15.4 Mean Deviation
In this section, we will learn how to calculate mean deviation about mean and median for various types of data.
Recall that the deviation of an observation \(x\) from a fixed value ‘ \(a\) ‘ is the difference \(x-a\). In order to find the dispersion of values of \(x\) from a central value ‘ \(a\) ‘, we find the deviations about \(a\). An absolute measure of dispersion is the mean of these deviations. To find the mean, we must obtain the sum of the deviations. But, we know that a measure of central tendency lies between the maximum and the minimum values of the set of observations. Therefore, some of the deviations will be negative and some positive. Thus, the sum of deviations may vanish. Moreover, the sum of the deviations from mean \((\bar{x})\) is zero.
Also \(\text { Mean of deviations }=\frac{\text { Sum of deviations }}{\text { Number of observations }}=\frac{0}{n}=0\)
Thus, finding the mean of deviations about mean is not of any use for us, as far as the measure of dispersion is concerned.
Remember that, in finding a suitable measure of dispersion, we require the distance of each value from a central tendency or a fixed number ‘ \(a\) ‘. Recall, that the absolute value of the difference of two numbers gives the distance between the numbers when represented on a number line. Thus, to find the measure of dispersion from a fixed number ‘ \(a\) ‘ we may take the mean of the absolute values of the deviations from the central value. This mean is called the ‘mean deviation’. Thus mean deviation about a central value ‘ \(a\) ‘ is the mean of the absolute values of the deviations of the observations from ‘ \(a\) ‘. The mean deviation from ‘ \(a\) ‘ is denoted as M.D. (a). Therefore,
\(
\text { M.D. }(a)=\frac{\text { Sum of absolute values of deviations from ‘ } a \text { ‘ }}{\text { Number of observations }} \text {. }
\)
Mean deviation for ungrouped data
Let \(n\) observations be \(x_1, x_2, x_3, \ldots, x_n\). The following steps are involved in the calculation of mean deviation about mean or median:
Step 1: Calculate the measure of central tendency about which we are to find the mean deviation. Let it be ‘ \(a\) ‘.
Step 2: Find the deviation of each \(x_i\) from \(a\), i.e., \(x_1-a, x_2-a, x_3-a, \ldots, x_n-a\)
Step 3: Find the absolute values of the deviations, i.e., drop the minus sign (-), if it is there, i.e., \(\left|x_1-a\right|,\left|x_2-a\right|,\left|x_3-a\right|, \ldots .,\left|x_n-a\right|\)
Step 4: Find the mean of the absolute values of the deviations. This mean is the mean deviation about \(a\), i.e.,
\(
\text { M.D. }(a)=\frac{\sum_{i=1}^n\left|x_i-a\right|}{n}==\frac{1}{n} \sum_{i=1}^n\left|x_i-a\right|=\frac{1}{n} \Sigma\left|d_i\right| \text {, where } d_i=x_i-a
\)
\(
\text { Thus } \quad \text { M.D. }(\bar{x})=\frac{1}{n} \sum_{i=1}^n\left|x_i-\bar{x}\right| \text {, where } \bar{x}=\text { Mean }
\)
\(
\text { and } \quad \text { M.D. }( M )=\frac{1}{n} \sum_{i=1}^n\left|x_i- M \right| \text {, where } M =\text { Median }
\)
Example 1: Find the mean deviation about the mean for the following data:
\(6,7,10,12,13,4,8,12\)
