Consider 10 observations such that and , where are positive integers. Let the mean and the variance of the observations be and respectively. Then is equal to

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Solution

Q.
Consider 10 observations $x_{1}, x_{2}, . . ., x_{10}$ such that $\sum_{i = 1}^{10} (x_{i} - α) = 2$ and $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 40$ , where $α, β$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. Then $\frac{β}{α}$ is equal to : [2024]

1 2

2 1

3 $\frac{5}{2}$

4 $\frac{3}{2}$

Ans.
(1)

We have, $\frac{\sum_{i = 1}^{10} x_{i}}{n} = \frac{6}{5} \Rightarrow \sum_{i = 1}^{10} x_{i} = 12$ $[∵ n = 10]$

Also, $\sum_{i = 1}^{10} (x_{i} - α) = 2 \Rightarrow \sum_{i = 1}^{10} x_{i} - 10 α = 2 \Rightarrow α = 1$

Now, $\frac{\sum_{i = 1}^{10} x_{i}^{2}}{n} - {(\bar{x})}^{2} = \frac{84}{25} \Rightarrow \frac{\sum_{i = 1}^{10} x_{i}^{2}}{10} = \frac{84}{25} + \frac{36}{25} = \frac{120}{25}$

$\Rightarrow \sum_{i = 1}^{10} x_{i}^{2} = 48$

Also, $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 40 \Rightarrow \sum_{i = 1}^{10} x_{i}^{2} + 10 β^{2} - 2 β \sum_{i = 1}^{10} x_{i} = 40$

$\Rightarrow 48 + 10 β^{2} - 24 β = 40 \Rightarrow 5 β^{2} - 12 β + 4 = 0$

$\Rightarrow (5 β - 2) \Rightarrow (β - 2) = 0 \Rightarrow β = \frac{2}{5} or β = 2$

If $β = 2$ , then $\frac{β}{α} = 2$

If $β = \frac{2}{5}$ , then $\frac{β}{α} = \frac{2}{5}$

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