Let,
n = 2k + 1
⇒ k = \(\frac{n-1}{2}\)
∴ \(\bar{x}=\frac{1+2+3+....+n}{n}\)
\(=\frac{n(n+1)}{2n}\)
\(=\frac{n+1}{2}\)
\(=\frac{2k+1+1}{2}\)
= k + 1.
\(\sum|x_i-\bar{x}|\)
= 2 (1 + 2 + 3 +…+ k)
\(= 2\{\frac{k(k+1)}{2}\}\)
= k (k + 1)
\(= (\frac{n-1}{2})(\frac{n-1}{2}+1)\)
\(= \frac{n-1}{2}.\frac{n+1}{2}\)
\(= \frac{n^2-1}{4}\)
∴ Read mean deviation about mean \((\bar{x})\)
i.e., M.D.\((\bar{x})\)
\(=\frac{1}{n}\sum|x_1-\bar{x}|\)
\(=\frac{1}{n}.\frac{n^2-1}{4}\)
\(=\frac{n^2-1}{4n}\)
