Question

Find the mean deviation about the median for the following data: $36,72,46,42,60,45,53,46,51,49.$

Answer 1

Given data: $36,72,46,42,60,45,53,46,51,49.$

Arranging in increasing order $36,42,45,46,46,49,51,53,60,72.$

Total number of terms is 10 which is odd.
$\text{Median}=\frac{({\left(\frac{10}{2}\right)}^{th}\text{observation}+{(\left(\frac{10}{2}\right)+1)}^{th}\text{observation})}{2}$

$\because$ The $5^{th}$ and $6^{th}$ observation which is equal to $46$ and $49$.

$\therefore M=\frac{(46+49)}{2}=\frac{95}{2}=47.5$

Median , $M=47.5$

Mean deviation about median $=\frac{\sum |x_i-M|}{n}M.D.\left(M\right)=\frac{\left(\begin{array}{c}|36-47.5|+|42-47.5|+|45-47.5|+|46-47.5|+|46-47.5|\\ +|49-47.5|+|51-47.5|+|53-47.5|+|60-47.5|+|72-47.5|\end{array}\right)}{10}$

$=\frac{\left(\begin{array}{c}|-11.5|+|-5.5|+|-2.5|+|-1.5|+|-1.5|\\ +|1.5|+|3.5|+|5.5|+|12.5|+|24.5|\end{array}\right)}{10}$

$=\frac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}$

$=\frac{70}{10}$

$=7$