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

1 Answer

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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