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

Asked by Pragya Singh | 11 months ago |  79

Solution :-

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 = 5th = 46

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

= 6th = 49

Median = $$\dfrac{(46 + 49)}{2}$$

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |xi – 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.

Answered by Abhisek | 11 months ago

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