Question

Find the mean and variance for the first $n$ natural numbers

Answer 1

First $n$ natural number are, $x=1,2,3,.............n$
So mean of first $n$ natural number is $\overline{x}=\frac{\sum x_i}{n}=\frac{n+1}{2}$
$\therefore$ variance, ${\sigma }^2=\frac{\sum x_i^2}{n}-(\overline{x}{)}^2$
$=\frac{\frac{1}{6}n(n+1)(2n+1)}{n}-{\left(\frac{n+1}{2}\right)}^2=\frac{1}{12}(n+1)(n-1)=\frac{n^2-1}{12}$