Calculate the mean deviation about the mean

Solution:

Given set of first $\mathrm{n}$ natural numbers when $\mathrm{n}$ is an odd number Now we have to find the mean deviation about the mean We know first n natural numbers are $1,2,3 \ldots . . \mathrm{n}$. And given $n$ is odd number.

So mean is,

$\overline{\mathrm{x}}=\frac{1+2+3+\cdots+\mathrm{n}}{\mathrm{n}}=\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\mathrm{n}}=\frac{(\mathrm{n}+1)}{2}$

The deviations of numbers from the mean are as shown below,

$\begin{aligned} 1-\frac{(n+1)}{2}, & 2-\frac{(n+1)}{2}, 3-\frac{(n+1)}{2}, \ldots \ldots,(n-2)-\frac{(n+1)}{2},(n-1) \\ &-\frac{(n+1)}{2}, n-\frac{(n+1)}{2} \end{aligned}$

Or,

$\frac{2-(n+1)}{2}, \frac{4-(n+1)}{2}, \frac{6-(n+1)}{2}, \ldots \frac{2(n-2)-(n+1)}{2}, \frac{2(n-1)-(n+1)}{2}, \frac{2 n-(n+1)}{2}$

$\frac{2-(\mathrm{n}+1)}{2}, \frac{4-(\mathrm{n}+1)}{2}, \frac{6-(\mathrm{n}+1)}{2}, \ldots \ldots \frac{2 \mathrm{n}-4-(\mathrm{n}+1)}{2}, \frac{2 \mathrm{n}-2-(\mathrm{n}+1)}{2}, \frac{2 \mathrm{n}-(\mathrm{n}+1)}{2}$

Or,

$\frac{1-n}{2}, \frac{3-n}{2}, \frac{5-n}{2}, \ldots \ldots \frac{n-5}{2}, \frac{n-3}{2}, \frac{n-1}{2}$

By taking negative common

$\frac{-(n-1)}{2}, \frac{-(n-3)}{2}, \frac{-(n-5)}{2}, \ldots \ldots, \frac{n-5}{2}, \frac{n-3}{2}, \frac{n-1}{2}$

So the absolute values of deviation from the mean is

$\left|x_{i}-\bar{x}\right|=\frac{(n-1)}{2}, \frac{(n-3)}{2}, \frac{(n-5)}{2}, \ldots \ldots, \frac{n-5}{2}, \frac{n-3}{2}, \frac{n-1}{2}$

The sum of absolute values of deviations from the mean, is

$\sum\left|x_{i}-\bar{x}\right|=\frac{(n-1)}{2}+\frac{(n-3)}{2}+\frac{(n-5)}{2}+\cdots+\frac{n-5}{2}+\frac{n-3}{2}+\frac{n-1}{2}$

$\sum\left|x_{i}-\bar{x}\right|=2\left(1+2+3+\cdots+\frac{(n-5)}{2}+\frac{(n-3)}{2}+\frac{(n-1)}{2}+\right)$

$\sum\left|x_{i}-\bar{x}\right|=2\left(1+2+3+\cdots+\frac{(n-5)}{2}+\frac{(n-3)}{2}+\frac{(n-1)}{2}\right)$

That is 2 times sum of $(n-1) 2$ terms, so it can be written as

$\sum\left|x_{i}-\bar{x}\right|=2\left(\frac{\frac{(n-1)}{2}\left(\frac{(n-1)}{2}+1\right)}{2}\right)$

Taking LCM we get

$\sum\left|x_{i}-\bar{x}\right|=2\left(\frac{\frac{(n-1)}{2}\left(\frac{(n-1)+2}{2}\right)}{2}\right)$

The above equation can be written as

$\sum\left|x_{i}-\bar{x}\right|=\left(\frac{(n-1)}{2}\left(\frac{n+1}{2}\right)\right)$

Multiplying the above equation we get

$\sum\left|x_{i}-\bar{x}\right|=\left(\frac{n^{2}-1}{4}\right)$

Therefore, mean deviation about the mean is

M. D $=\frac{\sum\left|x_{i}-\bar{x}\right|}{n}=\frac{\left(\frac{n^{2}-1}{4}\right)}{n}=\left(\frac{n^{2}-1}{4 n}\right)$

Hence the mean deviation about the mean of the set of first $n$ natural numbers when $\mathrm{n}$ is a odd number is

$\left(\frac{\mathrm{n}^{2}-1}{4 \mathrm{n}}\right)$