Solution: We proceed step-wise and get the following:
Step 1: Mean of the given data is
\(
\bar{x}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9
\)
Step 2: The deviations of the respective observations from the mean \(\bar{x}\), i.e., \(x_i-\bar{x}\) are
\(
\begin{aligned}
& 6-9,7-9,10-9,12-9,13-9,4-9,8-9,12-9 \text {, } \\
& \text { or } 3,-2,1,3,4,-5,-1,3
\end{aligned}
\)
Step 3: The absolute values of the deviations, i.e., \(\left|x_i-\bar{x}\right|\) are
\(
3,2,1,3,4,5,1,3
\)
Step 4: The required mean deviation about the mean is
M.D. \((\bar{x})=\frac{\sum_{i=1}^8\left|x_i-\bar{x}\right|}{8}\)
\(
=\frac{3+2+1+3+4+5+1+3}{8}=\frac{22}{8}=2.75
\)
Example 2: Find the mean deviation about the mean for the following data :
\(12,3,18,17,4,9,17,19,20,15,8,17,2,3,16,11,3,1,0,5\)
Solution: We have to first find the mean \((\bar{x})\) of the given data
\(
\bar{x}=\frac{1}{20} \sum_{i=1}^{20} x_i=\frac{200}{20}=10
\)
The respective absolute values of the deviations from mean, i.e., \(\left|x_i-\bar{x}\right|\) are
\(
\begin{gathered}
2,7,8,7,6,1,7,9,10,5,2,7,8,7,6,1,7,9,10,5 \\
\text { Therefore } \sum_{i=1}^{20}\left|x_i-\bar{x}\right|=124
\end{gathered}
\)
and M.D. \((\bar{x})=\frac{124}{20}=6.2\)
Example 3: Find the mean deviation about the median for the following data:
\(3,9,5,3,12,10,18,4,7,19,21 \text {. }\)
Solution: Here the number of observations is 11 which is odd. Arranging the data into ascending order, we have \(3,3,4,5,7,9,10,12,18,19,21\)
Now \(\text { Median }=\left(\frac{11+1}{2}\right)^{\text {th }} \text { or } 6^{\text {th }} \text { observation }=9\)
The absolute values of the respective deviations from the median, i.e., \(\left|x_i- M \right|\) are
\(
\begin{gathered}
6,6,5,4,2,0,1,3,9,10,12 \\
\text { Therefore } \sum_{i=1}^{11}\left|x_i- M \right|=58
\end{gathered}
\)
and \(\text { M.D. }( M )=\frac{1}{11} \sum_{i=1}^{11}\left|x_i- M \right|=\frac{1}{11} \times 58=5.27\)
Mean deviation for grouped data We know that data can be grouped into two ways :
- Case-I: Discrete frequency distribution,
- Case-II: Continuous frequency distribution.
Let us discuss the method of finding mean deviation for both types of the data.
(a) Discrete frequency distribution
Let the given data consist of \(n\) distinct values \(x_1, x_2, \ldots, x_n\) occurring with frequencies \(f_1, f_2, \ldots, f_n\) respectively. This data can be represented in the tabular form as given below, and is called discrete frequency distribution:
\(
\begin{array}{lll}
x: x_1 & x_2 & x_3 \ldots x_n \\
f: f_1 & f_2 & f_3 \ldots f_n
\end{array}
\)
(i) Mean deviation about mean
First of all we find the mean \(\bar{x}\) of the given data by using the formula
\(
\bar{x}=\frac{\sum_{i=1}^n x_i f_i}{\sum_{i=1}^n f_i}=\frac{1}{ N } \sum_{i=1}^n x_i f_i
\)
where \(\sum_{i=1}^n x_i f_i\) denotes the sum of the products of observations \(x_1\) with their respective frequencies \(f_{ i }\) and \(N =\sum_{i=1}^n f_i\) is the sum of the frequencies.
Then, we find the deviations of observations \(x_i\) from the mean \(\bar{x}\) and take their absolute values, i.e., \(\left|x_i-\bar{x}\right|\) for all \(i=1,2, \ldots, n\).
