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AP

What is Mean Absolute Deviation (MAD)?

The phrase ‘mean absolute deviation’ can be simply understood after separating this phrase into ‘mean’ and ‘absolute deviation’, which manifest the mean of the given data and the absolute value of the deviation of the given data from the calculated mean value, respectively.

MAD is always measured by the measure of the central tendency, which is mean, median, or mode. Therefore, the mean absolute deviation is a way to describe the variation of the given data set, and therefore, MAD is always a measure of the spread of the data set from a given reference central value. Therefore, if the given data set has the larger values, then they will show a larger spread, however, the smaller values of the given data set will show a smaller spread about the central value.

In comparison to the mean absolute deviation, there are other measures also like range, quartile deviation, standard deviation, etc. Interestingly, the mean absolute deviation is a good measure because of its accuracy and simplicity, however, it cannot be used for further estimates because of its absolute measure of the spread.

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How to Calculate MAD?

Step-by-step Introductions for calculating MAD for ungrouped data

  • Step 1: Calculate the mean{\rm{(}}\overline {\rm{x}} {\rm{)}} of the given data
  • Step 2: Calculate the deviation \left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right)and determine the absolute value

        \[\left( {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} \right)\]

  • Step 3: Add all the deviations to obtain the sum \left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} } \right) in step 2
  • Step 4: Divide the obtained sum in step 3 by the number of observations in the given data \left( {\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\rm{n}}}} \right)

Step-by-step Introductions for calculating MAD for grouped data

  • Step 1: Calculate the mean{\rm{(}}\overline {\rm{x}} {\rm{)}} of the given data 
    • First calculate midpoint of the class, it will be {\rm{(}}{{\rm{x}}_{\rm{i}}}{\rm{)}}
    • Multiply the obtained mid-point {\rm{(}}{{\rm{x}}_{\rm{i}}}{\rm{)}} with the frequency of the corresponding class {\rm{(}}{{\rm{f}}_{\rm{i}}}{\rm{)}}, it will give {\rm{(}}{{\rm{f}}_{\rm{i}}}{{\rm{x}}_{\rm{i}}}{\rm{)}}
    • Mean,

          \[\overline {\rm{x}} {\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}{{\rm{x}}_{\rm{i}}}} }}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}} }}\]

  • Step 2: Calculate the absolute deviation

        \[\left( {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} \right)\]

    and multiply the obtained absolute deviation with the corresponding frequency of the class {\rm{(}}{{\rm{f}}_{\rm{i}}}{\rm{)}}, it will give,

        \[\left( {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} \right)\]

  • Step 3: Add all the deviations to obtain the sum \left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} } \right) in step 2
  • Step 4: Divide the obtained sum in step 3 by the total number of observations in the given data \left( {\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}} }}} \right)
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Formula to calculate MAD

For ungrouped data

{\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\rm{n}}} 

For grouped data

{\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}} }}

Examples of MAD in action

Calculating MAD for a small set of data

Given data: 5, 7, 9, 11, 4, 6

Number of Observations, n = 6

  • Step 1: Calculate the mean of the given data

Mean,

    \[\overline {\rm{x}} {\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{42}}}}{{\rm{6}}}{\rm{ = 7}}\]

 

  • Step 2: Calculate the deviation and absolute deviation of the given data 

\begin{array}{*{20}{c}}{{\rm{Data}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&{{\rm{Deviation}}\;\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right)}&{{\rm{Absolute}}\;{\rm{Deviation}}\,\left( {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} \right)}\\{\rm{5}}&{{\rm{ - 2}}}&{\rm{2}}\\{\rm{7}}&{\rm{0}}&{\rm{0}}\\{\rm{9}}&{\rm{2}}&{\rm{2}}\\{{\rm{11}}}&{\rm{4}}&{\rm{4}}\\{\rm{4}}&{{\rm{ - 3}}}&{\rm{3}}\\{\rm{6}}&{{\rm{ - 1}}}&{\rm{1}}\end{array} 

  • Step 3: Determine the sum of the obtained absolute deviations

\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} {\rm{ = 12}}

  • Step 4: Calculate the mean absolute deviation

{\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{12}}}}{{\rm{6}}}{\rm{ = 2}}

Using MAD to compare data sets

There are two data sets A and B, which present the marks of the two classes of students for a test where the maximum score was 12. Calculate the MAD for these data sets and compare results.

Q. What are the marks that are only obtained for class A and not for class B?

Marks that are obtained for class A and not for class B are 3 and 8.

Q. What are the marks that are only obtained for class B and not for class A?

Marks that are obtained for class B and not for class A are 5 and 11.

