This measure of dispersion provides a good idea of the dispersion of the data using the mean as a reference point.
The mean deviation, generally denoted |MD,| is the mean (average) of all of the data values’ deviations from the distribution’s mean.
This means that the larger the mean deviation, the further away the data are from the mean. On the other hand, the smaller the mean deviation, the more concentrated the data are around the mean.
The calculation of the mean deviation involves several steps summarized by the following formula.
||MD=\dfrac{\sum\vert x_ {i}-\overline{x}\vert}{n}||
where
|\overline{x} :| mean of the distribution
|n :| size of the sample, the distribution, or the population
|\sum| refers to adding several elements in succession.
|x_i| represents the |i^\text{th}| value of the distribution
Note:
We use the symbol |\mu| instead of |\overline{x}| for the mean of a population. Deviations must always be positive. This is why the absolute value is used.
Here are the steps to follow to use the above formula properly.
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Determine the size of the distribution.
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Calculate the mean of the distribution.
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Calculate the deviation from the mean of each data value.
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Calculate the mean deviation.
Here are temperatures in degrees Celsius, in ascending order, recorded every hour during a day.
|-5,| |-4,| |-4,| |-3,| |-3,| |-2,| |-1,| |0,| |0,| |1,| |2,| |3,| |3,| |4,| |4,| |6,| |7,| |8,| |9,| |10,| |10,| |11,| |11,| |12|
Determine the mean deviation of this data set.
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Determine the size of the data set
Since there are |24| hours in a day, this distribution contains |24| data values |(n=24).| -
Calculate the mean of the data set
The arithmetic mean of all data is calculated.
||\begin{align}\overline{x}&=\dfrac{\left(\begin{alignat}{40}&-5&&-4&&-4&&-\ 3&&-\ 3&&-\ 2&&-\ 1&&+\ 0\\&+0&&+1&&+2&&+\ 3&&+\ 3&&+\ 4&&+\ 4&&+\ 6\\&+7&&+8&&+9&&+10&&+10&&+11&&+11&&+12\ \ \end{alignat}\right)}{24}\\&\approx3.29\end{align}|| -
Calculate the deviation from the mean of each data value
The deviations from the mean of each data value are calculated.
| Data value |x_i| |
Deviation from the mean |\vert x_{i}-\overline{x}\vert| |
Data value |x_i| |
Deviation from the mean |\vert x_{i}-\overline{x}\vert| |
|---|---|---|---|
| |-5| | |\vert -5-3.29\vert=8.29| | |3| | |\vert 3-3.29\vert=0.29| |
| |-4| | |\vert -4-3.29\vert=7.29| | |4| | |\vert 4-3.29\vert=0.71| |
| |-4| | |\vert -4-3.29\vert=7.29| | |4| | |\vert 4-3.29\vert=0.71| |
| |-3| | |\vert -3-3.29\vert=6.29| | |6| | |\vert 6-3.29\vert=2.71| |
| |-3| | |\vert -3-3.29\vert=6.29| | |7| | |\vert 7-3.29\vert=3.71| |
| |-2| | |\vert -2-3.29\vert=5.29| | |8| | |\vert 8-3.29\vert=4.71| |
| |-1| | |\vert -1-3.29\vert=4.29| | |9| | |\vert 9-3.29\vert=5.71| |
| |0| | |\vert 0-3.29\vert=3.29| | |10| | |\vert 10-3.29\vert=6.71| |
| |0| | |\vert 0-3.29\vert=3.29| | |10| | |\vert 10-3.29\vert=6.71| |
| |1| | |\vert 1-3.29\vert=2.29| | |11| | |\vert 11-3.29\vert=7.71| |
| |2| | |\vert 2-3.29\vert=1.29| | |11| | |\vert 11-3.29\vert=7.71| |
| |3| | |\vert 3-3.29\vert=0.29| | |12| | |\vert 12-3.29\vert=8.71| |
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Calculate the mean deviation
To calculate the mean deviation, all that is left is to add up all the deviations and divide this sum by the total number of data. In short, the mean (average) of the deviations from the mean must be calculated.
