QJEEM23Ma4d7e: JEE Main 2023 Mathematics - Class 11 JEEM23Ma4d7e Solution: Let be the mean and be the standard deviation of the…

Question

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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 :

Accepted Answer

We have,

\Sigma f_i=62

\begin{aligned}&amp; \left.(K+2)+2 K+\left(K^2-1\right)\right)+\left(K^2-1\right)+\left(K^2+1\right)+(K-3)=62 \\\\&amp; \Rightarrow 3 K^2+4 K-64=0 \\\\&amp; \Rightarrow (3 K+16)(K-4)=0 \\\\&amp; \Rightarrow K=4 \quad\end{aligned}

\left(\because k=\frac{-16}{3} \text { is not possible }\right)

\begin{array}{|r|c|c|c|}\hline x_i &amp; f_i &amp; f_i x_i &amp; f_i x_i^2 \\\hline 0 &amp; 6 &amp; 0 &amp; 0 \\1 &amp; 8 &amp; 8 &amp; 8 \\2 &amp; 15 &amp; 30 &amp; 60 \\3 &amp; 15 &amp; 45 &amp; 135 \\4 &amp; 17 &amp; 68 &amp; 272 \\5 &amp; 1 &amp; 5 &amp; 25 \\\hline \text { Total } &amp; 62 &amp; 156 &amp; 500 \\\hline\end{array}

\mu=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{0+8+30+45+68+5}{62}=\frac{156}{62}

\begin{aligned}\sigma^2= &amp; \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\\\= &amp; \frac{1 \times 8+4 \times 15+9 \times 15+16 \times 17+25 \times 1}{62}-\left(\frac{156}{62}\right)^2 \\\\&amp; =\frac{500}{62}-\left(\frac{156}{62}\right)^2=\frac{500}{62}-\mu^2\end{aligned}

\begin{aligned}&amp; \therefore \sigma^2+\mu^2=\frac{500}{62} \\\\&amp; \text { Hence, }\left[\sigma^2+\mu^2\right]=8\end{aligned}

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