After this, find the mean of the absolute values of the deviations, which is the required mean deviation about the mean. Thus
\(
\text { M.D. }(\bar{x})=\frac{\sum_{i=1}^n f_i\left|x_i-\bar{x}\right|}{\sum_{i=1}^n f_i}=\frac{1}{ N } \sum_{i=1}^n f_i\left|x_i-\bar{x}\right|
\)
(ii) Mean deviation about median
To find mean deviation about median, we find the median of the given discrete frequency distribution. For this the observations are arranged in ascending order. After this the cumulative frequencies are obtained. Then, we identify
the observation whose cumulative frequency is equal to or just greater than \(\frac{ N }{2}\), where \(N\) is the sum of frequencies. This value of the observation lies in the middle of the data, therefore, it is the required median. After finding median, we obtain the mean of the absolute values of the deviations from median.Thus,
\(
\text { M.D.(M) }=\frac{1}{ N } \sum_{i=1}^n f_i\left|x_i- M \right|
\)
Example 4: Find mean deviation about the mean for the following data :
\(
\begin{array}{rrrrrrr}
x_i & 2 & 5 & 6 & 8 & 10 & 12 \\
f_i & 2 & 8 & 10 & 7 & 8 & 5
\end{array}
\)
Solution: Let us make a Table below of the given data and append other columns after calculations.
\(
\begin{array}{|c|r|r|c|c|}
\hline x_i & f_i & f_i x_i & \left|x_i-\bar{x}\right| & f_i\left|x_i-\bar{x}\right| \\
\hline 2 & 2 & 4 & 5.5 & 11 \\
5 & 8 & 40 & 2.5 & 20 \\
6 & 10 & 60 & 1.5 & 15 \\
8 & 7 & 56 & 0.5 & 3.5 \\
10 & 8 & 80 & 2.5 & 20 \\
12 & 5 & 60 & 4.5 & 22.5 \\
\hline & 40 & 300 & & 92 \\
\hline
\end{array}
\)
\(
N =\sum_{i=1}^6 f_i=40, \quad \sum_{i=1}^6 f_i x_i=300, \sum_{i=1}^6 f_i\left|x_i-\bar{x}\right|=92
\)
\(
\text { Therefore } \quad \bar{x}=\frac{1}{ N } \sum_{i=1}^6 f_i x_i=\frac{1}{40} \times 300=7.5
\)
and M. D. \((\bar{x})=\frac{1}{ N } \sum_{i=1}^6 f_i\left|x_i-\bar{x}\right|=\frac{1}{40} \times 92=2.3\)
Example 5: Find the mean deviation about the median for the following data:
\(
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline x_i & 3 & 6 & 9 & 12 & 13 & 15 & 21 & 22 \\
\hline f_i & 3 & 4 & 5 & 2 & 4 & 5 & 4 & 3 \\
\hline
\end{array}
\)
Solution: The given observations are already in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get (Table below).
\(
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline x_i & 3 & 6 & 9 & 12 & 13 & 15 & 21 & 22 \\
\hline f_i & 3 & 4 & 5 & 2 & 4 & 5 & 4 & 3 \\
\hline \text { c.f. } & 3 & 7 & 12 & 14 & 18 & 23 & 27 & 30 \\
\hline
\end{array}
\)
Now, \(N =30\) which is even.
Median is the mean of the \(15^{\text {th }}\) and \(16^{\text {th }}\) observations. Both of these observations lie in the cumulative frequency 18 , for which the corresponding observation is 13 .
Therefore, Median \(M =\frac{15^{\text {th }} \text { observation }+16^{\text {th }} \text { observation }}{2}=\frac{13+13}{2}=13\)
Now, absolute values of the deviations from median, i.e., \(\left|x_i- M \right|\) are shown in Table below.
\(
\begin{array}{|c|r|r|r|r|r|r|r|r|}
\hline\left|x_i- M \right| & 10 & 7 & 4 & 1 & 0 & 2 & 8 & 9 \\
\hline f_{ i } & 3 & 4 & 5 & 2 & 4 & 5 & 4 & 3 \\
\hline f_{ i }\left|x_i- M \right| & 30 & 28 & 20 & 2 & 0 & 10 & 32 & 27 \\
\hline
\end{array}
\)
We have \(\quad \sum_{i=1}^8 f_i=30\) and \(\sum_{i=1}^8 f_i\left|x_i- M \right|=149\)
Therefore \(\text { M. D. }( M )=\frac{1}{ N } \sum_{i=1}^8 f_i\left|x_i- M \right|\)
\(
=\frac{1}{30} \times 149=4.97
\)
(b) Continuous frequency distribution
A continuous frequency distribution is a series in which the data are classified into different class-intervals without gaps alongwith their respective frequencies.
