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

First we have to arrange the given observations into ascending order,

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

The number of observations is 10

Then,

Median = \(\dfrac{(\dfrac{10}{2})^{th}observation+(\dfrac{10}{2})^{th}observation}{2} \)

(\( \dfrac{10}{2}\))^th observation = 5^th = 46

(\( \dfrac{10}{2}\))+ 1)^th observation = 5 + 1

= 6^th = 49

Median = \(\dfrac{(46 + 49)}{2}\)

= 95

= 47.5

So, the absolute values of the respective deviations 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

\( \displaystyle\sum_{i=1}^{10} f_i |X_i-M|=70\)

Mean Deviation,

\( MD(M) = \dfrac{1}{10} \displaystyle\sum_{i=1}^{10} |X_i-M|\)

= \( \dfrac{1}{10}\times 70\)

= 7

So, the mean deviation about the median for the given data is 7.