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Mean Absolute Deviation Formula

Mean Absolute deviation, in general, is the average deviation of the data points from a center point. The center point can be mean, median, mode, or any random point. Quite often the mean is taken as the center point. Here the mean absolute deviation formula helps in calculating mean absolute deviation (MAD), which is the average of the absolute deviation (distance) of the data points from the mean of the data set. Let us explore the mean absolute deviation formula in the following sections.

What is Mean Absolute Deviation Formula?

There are two formulas to find the mean absolute deviation. One is for the ungrouped data and another is for the grouped data. Let x₁, x₂, .... x_n be the data set and let μ be its mean of the ungrouped data. And, f is the frequency of the data point x_i,for the grouped data. The mean absolute deviation formulas for the two types of data are as follows.

For grouped data:

Mean Absolute Deviation = \( \dfrac{1}{n}\sum^n_{i=1} |x_i - μ| \)

For ungrouped data:

Mean Absolute Deviation = \( \dfrac{\Sigma f |x-x_i|}{\Sigma f} \)

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Let us try out the below-solved examples to learn how to use the mean absolute deviation formula.

Solved Examples using Mean Absolute Deviation Formula

Example 1: Find mean absolute deviation for the following data set: 302, 140, 352, 563, 455, 215, 213

Solution:

To find: Mean absolute deviation for the given data set.

Given:

Data set = {302, 140, 352, 563, 455, 215, 213}

Mean of the data (μ) = (302 + 140 + 352 + 563 + 455 + 215 + 213)/7 = 320

Using Mean Deviation Formula,

MAD =\( \dfrac{1}{n}\sum^n_{i=1} |x_i - μ| \)

= \( \dfrac{|302 - 320|+|140 - 320|+|352 - 320|+|563 - 320|+|455 - 320|+|215 - 320|+|213 - 320|}{7} \)

= 117.14

Answer: Mean absolute deviation of the data set is 117.14.
Example 2: For the given data set, if the mean of the dataset is 31. Find the mean absolute deviation of the data set.

30, 14, 35, 55, 45, 21, X

Solution:

To find: Mean absolute deviation of the data set

Given:

Data set = {30, 14, 35, 55, 45, 21, X }

Mean of the data set = 31

Mean of the data (μ) = (30 + 14 + 35 + 55 + 45 + 21 + X)/7 = 31

200 + X = 31 × 7 = 217

X = 17

Using Mean Deviation Formula,

MAD =\( \dfrac{1}{n}\sum^n_{i=1} |x_i - μ| \)

= \( \dfrac{|30 - 31|+|14 - 31|+|35 - 31|+|55 - 31|+|45 - 31|+|21 - 31|+|17 - 31|}{7} \)

= 12

Answer: Mean absolute deviation of the data set is 12.

Example 3: Find the mean absolute deviation for the following data set.

Class Interval	0-2	2-4	4-6	6-8	8-10
Frequency (f)	4	3	5	7	2

Solution:

Class Interval	f	Class Mark/Midpoint (x_i)	fx_i	Absolute Deviation \|x_i-x̄\|	f\|x_i-x̄\|
0-2	4	1	4	4	16
2-4	3	3	9	2	6
4-6	5	5	25	0	0
6-8	7	7	49	2	14
8-10	2	9	18	4	8
	\( \Sigma f = 21\)		\( \bar{x} = \frac{\Sigma f x_i}{\Sigma f} = \frac{105}{21} = 5\)		M.A.D. = \(\frac{\Sigma f \|x-x_i\|}{\Sigma f} = \frac{44}{21} = 2.09 \)

Answer: Mean absolute deviation of the data set is 2.09.

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