Question

Find the mean deviation about the median for the following data: $13,17,16,14,11,13,10,16,11,18,12,17.$

Answer 1

Given data: $13,17,16,14,11,13,10,16,11,18,12,17.$

Arranging in increasing order $10,11,11,12,13,13,14,16,16,17,17,18.$

Total number of terms is 12 which is odd.
$\text{Median}=\frac{({\left(\frac{12}{2}\right)}^{th}\text{observation}+{(\left(\frac{12}{2}\right)+1)}^{th}\text{observation})}{2}$

$=\frac{6^{th}\text{observation}+7^{th}\text{observation}}{2}$

$\therefore M=\frac{(13+14)}{2}=\frac{27}{2}=13.5$


Mean deviation about median = $\frac{\sum |x_i-M|}{n}$

M.D.$\left(M\right)$
$=\frac{\left(\begin{array}{c}|10-13.5|+|11-13.5|+|11-13.5|+|12-13.5|+|13-13.5|+|13-13.5|\\ +|14-13.5|+|16-13.5|+|16-13.5|+|17-13.5|+||17-13.5|+|18-13.5|\end{array}\right)}{12}$

$=\frac{\left(\begin{array}{c}|-13.5|+|-2.5|+|-2.5|+|-1.5|+|-0.5|+|-0.5|\\ +|0.5|+|2.5|+|2.5|+|3.5|+|3.5|+|4.5|\end{array}\right)}{12}$

$=\frac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12}$

$M.D.\left(M\right)=\frac{\left(28\right)}{12}$

$\therefore M.D.\left(M\right)=2.33$