Variance of the first 20 natural numbers

By definition, variance of $n$ numbers $x_1, x_2, x_3,...,x_i, ...,x_n$ is var $(x)=<x^2>-<x>^2$ , where $<x>$ is the mean and $<x^2>$ is the mean-squared value. For the first $n$ natural numbers, $<x>=\dfrac{n+1}{2}$ and $<x^2>=\dfrac{(n+1)(2n+1)}{6}$ . Hence the variance of first $n$ natural numbers is $\dfrac{(n+1)(2n+1)}{6}-(\dfrac{n+1}{2})^2=\dfrac{n^2-1}{12}$ . For $n=20$ , the variance is $\dfrac{399}{12}=\boxed {33.25}$

Variance of the first n natural numbers = (n^2 - 1)/12

= (400-1)/12 = 33.25