Question

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

Answer 1

Here x = 1, 2, 3, 4,.....,n

$\therefore \sum x=1+2+3+4+...+n=\frac{n(n+1)}{2}$

$\therefore \text{Mean}\left(\overline{x}\right)=\frac{n(n+1)}{2n}=\frac{(n+1)}{2}$

$\sum x^2=(1{)}^2+(2{)}^2+(3{)}^2+(4{)}^2+...+n^2$

=$\frac{n(n+1)(2n+1)}{6}$

$\text{Variance}(\sigma {)}^2=\frac{N\sum x^2-(\sum x{)}^2}{N^2}$

=$\frac{n\times \frac{n(n+1)(2n+1)}{6}-{\left[\frac{n(n+1)}{2}\right]}^2}{n^2}$

= $\frac{n\times \frac{n(n+1)(2n+1)}{6}-\left[\frac{(n+1{)}^2}{4}\right]}{n^2}$

=$\frac{4n^2+6n+2-3n^2-6n-3}{12}=\frac{n^2-1}{12}$