Variance of the first $11$ natural numbers is:

\sqrt{5}

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\sqrt{10}

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5 \sqrt{2}

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10

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Solution

The correct option is B $\sqrt{10}$
Sum of first $n$ natural numbers is $\frac{n^{2} - 1}{12}$
$= \frac{11^{2} - 1}{12}$
$= \frac{120}{12}$
$= 10$
Thus, variance is $\sqrt{\frac{n^{2} - 1}{12}}$ $= \sqrt{10}$ .

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