Q. What are the mean values of the marks for class A and class B, calculate and mark them on the scale.

The mean values of the marks for class A and class B are 8, and they is marked on the scale with an inverted solid triangle.

Q. What are the marks that are only obtained for class A and not for class B?

Marks that are obtained for class A and not for class B are 3 and 8.

Calculation of MAD for Class A

  • Step 1: Calculate the mean of the given data

    •     \[\overline {\rm{x}} {\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{96}}}}{{12}}{\rm{ = 8}}\]

  • Step 2: Calculate the deviation and absolute deviation of the given data

    •     \[\begin{array}{*{20}{c}}{{\rm{Marks}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&3&4&7&7&8&8&9&{10}&{10}&{10}&{\begin{array}{*{20}{c}}{10}&{10}\end{array}}\\{{\rm{Deviation}}\;\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right)}&{ - 5}&{ - 4}&{ - 1}&{ - 1}&0&0&1&2&2&2&{\begin{array}{*{20}{c}}2&2\end{array}}\\{{\rm{Absolute}}\;{\rm{Deviation}}\;\left( {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} \right)\;}&5&4&1&1&0&0&1&2&2&2&{\begin{array}{*{20}{c}}2&2\end{array}}\end{array}\]

  • Step 3: Determine the sum of the obtained absolute deviations
    • \sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} {\rm{ = 22}}
  • Step 4: Calculate the mean absolute deviation
    • {\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{22}}}}{{12}}{\rm{ = 1}}{\rm{.84}}

Thus, the MAD of the marks of the student of class A is 1.84.

Calculation of MAD for Class B: 

  • Step 1: Calculate the mean of the given data

    \[\overline {\rm{x}} {\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{96}}}}{{12}}{\rm{ = 8}}\]

  • Step 2: Calculate the deviation and absolute deviation of the given data

    \[\begin{array}{*{20}{c}}{{\rm{Marks}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&4&4&5&7&7&9&9&9&{10}&{10}&{\begin{array}{*{20}{c}}{11}&{11}\end{array}}\\{{\rm{Deviation}}\;\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right)}&{ - 4}&{ - 4}&{ - 3}&{ - 1}&{ - 1}&1&1&1&2&2&{\begin{array}{*{20}{c}}3&3\end{array}}\\{{\rm{Absolute}}\;{\rm{Deviation}}\;\left( {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} \right)\;}&4&4&3&1&1&1&1&1&2&2&{\begin{array}{*{20}{c}}3&3\end{array}}\end{array}\]

  • Step 3: Determine the sum of the obtained absolute deviations

\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} {\rm{ = 26}}

  • Step 4: Calculate the mean absolute deviation

{\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{26}}}}{{12}}{\rm{ = 2}}{\rm{.17}}

Thus, the MAD of the marks of the students of class B is 2.17.

Thus, from the above calculation of the MAD of the marks obtained for class A and class B, we can say that the MAD of class B is higher as compared to class A, and therefore, the marks obtained for the students of class B are more spread or deviated from the mean value as compared to the class A.

Advantages and disadvantages of MAD

Advantages

  • MAD is easy to calculate and very simple to understand.
  • MAD is calculated for each observation of the given data.
  • MAD does not depend upon the extreme values it is equally dependent on all the given data.
  • MAD manifests the deviation of the given data from a central value.
  • MAD is a good measure of the dispersion in the given data as compared to the other dispersion measures like range and quartile deviation.

Disadvantages

  • MAD is defined from the given central value and therefore it is not rigidly defined.
  • MAD can be very much affected by the scattered sample of the given data.
  • Calculation of the MAD for an open-end frequency distribution is not possible.
  • MAD is not much used for further algebraic and statistical calculations because it is an absolute measure of the deviation of the data, so the real information of the given data gets diluted.

Uses and properties

  • MAD is very useful in the field of commerce, trade, and business.
  • MAD is also widely used by economists owing to its simple and accurate output.
  • MAD is also very useful in the socio-economic scenario and calculation of the wealth in a society.
  • MAD calculated against the median has the smallest value as compared to the mean and mode.
  • MAD is not affected by the addition and subtraction of a constant value to the given data; however, it gets multiplied or divided by the same value as the value multiplied or divided by the given data, respectively.

Conclusion

Thus, mean average deviation is a simple, and accurate measure of the dispersion of the data as compared to the other dispersion of the data like quartile deviation, range, and standard deviation.

The mean absolute deviation manifests the spread of the data about the central value of the given data and gives an average estimate of the spread from a reference value. For the larger values in the data, the mean absolute deviation is large however it is smaller for the smaller values in the given data.

