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 x_{1}, x_{2}, .... 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 = \( \dfrac{\Sigma f xx_i}{\Sigma f} \)
Let us try out the belowsolved 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 02 24 46 68 810 Frequency (f) 4 3 5 7 2 Solution:
Class Interval f Class Mark/Midpoint (x_{i}) fx_{i} Absolute Deviation x_{i}x̄ fx_{i}x̄ 02 4 1 4 4 16 24 3 3 9 2 6 46 5 5 25 0 0 68 7 7 49 2 14 810 2 9 18 4 8 \( \Sigma f = 21\) \( \bar{x} = \frac{\Sigma f x_i}{\Sigma f} = \frac{105}{21} = 5\) M.A.D. = \(\frac{\Sigma f xx_i}{\Sigma f} = \frac{44}{21} = 2.09 \) Answer: Mean absolute deviation of the data set is 2.09.
visual curriculum