# What is Mean Absolute Deviation? - Definition & Examples

Instructor
Lynne Hampson

Lynne Hampson has a Masters in Instr. Design & Bach. in Elem./Spec. Educ. She taught 8 years in Elem. Core, Science, Coding, Microsoft, Internet Safety, and Life Skills.

Expert Contributor
Elaine Chan

Dr. Chan has taught computer and college level physics, chemistry, and math for over eight years. Dr. Chan has a Ph.D. in Chemistry from U. C. Berkeley, an M.S. Physics plus 19 graduate Applied Math credits from UW, and an A.B. with honors from U.C .Berkeley in Physics.

In this lesson, we will first learn how to use the mean absolute deviation to find not only the average of a set of numbers but the average distance each number is from the average number. Then, we will show a step-by-step, real-world problem, to practice your skills. Updated: 07/10/2020

## Mean Absolute Deviation

Finding the mean is essentially finding the average of a set of numbers. Once the mean is found, the next step is the absolute deviation, in other words, determining the distance between each of the original numbers from the mean you found in step one.

Mean absolute deviation is the average distance between the mean of a set of numbers.

## How Is It Used?

Many professionals use mean in their everyday lives. Teachers give tests to students and then average the results to see if the average score was high, in between, or too low. Each average tells a story.

Absolute deviation can further help to see the distance between each of the scores and the beginning average scores. Analyzing information in this way helps the teacher to see if the test was too hard, too easy, or just right, based upon the mathematical outcomes.

Biologists can use this concept to compare the differences in animal weight to decide what a healthy weight might be. Let's take a look at an example and walk through the whole process.

## Example

1. Finding the Mean First

First, find the mean (average) of a number group. Let's say the number group is the rounded-up weight of infants at birth at a specific hospital in the month of January. Five births were recorded, so there should be five numbers representing their weights:

5 lbs., 9 lbs., 6 lbs., 7 lb., 8 lbs.

To find the mean, first add up the numbers: 5 + 9 + 6 + 7 + 8. The total should come to 35. Now, divide the number of birth weights by the total weight. 35/5= the mean (average) of 7.

2. Now Use Absolute Deviation

Now, refer to your original 5 baby weights. Use addition or subtraction to find the difference in distance between each of those numbers and the average. If the difference is negative, ignore that it is negative. You are only looking for the distance apart, not the value regarding negative or positive. Keep record of your findings:

5 and 2 are 7
9 and 2 are 7
6 and 1 are 7
7 and 0 are 7
8 and 1 are 7

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## Practice Problems:

The degree to which numerical data tend to spread around the average value is called the dispersion or variation of the data.

The mean absolute deviation MAD is defined for a set X = x1, x2, ... xN as the sum over all N elements of the set

MAD = (1/N) Σ | xi - m(X) |

We have been using m(X) as the average or mean. The mean or average is found by adding up all the elements of X and dividing by the total number of elements, N.

### Problem 1:

Find the mean absolute deviation of the set of numbers 12, 6, 7, 3, 15, 10, 18, 5.

### Problem 2:

Find the mean absolute deviation of the set of numbers 9, 3, 8, 8, 9, 8, 9, 18.

### Problem 3:

Find the mean absolute deviation of the set 2, 3, 6, 8,11.

The mean or average is:

( 12 + 6 + & + 3 + 15 + 10 + 19 + 5)/8 = 76.8= 9.5

The mean absolute deviation is:

(2.5 + 3.5+ 2.5 + 6.5 + 5.5 + .5 + 8.5 + 4.5)/8 = 4.25

The mean or average is 9.

The mean absolute deviation is 2.25.

The mean is ( 2 + 3 + 6 + 8 + 11 )/5 = 6.

The mean absolute deviation is (4 + 3 + 0 + 2 + 5 )/5 = 2.8.

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