Find the mean deviation about the median for the data

 xi 15 21 27 30 35 fi 3 5 6 7 8

Asked by Pragya Singh | 11 months ago |  71

##### Solution :-

The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table

Now, N = 29, which is odd.

So $$\dfrac{29}{2}$$ = 14.5

The cumulative frequency greater than 14.5 is 21, for which the corresponding observation is 30.

Then,

Median = $$\dfrac{ (15^{th} observation + 16^{th} observation)}{2}$$

$$\dfrac{(30 + 30)}{2}$$

$$\dfrac{60}{2}$$

= 30

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.

Therefore,

$$\displaystyle\sum_{i=1}^{5} f_i = 29$$ and $$\displaystyle\sum_{i=1}^{5} f_i |X_i-M|$$

$$\overline{X} = \dfrac{1}{N} \displaystyle\sum_{i=1}^{6} f_i |X_i-M|$$

$$\dfrac{1}{29}\times 148$$

= 5.1

Answered by Abhisek | 11 months ago

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