||\begin{align}MD&= \dfrac{\sum\vert x_{i}-\overline{x}\vert}{n}\\&=\dfrac{\left(\begin{alignat}{40}&\ \ \ \ \ 8.29&&+ 7.29&&+ 7.29&&+ 6.29&&+ 6.29&&+ 5.29&&+ 4.29&&+ 3.29\\&+3.29&&+2.29&&+1.29&&+ 0.29&&+0.29&&+0.71&&+0.71&&+2.71\\&+3.71&&+4.71&&+5.71&&+6.71&&+6.71&&+7.71&&+7.71&&+8.71\ \ \end{alignat}\right)}{24}\\&\approx 4.65\end{align}||
Answer: The mean deviation of the distribution is approximately |4.65^\circ\text{C}.|
When there is more than one data set, their mean deviations can be compared to see which has more dispersed data, or more concentrated data, around the mean.
The scores, in percent, of |3| students on their last |5| science exams are compared.
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Victor: |95,| |83,| |71,| |91,| and |85|
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Elyanne: |75,| |65,| |67,| |84,| and |89|
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Maxime: |62,| |70,| |54,| |58,| and |66|
Which student is the most consistent in his or her scores? Which student is the least consistent?
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Determine the size of the distributions
All 3 distributions have |5| data values each. -
Calculate the mean of the distributions
Victor||\begin{align}\overline{x}_V&=\dfrac{95+83+71+91+85}{5}\\&=\dfrac{425}{5}\\&=85\ \%\end{align}||
Elyanne
||\begin{align}\overline{x}_E&=\dfrac{75+65+67+84+89}{5}\\&=\dfrac{380}{5}\\&=76\ \%\end{align}||
Maxime
||\begin{align}\overline{x}_M&=\dfrac{62+70+54+58+66}{5}\\&=\dfrac{310}{5}\\&=62\ \%\end{align}||
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Calculate the deviation from the mean of each data value
Victor
||\begin{align}\vert95-85\vert&=10\\\vert83-85\vert&=2\\\vert71-85\vert&=14\\\vert91-85\vert&=6\\\vert85-85\vert&=0\end{align}||
Elyanne
||\begin{align}\vert75-76\vert&=1\\\vert65-76\vert&=11\\\vert67-76\vert&=9\\\vert84-76\vert&=8\\\vert89-76\vert&=13\end{align}||
Maxime
||\begin{align}\vert62-62\vert&=0\\\vert70-62\vert&=8\\\vert54-62\vert&=8\\\vert58-62\vert&=4\\\vert66-62\vert&=4\end{align}||
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Calculate the mean deviation of each distribution
Victor
||\begin{align}MD_V&= \dfrac{\sum\vert x_{i}-\overline{x}\vert}{n}\\&=\dfrac{10+2+14+6+0}{5}\\&= 6.4\end{align}||
Elyanne
||\begin{align}MD_E&= \dfrac{\sum\vert x_{i}-\overline{x}\vert}{n}\\&=\dfrac{1+11+9+8+13}{5}\\&= 8.4\end{align}||
Maxime
||\begin{align}MD_M&= \dfrac{\sum\vert x_{i}-\overline{x}\vert}{n}\\&=\dfrac{0+8+8+4+4}{5}\\&= 4.8\end{align}||
Answer: Maxime is the most consistent student, since he has the smallest mean deviation |(4.8).| His marks are more concentrated around his average score of |62\ \%.| On the other hand, Elyanne's marks are the most dispersed, so she is the least consistent student.
When the data are represented differently, the approach to determine the mean deviation is similar, with the exception of the calculation of the mean. Depending on the situation, the mean of condensed data or the mean of data grouped into classes is calculated instead.