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

AP EAMCETAP EAMCET 2020

Solution:

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