Let be the mean and be the standard deviation of the distribution where f i = 62 . If [ x ] denotes the greatest integer x , then [ 2 + 2 ] is equal to [2023]

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Solution

Q.
Let $μ$ be the mean and $σ$ be the standard deviation of the distribution

$x_{i}$ 0 1 2 3 4 5

$f_{i}$ $k + 2$ $2 k$ $k^{2} - 1$ $k^{2} - 1$ $k^{2} + 1$ $k - 3$

where $\sum f_{i} = 62$ . If $[x]$ denotes the greatest integer $\leq x$ , then $[μ^{2} + σ^{2}]$ is equal to [2023]

1 8

2 6

3 9

4 7

Ans.
(1)

We have, $\sum f_{i} = 62$

$\Rightarrow$ $3 k^{2} + 4 k - 2 = 62 \Rightarrow 3 k^{2} + 16 k - 12 k - 64 = 0$

$\Rightarrow k (3 k + 16) - 4 (3 k + 16) = 0$

$\Rightarrow k = 4, \frac{- 16}{3} (not possible)$

$∴$ $k = 4, σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{\sum f_{i}} - μ^{2}$

$\Rightarrow σ^{2} = \frac{8 \times 1^{2} + 15 \times 2^{2} + 15 \times 3^{2} + 17 \times 4^{2} + 5^{2}}{62} - μ^{2}$

$\Rightarrow σ^{2} + μ^{2} = \frac{500}{62} \Rightarrow [σ^{2} + μ^{2}] = 8$

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