For example, marks obtained by 100 students are presented in a continuous frequency distribution as follows :
\(
\begin{array}{|l|c|c|c|c|c|c|}
\hline \text { Marks obtained } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\
\hline \text { Number of Students } & 12 & 18 & 27 & 20 & 17 & 6 \\
\hline
\end{array}
\)
(i) Mean deviation about mean
While calculating the mean of a continuous frequency distribution, we had made the assumption that the frequency in each class is centred at its mid-point. Here also, we write the mid-point of each given class and proceed further as for a discrete frequency distribution to find the mean deviation.
Let us take the following example.
Example 6: Find the mean deviation about the mean for the following data.
\(
\begin{array}{|l|c|c|c|c|c|c|c|}
\hline \text { Marks obtained } & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\
\hline \text { Number of students } & 2 & 3 & 8 & 14 & 8 & 3 & 2 \\
\hline
\end{array}
\)
Solution: We make the following Table below from the given data :
\(\begin{array}{|c|c|c|c|c|c|}
\hline \begin{array}{l}
\text { Marks } \\
\text { obtained }
\end{array} & \begin{array}{l}
\text { Number of } \\
\text { students } \\
f_i
\end{array} & \begin{array}{l}
\text { Mid-points } \\
x_i
\end{array} & f_i x_i & \left|x_i-\bar{x}\right| & f_{ i }\left|x_i-\bar{x}\right| \\
\hline 10-20 & 2 & 15 & 30 & 30 & 60 \\
\hline 20-30 & 3 & 25 & 75 & 20 & 60 \\
\hline 30-40 & 8 & 35 & 280 & 10 & 80 \\
\hline 40-50 & 14 & 45 & 630 & 0 & 0 \\
\hline 50-60 & 8 & 55 & 440 & 10 & 80 \\
\hline 60-70 & 3 & 65 & 195 & 20 & 60 \\
\hline 70-80 & 2 & 75 & 150 & 30 & 60 \\
\hline & 40 & & 1800 & & 400 \\
\hline
\end{array}
\)
Here \(N =\sum_{i=1}^7 f_i=40, \sum_{i=1}^7 f_i x_i=1800, \sum_{i=1}^7 f_i\left|x_i-\bar{x}\right|=400\)
Therefore \(\bar{x}=\frac{1}{ N } \sum_{i=1}^7 f_i x_i=\frac{1800}{40}=45\)
\(
\text { and } \quad \text { M.D. }(\bar{x})=\frac{1}{ N } \sum_{i=1}^7 f_i\left|x_i-\bar{x}\right|=\frac{1}{40} \times 400=10
\)
Shortcut method for calculating mean deviation about mean
We can avoid the tedious calculations of computing \(\bar{x}\) by following step-deviation method. Recall that in this method, we take an assumed mean which is in the middle or just close to it in the data. Then deviations of the observations (or mid-points of classes) are taken from the assumed mean. This is nothing but the shifting of origin from zero to the assumed mean on the number line, as shown in Fig below.
If there is a common factor of all the deviations, we divide them by this common factor to further simplify the deviations. These are known as step-deviations. The process of taking step-deviations is the change of scale on the number line as shown in Figure below.
The deviations and step-deviations reduce the size of the observations, so that the computations viz. multiplication, etc., become simpler. Let, the new variable be denoted by \(d_i=\frac{x_i-a}{h}\), where ‘ \(a\) ‘ is the assumed mean and \(h\) is the common factor. Then, the mean \(\bar{x}\) by step-deviation method is given by
\(
\bar{x}=a+\frac{\sum_{i=1}^n f_i d_i}{ N } \times h
\)
Let us take the data of Example 6 and find the mean deviation by using stepdeviation method.
