JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 8)
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 :
9
8
6
7
Explanation
We have, $\Sigma f_i=62$
$$ \begin{aligned} & \left.(K+2)+2 K+\left(K^2-1\right)\right)+\left(K^2-1\right)+\left(K^2+1\right)+(K-3)=62 \\\\ & \Rightarrow 3 K^2+4 K-64=0 \\\\ & \Rightarrow (3 K+16)(K-4)=0 \\\\ & \Rightarrow K=4 \quad \end{aligned} $$
$$ \left(\because k=\frac{-16}{3} \text { is not possible }\right) $$
$$ \begin{array}{|r|c|c|c|} \hline x_i & f_i & f_i x_i & f_i x_i^2 \\ \hline 0 & 6 & 0 & 0 \\ 1 & 8 & 8 & 8 \\ 2 & 15 & 30 & 60 \\ 3 & 15 & 45 & 135 \\ 4 & 17 & 68 & 272 \\ 5 & 1 & 5 & 25 \\ \hline \text { Total } & 62 & 156 & 500 \\ \hline \end{array} $$
$$ \mu=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{0+8+30+45+68+5}{62}=\frac{156}{62} $$
$$ \begin{aligned} \sigma^2= & \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\\\ = & \frac{1 \times 8+4 \times 15+9 \times 15+16 \times 17+25 \times 1}{62}-\left(\frac{156}{62}\right)^2 \\\\ & =\frac{500}{62}-\left(\frac{156}{62}\right)^2=\frac{500}{62}-\mu^2 \end{aligned} $$
$$ \begin{aligned} & \therefore \sigma^2+\mu^2=\frac{500}{62} \\\\ & \text { Hence, }\left[\sigma^2+\mu^2\right]=8 \end{aligned} $$
$$ \begin{aligned} & \left.(K+2)+2 K+\left(K^2-1\right)\right)+\left(K^2-1\right)+\left(K^2+1\right)+(K-3)=62 \\\\ & \Rightarrow 3 K^2+4 K-64=0 \\\\ & \Rightarrow (3 K+16)(K-4)=0 \\\\ & \Rightarrow K=4 \quad \end{aligned} $$
$$ \left(\because k=\frac{-16}{3} \text { is not possible }\right) $$
$$ \begin{array}{|r|c|c|c|} \hline x_i & f_i & f_i x_i & f_i x_i^2 \\ \hline 0 & 6 & 0 & 0 \\ 1 & 8 & 8 & 8 \\ 2 & 15 & 30 & 60 \\ 3 & 15 & 45 & 135 \\ 4 & 17 & 68 & 272 \\ 5 & 1 & 5 & 25 \\ \hline \text { Total } & 62 & 156 & 500 \\ \hline \end{array} $$
$$ \mu=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{0+8+30+45+68+5}{62}=\frac{156}{62} $$
$$ \begin{aligned} \sigma^2= & \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\\\ = & \frac{1 \times 8+4 \times 15+9 \times 15+16 \times 17+25 \times 1}{62}-\left(\frac{156}{62}\right)^2 \\\\ & =\frac{500}{62}-\left(\frac{156}{62}\right)^2=\frac{500}{62}-\mu^2 \end{aligned} $$
$$ \begin{aligned} & \therefore \sigma^2+\mu^2=\frac{500}{62} \\\\ & \text { Hence, }\left[\sigma^2+\mu^2\right]=8 \end{aligned} $$
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