Question

The mean deviation about the mean of the set of first ' $n$ ' natural numbers, when ' $n$ ' is an even number, is equal to

• A. $n$
• B. $\frac{n}{2}$
• C. $\frac{n}{3}$
• D. $\frac{n}{4}$

Answer 1

Answer: (D)

Mean
$\bar{x}=\frac{1+2+\dots n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}$
Here, $n=$ even number Now mean deviation is
$M\cdot \overline{D(x)}=\left[1-\frac{n+1}{2}|+|2-\frac{n+1}{2}\left|+\dots .+n-\frac{n+1}{2}\right|\right]\times \frac{1}{n}$
$=\frac{2}{n}\left\{1+2+\dots +\left(\frac{n}{2}\right)\right\}\frac{n}{2}$ terms