Let,
n = 2k
⇒ \(k=\frac{n}{2}\)
\(∴ \bar{x}=\frac{1+2+3+...+n}{n}\)
\(= \frac{n(n+1)}{2n}\)
\(= \frac{n+1}{2}\)
\(= \frac{2k+1}{2}\)
∴ Mean deviation about mean, M.D \((\bar{x})\)
\(=\frac{1}{n}\sum |x-\bar{x}|\)
\(=\frac{1}{n}.\frac{n^2}{4}\)
\(=\frac{n}{4}\)
