Question
Find the variance of first $n$ natural numbers.
✓
Answer
Here, the variables are 1, 2, 3, ........, n
$\begin{aligned} \therefore \text { Mean, } \bar{x} & =\frac{\sum x_i}{n} \\ & =\frac{1+2+3+\ldots .+n}{n} \\ & =\frac{n(n+1)}{2 n} \\ & =\frac{n+1}{2} \\ \text { Variance } & =\frac{\sum x_i^2}{n}-(\bar{x})^2 \\ & =\frac{1^2+2^2+3^2+\ldots+n^2}{n}-\left(\frac{n+1}{2}\right)^2 \\ & =\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n+1}{2}\right)^2 \\ & =(n+1)\left\{\frac{4 n+2-3 n-3}{12}\right\} \\ & =\frac{(n+1)(n-1)}{12} \\ & =\frac{n^2-1}{12} .\end{aligned}$
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