The mean deviation from the median for the fo | vedclass

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(A) Step $1$: Arrange the data in ascending order of $x_i$: $x_i: 2, 3, 5, 7, 8, 9$ and $f_i: 5, 6, 6, 1, 1, 3$.Step $2$: Calculate the cumulative fre

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$2$

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(A) Step $1$: Arrange the data in ascending order of $x_i$: $x_i: 2, 3, 5, 7, 8, 9$ and $f_i: 5, 6, 6, 1, 1, 3$.Step $2$: Calculate the cumulative frequency $(cf)$: $5, 11, 17, 18, 19, 22$. Total $N = 22$.Step $3$: The median is the value corresponding to the $(\frac{N}{2})^{th}$ and $(\frac{N}{2} + 1)^{th}$ observation,which are the $11^{th}$ and $12^{th}$ observations. The $11^{th}$ observation is $3$ and the $12^{th}$ observation is $5$. Median $M = \frac{3+5}{2} = 4$.Step $4$: Calculate $|x_i - M|$: $|2-4|=2, |3-4|=1, |5-4|=1, |7-4|=3, |8-4|=4, |9-4|=5$.Step $5$: Calculate $\sum f_i |x_i - M| = (5 	imes 2) + (6 	imes 1) + (6 	imes 1) + (1 	imes 3) + (1 	imes 4) + (3 	imes 5) = 10 + 6 + 6 + 3 + 4 + 15 = 44$.Step $6$: Mean deviation from median = $\frac{\sum f_i |x_i - M|}{N} = \frac{44}{22} = 2$.

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

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

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