The mean deviation from the median of the dat | vedclass

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Question

(C) Given data: $16, 22, 3, 14, 5, 10, 8, 6, 11, 4$. Arranging in ascending order: $3, 4, 5, 6, 8, 10, 11, 14, 16, 22$. Number of observations $n = 10

Vedclass · Accepted Answer

$4.7$

Vedclass · Answer

(C) Given data: $16, 22, 3, 14, 5, 10, 8, 6, 11, 4$. Arranging in ascending order: $3, 4, 5, 6, 8, 10, 11, 14, 16, 22$. Number of observations $n = 10$. Since $n$ is even,the median is the average of the $5^{	ext{th}}$ and $6^{	ext{th}}$ observations: $	ext{Median} = \frac{8 + 10}{2} = 9$. Mean deviation from the median is given by $\frac{1}{n} \sum |x_i - 	ext{Median}|$: $	ext{Mean Deviation} = \frac{|3-9| + |4-9| + |5-9| + |6-9| + |8-9| + |10-9| + |11-9| + |14-9| + |16-9| + |22-9|}{10}$ $= \frac{6 + 5 + 4 + 3 + 1 + 1 + 2 + 5 + 7 + 13}{10}$ $= \frac{47}{10} = 4.7$.

$x_i$	$15$	$21$	$27$	$30$	$35$
$f_i$	$3$	$5$	$6$	$7$	$8$

Income per day	Number of persons
$0-100$	$4$
$100-200$	$8$
$200-300$	$9$
$300-400$	$10$
$400-500$	$7$
$500-600$	$5$
$600-700$	$4$
$700-800$	$3$

The mean deviation from the median of the data $16, 22, 3, 14, 5, 10, 8, 6, 11, 4$ is

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