The mean deviation of the numbers a, a+d, a+2 | vedclass

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(D) The given numbers are $a, a+d, a+2d, \ldots, a+2nd$. This is an arithmetic progression with $N = 2n+1$ terms. The mean $\bar{x}$ is the middle ter

Vedclass · Accepted Answer

$\frac{n(n+1)|d|}{2n+1}$

Vedclass · Answer

(D) The given numbers are $a, a+d, a+2d, \ldots, a+2nd$. This is an arithmetic progression with $N = 2n+1$ terms. The mean $\bar{x}$ is the middle term: $\bar{x} = a + nd$. The mean deviation about the mean is given by $MD = \frac{1}{N} \sum_{i=0}^{2n} |x_i - \bar{x}|$. $MD = \frac{1}{2n+1} \sum_{k=0}^{2n} |(a+kd) - (a+nd)| = \frac{1}{2n+1} \sum_{k=0}^{2n} |k-n||d|$. Let $j = k-n$. As $k$ goes from $0$ to $2n$,$j$ goes from $-n$ to $n$. $MD = \frac{|d|}{2n+1} \sum_{j=-n}^{n} |j| = \frac{|d|}{2n+1} \left( 2 \sum_{j=1}^{n} j \right) = \frac{|d|}{2n+1} \cdot 2 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)|d|}{2n+1}$.

$x_i$	$5$	$7$	$9$	$10$	$12$	$15$
$f_i$	$8$	$6$	$2$	$2$	$2$	$6$

$\text{Marks}$	$10$	$15$	$20$	$25$	$30$
$\text{Number of students}$	$2$	$4$	$6$	$8$	$5$

The mean deviation of the numbers $a, a+d, a+2d, \ldots, a+2nd$ from their mean is equal to

Similar Questions

The mean deviation about the mean for the data:

$x_i$ $5$ $7$ $9$ $10$ $12$ $15$

$f_i$ $8$ $6$ $2$ $2$ $2$ $6$

is equal to: (in /$13$)

Let the mean of $6$ observations $1, 2, 4, 5, x,$ and $y$ be $5$ and their variance be $10$. Then their mean deviation about the mean is equal to $........$.

The following data represents the frequency distribution of $20$ observations.
$x_i$ $3$ $4$ $5$ $8$ $10$ $11$
$f_i$ $\alpha+2$ $(\alpha-1)^2$ $4$ $\alpha-1$ $2$ $\alpha$
Then its mean deviation about the mean is

The mean deviation from the mean of the data given below is
$\text{Marks}$ $10$ $15$ $20$ $25$ $30$
$\text{Number of students}$ $2$ $4$ $6$ $8$ $5$

Mean deviation is least if it is taken about :-

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