In a series of 3n observations,if n observati | vedclass

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(C) The total number of observations is $3n$.There are $n$ observations equal to $a$ and $2n$ observations equal to $-2a$.The mean $(\bar{X})$ is calc

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$\frac{4a}{3}$

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(C) The total number of observations is $3n$.There are $n$ observations equal to $a$ and $2n$ observations equal to $-2a$.The mean $(\bar{X})$ is calculated as:$\bar{X} = \frac{n(a) + 2n(-2a)}{3n} = \frac{na - 4na}{3n} = \frac{-3na}{3n} = -a$.The mean deviation about the mean is given by $\frac{1}{3n} \sum_{i=1}^{3n} |x_i - \bar{X}|$.Substituting the values:Mean deviation $= \frac{n|a - (-a)| + 2n|-2a - (-a)|}{3n}$$= \frac{n|2a| + 2n|-a|}{3n}$$= \frac{2na + 2na}{3n} = \frac{4na}{3n} = \frac{4a}{3}$.

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

In a series of $3n$ observations,if $n$ observations are equal to $a$ and the remaining $2n$ observations are equal to $-2a$,then the mean deviation of the observations about their mean is:

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