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The variance of the first $n$ natural numbers is
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Verified Answer
The correct answer is:
$\frac{n^2-1}{12}$
Variance $=(\text { S.D. })^2=\frac{1}{n} \Sigma x^2-\left(\frac{\Sigma x}{n}\right)^2,\left(\because \bar{x}=\frac{\Sigma x}{n}\right)$
$=\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n(n+1)}{2 n}\right)^2=\frac{n^2-1}{12}$.
$=\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n(n+1)}{2 n}\right)^2=\frac{n^2-1}{12}$.
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