Question

Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.

Answer 1

Consider the first natural no. where n is an odd i.e., $1,2,3,4,......n$(odd)

Mean $\left(\overline{x}\right)=\frac{1+2+\mathrm{3...}\dots .n}{n}=\frac{n(n+1)}{2n}$

$=\frac{n+1}{2}$

$Md=\frac{[1-\frac{n+1}{2}]+[2-\frac{n+1}{2}]+[3-\frac{n+1}{2}]+...\dots [n-\frac{n+1}{2}]}{n}$

$=|-\frac{n+1}{2}|+|2-\frac{n+1}{2}|+\dots \dots .|\frac{n-1}{2}-\frac{n+1}{2}|$

$=+|\frac{n+1}{2}-\frac{n+1}{2}|+|\frac{n+3}{2}-\frac{n+1}{2}|...\dots |\frac{2n-2}{2}-\frac{n+1}{2}|+|n-\frac{n+1}{2}|$

$=\frac{2}{n}[1+\mathrm{2...}\dots +\frac{n-3}{2}+\frac{n-1}{2}]\left(\frac{n-1}{2}\right)$ term

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

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

$=\frac{1}{n}\left[\frac{n^2-1}{4}\right]$

$=\frac{n^2-1}{4n}$.