Solved Examples

Example 1. Find the mean absolute deviation of the following data set:

10, 20, 30, 40, 50, 60, 70, 80

Solution:

  • Step 1: Calculate the mean of the given data

    \[\overline {\rm{x}} {\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{360}}}}{{\rm{8}}}{\rm{ = 45}}\]

  • Step 2: Calculate the deviation and absolute deviation of the given data

    \[\begin{array}{*{20}{c}}{{\rm{Data}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&{{\rm{Deviation}}\;\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right)}&{{\rm{Absolute}}\;{\rm{Deviation}}\,\left( {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} \right)}\\{{\rm{10}}}&{{\rm{ - 35}}}&{{\rm{35}}}\\{{\rm{20}}}&{{\rm{ - 25}}}&{{\rm{25}}}\\{{\rm{30}}}&{{\rm{ - 15}}}&{{\rm{15}}}\\{{\rm{40}}}&{{\rm{ - 5}}}&{\rm{5}}\\{{\rm{50}}}&{\rm{5}}&{\rm{5}}\\\begin{array}{l}{\rm{60}}\\{\rm{70}}\\{\rm{80}}\end{array}&\begin{array}{l}{\rm{15}}\\{\rm{25}}\\{\rm{35}}\end{array}&\begin{array}{l}{\rm{15}}\\{\rm{25}}\\{\rm{35}}\end{array}\end{array}\]

  • Step 3: Determine the sum of the obtained absolute deviations

\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} {\rm{ = 160}}

  • Step 4: Calculate the mean absolute deviation

{\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{160}}}}{{\rm{8}}}{\rm{ = 20}}

Thus, the MAD of the given data is 20.

Example 2. Find the mean absolute deviation of the following data set:

5, 5, 7, 7, 9, 6, 8, 1

Solution:

  • Step 1: Calculate the mean of the given data

    \[\overline {\rm{x}} {\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{48}}}}{{\rm{8}}}{\rm{ = 6}}\]

  • Step 2: Calculate the deviation and absolute deviation of the given data

    \[\begin{array}{*{20}{c}}{{\rm{Data}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&{{\rm{Deviation}}\;\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right)}&{{\rm{Absolute}}\;{\rm{Deviation}}\,\left( {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} \right)}\\{\rm{5}}&{{\rm{ - 1}}}&{\rm{1}}\\{\rm{5}}&{{\rm{ - 1}}}&{\rm{1}}\\{\rm{7}}&{\,{\rm{1}}}&{\rm{1}}\\{\rm{7}}&{\,{\rm{1}}}&{\rm{1}}\\{\rm{9}}&{\,{\rm{3}}}&{\rm{3}}\\\begin{array}{l}{\rm{6}}\\{\rm{8}}\\{\rm{1}}\end{array}&\begin{array}{l}\,\,\,{\rm{0}}\\\,\,\,{\rm{2}}\\{\rm{ - 5}}\end{array}&\begin{array}{l}{\rm{0}}\\{\rm{2}}\\{\rm{5}}\end{array}\end{array}\]

  • Step 3: Determine the sum of the obtained absolute deviations

\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} {\rm{ = 14}}

  • Step 4: Calculate the mean absolute deviation

{\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{14}}}}{{\rm{8}}}{\rm{ = 1}}{\rm{.75}}

Thus, the MAD of the given data set is 1.75

Example 3. Find the mean absolute deviation of the given marks of a student.

    \[\begin{array}{*{20}{c}}{{\rm{Subject}}}&{{\rm{Marks}}}\\{{\rm{Hindi}}}&{{\rm{40}}}\\{{\rm{English}}}&{{\rm{37}}}\\{{\rm{Maths}}}&{{\rm{45}}}\\{{\rm{Statistics}}}&{{\rm{48}}}\\{{\rm{Science}}}&{{\rm{46}}}\\{{\rm{Arts}}}&{{\rm{30}}}\end{array}\]

Solution:

  • Step 1: Calculate the mean of the marks of the student

    \[\overline {\rm{x}} {\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} }}{{\rm{n}}}{\rm{ = }}\frac{{246}}{6}{\rm{ = 41}}\]

