The mean deviation about the mean of the set | vedclass

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Question

(D) The mean $\bar{x}$ of the first $n$ natural numbers is given by $\bar{x} = \frac{1+2+\ldots+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.Given that

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$\frac{n^2-1}{4n}$

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(D) The mean $\bar{x}$ of the first $n$ natural numbers is given by $\bar{x} = \frac{1+2+\ldots+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.Given that $n$ is an even number,the mean deviation about the mean is calculated as:$M.D.(\bar{x}) = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| = \frac{1}{n} \sum_{i=1}^{n} |i - \frac{n+1}{2}|$.Since $n$ is even,the sum splits into two equal parts of absolute differences:$M.D.(\bar{x}) = \frac{1}{n} \left[ \sum_{i=1}^{n/2} (\frac{n+1}{2} - i) + \sum_{i=n/2+1}^{n} (i - \frac{n+1}{2}) \right]$.Evaluating this sum,we get $M.D.(\bar{x}) = \frac{n^2-1}{4n}$.

$x_i$	$3$	$6$	$9$	$12$	$13$	$15$	$21$	$22$
$f_i$	$3$	$4$	$5$	$2$	$4$	$5$	$4$	$3$

The mean deviation about the mean of the set of first $n$ natural numbers,when $n$ is an even number,is equal to:

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