Let be ten observations such that , , , and their variance is . If and are respectively the mean and the variance of , then is equal to : [2025]

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Solution

Q.
Let $x_{1}, x_{2}, . . ., x_{10}$ be ten observations such that $\sum_{i = 1}^{10} (x_{i} - 2) = 30$ , $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 98$ , $β > 2$ , and their variance is $\frac{4}{5}$ . If $μ$ and $σ^{2}$ are respectively the mean and the variance of $2 (x_{1} - 1) + 4 β, 2 (x_{2} - 1) + 4 β, . . ., 2 (x_{10} - 1) + 4 β$ , then $\frac{β μ}{σ^{2}}$ is equal to : [2025]

1 100

2 110

3 120

4 90

Ans.
(1)

We have, $\sum_{i = 1}^{10} (x_{i} - 2) = 30 \Rightarrow \sum_{i = 1}^{10} x_{i} - 2 \times 10 = 30$

$\Rightarrow \sum_{i = 1}^{10} x_{i}^{} = 50$

Now, variance $= \frac{4}{5} \Rightarrow \frac{4}{5} = \frac{1}{10} \sum_{}^{} x_{i}^{2} - {(\frac{\sum_{}^{} x_{i}}{10})}^{2}$

$\Rightarrow \frac{4}{5} = \frac{1}{10} \sum_{}^{} x_{i}^{2} - 25 \Rightarrow \sum_{}^{} x_{i}^{2} = 258$

Now, $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 98 \Rightarrow \sum_{i = 1}^{10} (x_{i}^{2} - 2 β x_{i} + β^{2}) = 98$

$\Rightarrow 258 - 2 β \cdot 50 + 10 β^{2} = 98$

$\Rightarrow 10 β^{2} - 100 β + 160 = 0$

$\Rightarrow (β - 8) (β - 2) = 0 \Rightarrow β = 8$ [ $∵ β > 2$ ]

Now, $μ = \frac{1}{10} \sum_{i = 1}^{10} [2 (x_{i} - 1) + 4 β]$

$= \frac{1}{10} [2 \sum_{i = 1}^{10} x_{i} - 20 + 40 β]$

$= \frac{1}{10} [2 \times 50 - 20 + 320] = 40$

$σ^{2} = 2^{2} (\frac{4}{5}) = \frac{16}{5}$

$∴ \frac{β μ}{σ^{2}} = \frac{8 \times 40 \times 5}{6} = 100$ .

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