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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(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 = $\frac{\sum f_i |x_i - M|}{N} = \frac{44}{22} = 2$.

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

$x$	$6$	$12$	$18$	$24$	$30$	$36$	$42$
$f$	$4$	$7$	$9$	$18$	$15$	$10$	$5$

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

The mean deviation about the median of the following distribution is:
$x$ $6$ $12$ $18$ $24$ $30$ $36$ $42$
$f$ $4$ $7$ $9$ $18$ $15$ $10$ $5$

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

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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$