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

1 Answer

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

Related Questions

Find the mean deviation from the mean and from a median of the following distribution:

Marks 0-10 10-20 20-30 30-40 40-50
No. of students 5 8 15 16 6

 

Class 11 Maths Statistics View Answer

The age distribution of 100 life-insurance policy holders is as follows

Age (on nearest birthday) 17-19.5 20-25.5 26-35.5 36-40.5 41-50.5 51-55.5 56-60.5 61-70.5
No. of persons 5 16 12 26 14 12 6 5

 

Class 11 Maths Statistics View Answer

Compute mean deviation from mean of the following distribution:

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No. of students 8 10 15 25 20 18 9 5

 

Class 11 Maths Statistics View Answer

Find the mean deviation from the mean for the following data:

Classes 95-105 105-115 115-125 125-135 135-145 145-155
Frequencies 9 13 16 26 30 12

 

Class 11 Maths Statistics View Answer

Find the mean deviation from the mean for the following data:

Classes 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
Frequencies 4 8 9 10 7 5 4 3

 

Class 11 Maths Statistics View Answer