If M_1 is the mean deviation from the mean of | vedclass

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(D) The given data is $44, 5, 27, 20, 8, 54, 9, 14, 35$. The number of observations $N = 9$.First,calculate the mean $\bar{x} = \frac{44+5+27+20+8+54+

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$\frac{4}{9}$

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(D) The given data is $44, 5, 27, 20, 8, 54, 9, 14, 35$. The number of observations $N = 9$.First,calculate the mean $\bar{x} = \frac{44+5+27+20+8+54+9+14+35}{9} = \frac{216}{9} = 24$.The mean deviation about the mean $M_1 = \frac{1}{N} \sum |x_i - \bar{x}|$.$|44-24| + |5-24| + |27-24| + |20-24| + |8-24| + |54-24| + |9-24| + |14-24| + |35-24| = 20 + 19 + 3 + 4 + 16 + 30 + 15 + 10 + 11 = 128$.So,$M_1 = \frac{128}{9}$.Now,arrange the data in ascending order: $5, 8, 9, 14, 20, 27, 35, 44, 54$.The median $M$ is the $\left(\frac{9+1}{2}ight)^{	ext{th}} = 5^{	ext{th}}$ term,which is $20$.The mean deviation about the median $M_2 = \frac{1}{N} \sum |x_i - M|$.$|5-20| + |8-20| + |9-20| + |14-20| + |20-20| + |27-20| + |35-20| + |44-20| + |54-20| = 15 + 12 + 11 + 6 + 0 + 7 + 15 + 24 + 34 = 124$.So,$M_2 = \frac{124}{9}$.Therefore,$M_1 - M_2 = \frac{128}{9} - \frac{124}{9} = \frac{4}{9}$.

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

Height in $cm$	Number of boys
$95-105$	$9$
$105-115$	$13$
$115-125$	$26$
$125-135$	$30$
$135-145$	$12$
$145-155$	$10$

If $M_1$ is the mean deviation from the mean of the discrete data $44, 5, 27, 20, 8, 54, 9, 14, 35$ and $M_2$ is the mean deviation from the median of the same data,then $M_1 - M_2 =$

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