  • Step 2: Calculate the deviation and absolute deviation of the marks of the student

    \[\begin{array}{*{20}{c}}{{\rm{Subject}}}&{{\rm{Marks}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&{{\rm{Deviation}}\;\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right)}&{{\rm{Absolute}}\;{\rm{Deviation}}\;\left( {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} \right)}\\{{\rm{Hindi}}}&{{\rm{40}}}&{{\rm{ - 1}}}&{\rm{1}}\\{{\rm{English}}}&{{\rm{37}}}&{{\rm{ - 4}}}&{\rm{4}}\\{{\rm{Maths}}}&{{\rm{45}}}&{\rm{4}}&{\rm{4}}\\{{\rm{Statistics}}}&{{\rm{48}}}&{\rm{7}}&{\rm{7}}\\{{\rm{Science}}}&{{\rm{46}}}&{\rm{5}}&{\rm{5}}\\{{\rm{Arts}}}&{{\rm{30}}}&{{\rm{ - 11}}}&{{\rm{11}}}\end{array}\]

  • Step 3: Determine the sum of the obtained absolute deviations

\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} {\rm{ = 32}}

  • Step 4: Calculate the mean absolute deviation

{\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{32}}}}{{\rm{6}}}{\rm{ = 5}}{\rm{.33}}

Thus, the MAD of the given data set is 5.33

Example 4. Calculate the mean absolute deviation of the following data showing the age group of employees of a given company.

    \[\begin{array}{*{20}{c}}{{\rm{Age Group}}\;\left( {{\rm{in}}\;{\rm{years}}} \right)}&{{\rm{Number of Employees}}}\\{20 - 25}&{{\rm{10}}}\\{25 - 30}&{12}\\{35 - 40}&{15}\\{40 - 45}&{\rm{8}}\\{45 - 50}&{{\rm{16}}}\\{50 - 55}&6\end{array}\]

Solution:

  • Step 1: Calculate the mean age of the employees 

    \[\begin{array}{*{20}{c}}{{\rm{Age Group}}}&{{\rm{Employees}}\;\left( {{{\rm{f}}_{\rm{i}}}} \right)}\\{20 - 25}&{{\rm{10}}}\\{25 - 30}&{12}\\{35 - 40}&{15}\\{40 - 45}&{\rm{8}}\\{45 - 50}&{{\rm{16}}}\\{50 - 55}&6\end{array}\begin{array}{*{20}{c}}{\;{\rm{Mid - point}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&{{{\rm{f}}_{\rm{i}}}{{\rm{x}}_{\rm{i}}}}\\{22.5}&{225}\\{27.5}&{330}\\{32.5}&{487.5}\\{37.5}&{300}\\{42.5}&{680}\\{47.5}&{285}\end{array}\]

    \[\overline {\rm{x}} {\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}{{\rm{x}}_{\rm{i}}}} }}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}} }}{\rm{ = }}\frac{{2307.5}}{{67}}{\rm{ = 34}}{\rm{.5}}\]

  • Step 2: Calculate the deviation and absolute deviation of the marks of the student

    \[\begin{array}{*{20}{c}}{\;{\rm{Mid - point}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&{{\rm{Deviation}}\;\left( {{{\rm{x}}_{\rm{i}}} - \overline {\rm{x}} } \right)}\\{22.5}&{ - 12}\\{27.5}&{ - 7}\\{32.5}&{ - 2}\\{37.5}&3\\{42.5}&8\\{47.5}&{13}\end{array}\begin{array}{*{20}{c}}{\;{\rm{Absolute Deviation}}\;\left( {\left| {{{\rm{x}}_{\rm{i}}} - \overline {\rm{x}} } \right|} \right)}&{{\rm{Employees}}\;\left( {{{\rm{f}}_{\rm{i}}}} \right)}&{\left( {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}} - \overline {\rm{x}} } \right|} \right)}\\{12}&{10}&{120}\\7&{12}&{84}\\2&{15}&{30}\\3&8&{24}\\8&{16}&{128}\\{13}&6&{78}\end{array}\]

  • Step 3: Determine the sum of the obtained absolute deviations

\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} {\rm{ = 464}}

  • Step 4: Calculate the mean absolute deviation

{\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}} }}{\rm{ = }}\frac{{464}}{{{\rm{67}}}}{\rm{ = 6}}{\rm{.93}}

Thus, the MAD of the given data set is 6.93

Example 5. Calculate the mean absolute deviation of the following data showing the average mealtime of the employees of a given company.

    \[\begin{array}{*{20}{c}}{{\rm{Meal Time }}\left( {{\rm{Minutes}}} \right)}&{{\rm{Number of Employees}}}\\{0 - 5}&{12}\\{5 - 10}&{16}\\{10 - 15}&{10}\\{15 - 20}&6\\{20 - 25}&{{\rm{12}}}\\{25 - 30}&{16}\end{array}\]

Solution:

