Question

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

Answer 1

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 M= $\frac{(\frac{10}{2})^{th}observation+(\frac{10}{2}+1)^{th} \text{observation}}{2}$

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

$\frac{46+49}{4}=\frac{95}{2}=47.5$

The deviations of the respective observations from the median, i.e. $x_i-M,$ 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.(M)=$\frac{\sum_{I=1}^{10}|x_i-M|}{10}$

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

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