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