  • Step 1: Calculate the mean of the mealtime for the employees

    \[\begin{array}{*{20}{c}}{{\rm{Meal Time}}}&{{\rm{Employees}}\;\left( {{{\rm{f}}_{\rm{i}}}} \right)}\\{0 - 5}&{{\rm{12}}}\\{5 - 10}&{16}\\{10 - 15}&{10}\\{15 - 20}&6\\{20 - 25}&{{\rm{12}}}\\{25 - 30}&{16}\end{array}\begin{array}{*{20}{c}}{\;{\rm{Mid - point}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&{{{\rm{f}}_{\rm{i}}}{{\rm{x}}_{\rm{i}}}}\\{2.5}&{30}\\{7.5}&{120}\\{12.5}&{125}\\{17.5}&{105}\\{22.5}&{270}\\{27.5}&{440}\end{array}\]

    \[\overline {\rm{x}} {\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}{{\rm{x}}_{\rm{i}}}} }}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}} }}{\rm{ = }}\frac{{1090}}{{72}}{\rm{ = 15}}{\rm{.2}}\]

  • Step 2: Calculate the deviation and absolute deviation of the marks of the student

    \[\begin{array}{*{20}{c}}{\;{\rm{Mid - point}}\;\left( {{{\rm{x}}_{\rm{i}}}} \right)}&{{\rm{Deviation}}\;\left( {{{\rm{x}}_{\rm{i}}} - \overline {\rm{x}} } \right)}\\{2.5}&{ - 12.7}\\{7.5}&{ - 7.7}\\{12.5}&{ - 2.7}\\{17.5}&{2.3}\\{22.5}&{7.3}\\{27.5}&{12.3}\end{array}\begin{array}{*{20}{c}}{\;{\rm{Absolute Deviation}}\;\left( {\left| {{{\rm{x}}_{\rm{i}}} - \overline {\rm{x}} } \right|} \right)}&{{\rm{Employees}}\;\left( {{{\rm{f}}_{\rm{i}}}} \right)}&{\left( {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}} - \overline {\rm{x}} } \right|} \right)}\\{12.7}&{12}&{152.4}\\{7.7}&{16}&{123.2}\\{2.7}&{10}&{27}\\{2.3}&6&{13.8}\\{7.3}&{12}&{87.6}\\{12.3}&{16}&{196.8}\end{array}\]

  • Step 3: Determine the sum of the obtained absolute deviations

\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} {\rm{ = 600}}{\rm{.8}}

  • Step 4: Calculate the mean absolute deviation

{\rm{MAD = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}\left| {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\overline {\rm{x}} } \right|} }}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{f}}_{\rm{i}}}} }}{\rm{ = }}\frac{{600.8}}{{72}}{\rm{ = 8}}{\rm{.35}}

Thus, the MAD of the given data set is 8.35.

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Frequently Asked Questions (FAQs)

What is the purpose of calculating the mean absolute deviation?

For a given data set mean absolute deviation is calculated from the central value of the data to obtain the average distribution of the given data set. Therefore, MAD is useful to obtain the estimate of the variation of the given data.

What are the possible values of the mean absolute deviation?

The mean absolute deviation of any given data can have positive or zero values, it can not be negative since we are calculating the absolute measures of the deviation. Moreover, the obtained value of MAD can be an integer or in fraction. 

Which central value is considered the best measure to calculate the mean absolute deviation?

For a given data set, one can define the three central values, namely, mean, median, and mode. The calculation of MAD always considers a central value as a reference point. Since, the mean of a given data is based on all values in the data as compared to the median and mode, which are based on the limited values from the given data, therefore, the mean provides the best estimate of the MAD.

What are the other measures of dispersion and compare them with MAD?

The other measures of the dispersion of a given data set are range (calculated from the difference between the higher and lower limit of the given data), quartile deviation (calculated after dividing the data set into four equal halves), standard deviation (squared average of the deviation of the given data from the central value). However, MAD is considered best among all because it’s based on all values of the given data and it’s easy to calculate and accurate.

What are the different uses of MAD in practical life?

MAD can be used in different fields where one needs to have an estimate of the variation of the given data set; therefore, meteorologists use it to estimate the errors in the forecasting data, and teachers or schools use it to calculate the score of the class in the test, economists use it to observe the variation of the prices with respect to the time, etc.

References

  1. Mean Absolute Deviation. (2021, February 21). GeeksforGeeks. Retrieved January 12, 2023, from https://www.geeksforgeeks.org/mean-absolute-deviation/
  2. Mean Absolute Deviation (MAD): What It Means and How To Find It | Outlier. (n.d.). Outlier Articles. Retrieved January 12, 2023, from https://articles.outlier.org/mean-absolute-deviation-meaning
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