Here, the variable are1,2,3,…, n
∴ Mean, \(\bar{x}=\frac{\sum x_i}{n}\)
\(=\frac{1+2+3+...+n}{n}\)
\(=\frac{n(n+1)}{2n}\)
\(=\frac{n+1}{2}\)
Variance = \(\sum x_i^2-(\bar{x})^2\)
= \(\frac{1^2+2^2+3^2+...+n^2}{n}-(\frac{n+1}{2})^2\)
= \(\frac{n(n+1)(2n+1)}{6n}-(\frac{n+1}{2})\)
= \((n+1)\{\frac{4n+2-3n-3}{12}\}\)
= \(\frac{(n+1)(n-1)}{12}\)
= \(\frac{n^2-1}{12}\)
