Find the mean deviation about the median for the data

Solution:

The given data is

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Median $\mathrm{M}=\frac{\left(\frac{10}{2}\right)^{t h} \text { observation }+\left(\frac{10}{2}+1\right)^{m} \text { observation }}{2}$

$=\frac{5^{\text {th }} \text { observation }+6^{\text {th }} \text { observation }}{2}$

$=\frac{46+49}{2}=\frac{95}{2}=47.5$

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

–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

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

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Thus, the required mean deviation about the median is

M.D. $(\mathrm{M})=\frac{\sum_{i=1}^{10}\left|x_{i}-\mathrm{M}\right|}{10}=\frac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}$

$=\frac{70}{10}=7$