Let be 10 observations such that and Then the standard deviation of is equal to

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Solution

Q.
Let $a_{1}, a_{2}, \dots, a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100 .$ Then the standard deviation of $a_{1}, a_{2}, \dots, a_{10}$ is equal to [2024]

1 5

2 $\sqrt{115}$

3 $\sqrt{5}$

4 10

Ans.
(3)

Given, $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100$

$σ^{2} = \frac{\sum a_{k}^{2}}{10} -$ $({Mean)}^{2}$ $= \frac{\sum a_{k}^{2}}{10} - {(\frac{\sum a_{k}}{10})}^{2} = \frac{\sum a_{k}^{2}}{10} - {(\frac{50}{10})}^{2}$

$= \frac{\sum a_{k}^{2}}{10} - 25$ ...(i)

Also, ${(\sum a_{k})}^{2} = \sum a_{k}^{2} + 2 \sum_{\forall k < j} a_{k} a_{j}$

$\Rightarrow 2500 = \sum a_{k}^{2} + 2 (1100) \Rightarrow \sum a_{k}^{2} = 2500 - 2200 = 300$

From (i), we get

$σ^{2} = \frac{300}{10} - 25 = 30 - 25 = 5 \Rightarrow σ = \sqrt{5}$

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