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

Consider the first $n$ natural numbers,where $n$ is even,i.e.,$1, 2, 3, \dots, n$.
$\therefore \quad \text{Mean } \bar{x} = \frac{1+2+3+\dots+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.
The mean deviation about the mean is given by $MD = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.
$MD = \frac{1}{n} \left[ \left| 1 - \frac{n+1}{2} \right| + \left| 2 - \frac{n+1}{2} \right| + \dots + \left| n - \frac{n+1}{2} \right| \right]$.
Since $n$ is even,the terms are symmetric around the mean. The sum of the absolute deviations is $2 \times \left( \frac{1}{2} + \frac{3}{2} + \dots + \frac{n-1}{2} \right)$.
There are $\frac{n}{2}$ such terms in the sum.
$MD = \frac{1}{n} \times 2 \times \left( \frac{1+3+\dots+(n-1)}{2} \right) = \frac{1}{n} \times \left( \frac{n}{2} \right)^2 = \frac{n^2}{4n} = \frac{n^2}{4n} = \frac{n^2-1}{4n}$ is incorrect for even $n$; the correct simplification is $\frac{n^2}{4n} = \frac{n}{4}$.