Find the mean deviation about the median for the data.

Solution:

The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

Median, $\mathrm{M}=\frac{\left(\frac{12}{2}\right)^{\text {min }} \text { observation }+\left(\frac{12}{2}+1\right)^{\text {mi }} \text { observation }}{2}$

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

$=\frac{13+14}{2}=\frac{27}{2}=13.5$

The deviations of the respective observations from the median, i.e. $x_{i}-\mathrm{M}$, are

$-3.5,-2.5,-2.5,-1.5,-0.5,-0.5,0.5,2.5,2.5,3.5,3.5,4.5$

The absolute values of the deviations, $\left|x_{i}-\mathrm{M}\right|$, are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is

M.D. $(M)=\frac{\sum_{i=1}^{12}\left|x_{i}-M\right|}{12}$

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

$=\frac{28}{12}=2.33$