# MEAN ABSOLUTE DEVIATION

Mean Absolute Deviation (MAD) :

A measure of variability is a single number used to describe the spread of a data set. It can also be called a measure of spread. One measure of variability is the mean absolute deviation (MAD), which is the mean of the distances between the data values and the mean of the data set.

For a given set of observation, MAD is defined as the arithmetic mean of the absolute deviations of the observations from an appropriate measure of central tendency.

Let the  variable "x" assume "n" values  as given below

Then the MAD of x is given by

For a grouped frequency distribution, MAD is given by

Where, N  =  ∑f

## Coefficient of MAD

A relative measure of dispersion applying MAD is given by

MAD takes its minimum value when the deviations are taken from the median.

Also MAD remains unchanged due to a change of origin but changes in the same ratio due to a change in scale

i.e. if y = a + bx, a and b being constants,

then MD of y = |b| × MD of x

## Properties of MAD

1)  MAD takes its minimum value when the deviations are taken from the median.

2)  MAD remains unchanged due to a change of origin but changes in the same ratio due to a change in scale

i.e. if y = a + bx, a and b being constants,

then MD of y = |b| × MD of x

3)  It is rigidly defined

4) It is based on all the observations and not much affected by sampling fluctuations.

5)  It is difficult to comprehend and its computation

6) Furthermore, unlike SD, MAD does not possess mathematical properties.

## Mean Absolute Deviation - Examples

Example 1 :

What is the MAD for the following numbers?

5, 8, 10, 10, 12, 9

Solution :

The mean is given by

x̄  =  (5 + 8 + 10 + 10 + 12 + 9) / 6

x̄  =   54 / 6

x̄  =   9

Thus MAD is given by

∑|x - x̄ | / n  =  10 / 6  =  1.67

Hence, MAD for the given data is 1.67

Example 2 :

The data represent the height, in feet, of various buildings. Find the mean absolute deviation.

60, 58, 54, 56, 63, 65, 62, 59, 56, 57

Solution :

Let x  =    60, 58, 54, 56, 63, 65, 62, 59, 56, 57

The mean is given by

x̄  =  (60+58+54+56+63+65+62+59+56+57) / 10

x̄  =   590 / 10

x̄  =   59

Absolute deviations of observations from mean :

|60 - 59 |  =  |1|  =  1

|58 - 59 |  =  |-1|  =  1

|54 - 59 |  =  |-5|  =  5

|56 - 59 |  =  |-3|  =  3

|63 - 59 |  =  |4|  =  4

|65 - 59 |  =  |6|  =  6

|62 - 59 |  =  |3|  =  3

|59 - 59 |  =  |0|  =  0

|56 - 59 |  =  |-3|  =  3

|57 - 59 |  =  |-2|  =  2

Calculate the MAD by finding the mean of the above absolute deviations of observations from mean.  Round to the nearest whole number.

MAD  =  (1+1+5+3+4+6+3+0+3+2) / 10

MAD  =  28 / 10

MAD  =  2.8 ≈  3

After having gone through the stuff given above, we hope that the students would have understood mean absolute deviation.

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