 \\ \hline \text { Total } & 62 & 156 & 500 \\ \hline \end{array} $$ <br/><br/>$\mu=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{0+8+30+45+68+5}{62}=\frac{156}{62}$ <br/><br/>$$ \begin{aligned} \sigma^2= & \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\\\ = & \frac{1 \times 8+4 \times 15+9 \times 15+16 \times 17+25 \times 1}{62}-\left(\frac{156}{62}\right)^2 \\\\ & =\frac{500}{62}-\left(\frac{156}{62}\right)^2=\frac{500}{62}-\mu^2 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \therefore \sigma^2+\mu^2=\frac{500}{62} \\\\ & \text { Hence, }\left[\sigma^2+\mu^2\right]=8 \end{aligned} $$