Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution \begin{tabular}{|l|l|l|l|l|l|l|} \hline $\mathrm{x_{i}}$ & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline $\mathrm{f_{i}}$ & $\mathrm{k}+2$ & 2 k & $\mathrm{k}^{2}-1$ & $\mathrm{k}^{2}-1$ & $\mathrm{k}^{2}+1$ & $\mathrm{k}-3$ \\ \hline \end{tabular} where $\sum f_{i}=62$. if $[x]$ denotes the greatest integer $\leq \mathrm{x}$, then $\left[\mu^{2}+\sigma^{2}\right]$ is equal

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Statistics

Let

μ

be the mean and

σ

be the standard deviation of the distribution

\begin{tabular}{|l|l|l|l|l|l|l|}
\hline

x_{i}

& 0 & 1 & 2 & 3 & 4 & 5 \\
\hline

f_{i}

k + 2

& 2 k &

k^{2} - 1

k^{2} - 1

k^{2} + 1

k - 3

\\
\hline
\end{tabular}
where

\sum f_{i} = 62

. if

[x]

denotes the greatest integer

\leq x

, then

[μ^{2} + σ^{2}]

is equal

Answer: (A)

From

\sum f_{i} = 62

(k + 2) + 2 k + (k^{2} - 1) + (k^{2} - 1) + (k^{2} + 1) + (k - 3) = 3 k^{2} + 4 k - 2 = 62

, so

3 k^{2} + 4 k - 64 = 0

and

k = 4

. The frequencies are then

6, 8, 15, 15, 17, 1

. Since

μ^{2} + σ^{2} = \frac{1}{N} \sum f_{i} x_{i}^{2}

, and

\sum f_{i} x_{i}^{2} = 0 + 8 + 60 + 135 + 272 + 25 = 500

, it follows that

μ^{2} + σ^{2} = \frac{500}{62} \approx 8.06

. Therefore

[μ^{2} + σ^{2}] = 8

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Single correct option