Question

If the value observed are $1,2,3,...,n$ each with frequency $1$ and $n$ is even, then the mean deviation from mean equals to

• A. $n$
• B. $\frac{n}{2}$
• C. $\frac{n}{4}$
• D. None of these

Answer 1

The correct option is B

$\frac{n}{2}$

Explanation for the correct option :

Step 1. Find the mean.

The mean of the values $1,2,3,...,n$ is given as:

$\begin{array}{rcl}\overline{X}& =& \frac{1+2+3+...+n}{n}\left[1+2+3+...+n=\frac{n(n+1)}{2}\right]\mathbf{}\\ & =& \frac{{\displaystyle \frac{n(n+1)}{2}}}{n}\\ & =& \frac{n+1}{2}\end{array}$

Step 2. Find the mean deviation from the mean

The mean deviation from the mean is given as: $\frac{{\displaystyle \sum _{i=1}^n}{f}_i\left|x_i-\overline{X}\right|}{\sum _{i=1}^n{f}_i}$

Now the variables under consideration $1,2,3,...,n$ all have frequency $1$. So,

$\begin{array}{rcl}\sum _{i=1}^n{f}_i& =& 1+1+1+...n\mathrm{times}\\ & =& n\end{array}$

Now the mean deviation is found as:

$\begin{array}{rcl}\frac{{\displaystyle \sum _{i=1}^n{f}_i\left|x_i-\overline{X}\right|}}{\sum _{i=1}^n{f}_i}& =& \frac{{\displaystyle 1\left|1-\frac{n+1}{2}\right|+1\left|2-\frac{{\displaystyle n+1}}{{\displaystyle 2}}\right|+1\left|3-\frac{{\displaystyle n+1}}{{\displaystyle 2}}\right|+...+1\left|n-\frac{{\displaystyle n+1}}{{\displaystyle 2}}\right|}}{n}\\ & =& \frac{\left|\frac{1-n}{2}\right|+\left|\frac{3-n}{2}\right|+\left|\frac{5-n}{2}\right|+...+\left|-\frac{1}{2}\right|+....+\left|\frac{n-3}{2}\right|+\left|\frac{n-1}{2}\right|}{n}\\ & =& \frac{1}{n}\times 2\left[\frac{n-1}{2}+\frac{n-3}{2}+...+\frac{1}{2}\right]\\ & =& \frac{1}{n}\times 2\times \frac{1}{2}\left[\left(n-1\right)+\left(n-3\right)+...+1\right]\\ & =& \frac{1}{n}\left[1+3+5+....+\left(n-1\right)\right]\\ & =& \frac{1}{n}\left[\frac{n}{2}\left(1+n-1\right)\right]\\ & =& \frac{1}{n}\times \frac{n^2}{2}\\ & =& \frac{n}{2}\end{array}$

So, the mean deviation from the mean is $\frac{n}{2}$.

Hence, the correct option is B.