The mean deviation about the mean of the set | vedclass

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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

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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 $M.D.(\bar{x}) = \frac{n^2-1}{4n}$.

$\text{Marks obtained}$	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
$\text{Number of Boys}$	$6$	$8$	$10$	$4$	$2$

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

Similar Questions

The mean deviation about the mean for the following data is:
$\text{Marks obtained}$ $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
$\text{Number of Boys}$ $6$ $8$ $10$ $4$ $2$

What is the coefficient of mean deviation about the median for the observations $40, 62, 54, 90, 68, 76$?

The mean deviation from the median for the following data is
$x_i$ $2$ $3$ $5$ $7$ $9$
$f_i$ $8$ $6$ $4$ $2$ $1$

Find the mean deviation about the median for the data $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$.

Find the mean deviation about the median for the given data: $340, 150, 210, 240, 300, 310, 320$.

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