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Calculate the mean deviation about the mean of the set of first `n`natural numbers when `n`is odd natural number. |
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Answer» Mean of first `n` natural numbers when `n` is odd `= (n(n+1))/2` `:.` Mean deviation `= (sum |x_i-barx|)/n` `=1/n[(1-(n(n+1))/2)+(2-(n(n+1))/2)+...(n - (n(n+1))/2)]` `=2/n[1+2+...(n-1)/2]` This is a series of first `(n-1)/2` numbers, `:. 2/n[1+2+...(n-1)/2] = 2/n((((n-1)/2)((n-1)/2+1))/2)` `=2/n((n^2-1)/8)` `=(n^2-1)/(4n)`,which is the required mean deviation. |
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