# Mean Absolute Deviation: Formula and Examples

Joao Amadeu, Lynne Hampson, Elaine Chan
• Author

Joao Amadeu has more than 10 years of experience in teaching physics and mathematics at different educational levels. Joao earned two degrees at Londrina State University: B.S. in Physics and M.S. in Science and Mathematics Education. He is currently working on his PhD in Science Education at Western Michigan University.

• 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.

Learn about what mean absolute deviation is. Explore the mean absolute deviation formula and how to find mean absolute deviation, and see examples of it. Updated: 09/09/2021

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## What is Mean Absolute Deviation?

The Mean Absolute Deviation (MAD) is a descriptive statistic. Like the Average, it is used to summarize a given data set through one value. The method of calculating the average of two or more numbers is known as Mean. However, average and mean are commonly used interchangeably, as applied in this lesson. The Mean Absolute Deviation takes the distance between each item of the data set and its average, creates a new data set with these distance values, and indicates the average of these distances. In other terms, the mean absolute deviation summarizes how each number from a data set spreads out from the average.

## Examples and Applications

Meteorologists use Mean Absolute Deviation as part of the process to find errors of measurement in forecasting. Biologists use it as a mean to analyze a data set composed by animal weights and find out information healthy animals weights. Teachers may use MAD to investigate how students' test scores are distributed around the class average. For example, only the average may not suffice to gather information about students' test scores. A class average score can be 8/10 with each student scoring 8/10, or with some students scoring 0/10 and other ones scoring 10/10. To exemplify it, the lists below present tests scores (out of 10) from 2 hypothetical class groups with 5 students each.

Class A: 8, 8, 8, 7, 9

Class B: 2, 8, 10, 10, 10.

Using the mean calculation,

{eq}Mean (class A) = (8+8+8+7+9) / 5 {/eq}

{eq}Mean (class B) = (2+8+10+10+10) / 5 {/eq},

we obtain:

Average score of class A is 8.

Average score of class B is 8.

For each score set, the absolute deviation can be calculated by finding out the distance between each value of the set and the set's average. For example, The first value of the score set (class B) is 2. The distance between 2 and 8 (average score of class B) is 6. Calculating all the absolute distances, we have created a new data set for each group: the absolute deviation.

Absolute deviation (class A): 0, 0, 0, 1, 1.

Absolute deviation (class B): 6, 0, 2, 2, 2.

Calculating the mean for each absolute deviation set, we obtain the mean absolute deviation for each class:

{eq}Mean Absolute Deviation (Class A) = (0+0+0+1+1) / 5 = 0.4 {/eq}

{eq}Mean Absolute Deviation (Class B) = (6+0+2+2+2) / 5 = 2.4 {/eq}

A higher Mean Absolute Deviation value indicates that the numbers in a set have a higher deviation from the average.

Even without the MAD, a teacher could easily identify deviations from the mean by looking at students' scores one-by-one in a small class context. However, when dealing with larger classes, further statistical analysis become essential. The two histograms below illustrate test's scores from two other hypothetical classes (C and D) with 30 students each.

The difference in distribution around the mean is clear in the images above. Both score sets have the same average of 8. However, the score values in class C have a mean absolute deviation of 0.53, while the MAD for the values in class D is 2.47. The MAD value can also be important in technology and communication. That can be illustrate with an example that, at first, may be seen as superficial.

A weight loss smartphone app updates weekly the daily average mileage for joggers. This number can help the app users to know if their are on track to achieve their exercising objectives. If the daily average mileage between two joggers is the same, it indicates that they are maintaining a similar amount of exercising in the course of 7 days. However, the average hides information about their regularity when distributing their jogging for each day of the week. The Mean Absolute Deviation could play an important role for the users to assess their exercising regularity.

## Finding Mean Absolute Deviation: Formula and Process

The previous example introduced some steps of the Mean Absolute Deviation calculation in a specific context. This section will present detailed steps of how to calculate the MAD of a set of numbers.

Given the set defined by {eq}X = X1, X2, ..., XN {/eq} ,

and its average calculated by the mean:

{eq}M(X) = (X1+X2+...+XN) / N {/eq} ,

The Mean Absolute Deviation of the set is calculated by:

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• FAQs

## 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.

#### What is the difference between standard deviation and mean absolute deviation?

Both standard deviation and mean absolute deviation (MAD) are calculations of average between values that represent deviation. While MAD deals with deviation by finding its absolute values, standard deviation relies on the squared deviations from the mean.

#### What is the mean absolute deviation of the data set?

The mean absolute deviation (mad) of a data set indicates how each value of the set is distributed in relation to the set's average. The value of mad is found by, first, finding the average of the set, calculating the absolute distance between each value of the set and its average, and finding the average between these values for absolute distance.

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