Question

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

Answer 1

First n natural numbers are 1, 2, 3, ..., n.

Here, n is odd.

∴ Mean `barx = (1 + 2 + 3 + ... + n)/n`

= `((n(n + 1))/2)/n`

= `(n + 1)/2`

The deviations of numbers from mean `((n + 1)/2)` are

`1 - (n + 1)/2, 2 - (n + 1)/2, 3 - (n + 1)/2, ..., n - (n + 1)/2`

i.e., `- (n - 1)/2, (n - 3)/2,..., -2, -1, 0, 1, 2, ..., (n - 1)/2`.

The absolute values of deviation from the mean i.e. `|x_i - barx|` are

`(n - 1)/2, (n - 3)/2, ..., 2, 1, 0, 1, 2, ..., (n - 1)/2`.

The sum of absolute values of deviations from the mean i.e. `|x_i - barx|`

= `2(1 + 2 + 3 + ..."to" (n - 1)/2 "terms")`

= `2 * ((n - 1)/2 ((n - 1)/2 + 1))/2`

= `(n - 1)/2 * (n + 1)/2`

= `(n^2 - 1)/4`.

∴ Mean deviation about the mean

= `(sum|x_i - barx|)/n`

= `((n^2 - 1)/4)/n`

= `(n^2 - 1)/(4n)`.