Take the assumed mean \(a=45\) and \(h=10\), and form the following Table below.
\(
\begin{array}{|c|c|c|c|c|c|c|}
\hline \begin{array}{l}
\text { Marks } \\
\text { obtained }
\end{array} & \begin{array}{l}
\text { Number of } \\
\text { students }
\end{array} & \text { Mid-points } & d_i=\frac{x_i-45}{10} & f_i d_i & \left|x_i-\bar{x}\right| & f_{ i }\left|x_i-\bar{x}\right| \\
\hline & f_i & x_i & & & & \\
10-20 & 2 & 15 & -3 & -6 & 30 & 60 \\
20-30 & 3 & 25 & -2 & -6 & 20 & 60 \\
30-40 & 8 & 35 & -1 & -8 & 10 & 80 \\
40-50 & 14 & 45 & 0 & 0 & 0 & 0 \\
50-60 & 8 & 55 & 1 & 8 & 10 & 80 \\
60-70 & 3 & 65 & 2 & 6 & 20 & 60 \\
70-80 & 2 & 75 & 3 & 6 & 30 & 60 \\
\hline
& 40 & & & 0 & & 400 \\
\hline
\end{array}
\)
Therefore
\(
\begin{aligned}
& \bar{x}=a+\frac{\sum_{i=1}^7 f_i d_i}{ N } \times h \\
& =45+\frac{0}{40} \times 10=45
\end{aligned}
\)
and M.D. \((\bar{x})=\frac{1}{ N } \sum_{i=1}^7 f_i\left|x_i-\bar{x}\right|=\frac{400}{40}=10\)
(ii) Mean deviation about median
The process of finding the mean deviation about median for a continuous frequency distribution is similar as we did for mean deviation about the mean. The only difference lies in the replacement of the mean by median while taking deviations.
Let us recall the process of finding median for a continuous frequency distribution.
The data is first arranged in ascending order. Then, the median of continuous frequency distribution is obtained by first identifying the class in which median lies (median class) and then applying the formula
\(
\text { Median }=l+\frac{\frac{ N }{2}- C }{f} \times h
\)
where median class is the class interval whose cumulative frequency is just greater than or equal to \(\frac{ N }{2}, N\) is the sum of frequencies, \(l, f, h\) and \(C\) are, respectively the lower limit, the frequency, the width of the median class and \(C\) the cumulative frequency of the class just preceding the median class. After finding the median, the absolute values of the deviations of mid-point \(x_i\) of each class from the median i.e., \(\left|x_i- M \right|\) are obtained.
\(
\text { Then } \quad \text { M.D. }( M )=\frac{1}{ N } \sum_{i=1}^n f_i\left|x_i- M \right|
\)
The process is illustrated in the following example:
Example 7: Calculate the mean deviation about median for the following data :
\(
\begin{array}{|l|c|c|c|c|c|c|}
\hline \text { Class } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\
\hline \text { Frequency } & 6 & 7 & 15 & 16 & 4 & 2 \\
\hline
\end{array}
\)
Solution: Form the following Table below from the given data :
\(
\begin{array}{|l|c|c|c|c|c|}
\hline \text { Class } & \text { Frequency } & \begin{array}{c}
\text { Cumulative } \\
\text { frequency }
\end{array} & \text { Mid-points } & \mid x_i-\text { Med. } \mid & f_i \mid x_i-\text { Med. } \mid \\
\hline & f_i & (c . f .) & x_i & & \\
0-10 & 6 & 6 & 5 & 23 & 138 \\
10-20 & 7 & 13 & 15 & 13 & 91 \\
20-30 & 15 & 28 & 25 & 3 & 45 \\
30-40 & 16 & 44 & 35 & 7 & 112 \\
40-50 & 4 & 48 & 45 & 17 & 68 \\
50-60 & 2 & 50 & 55 & 27 & 54 \\
\hline
& 50 & & & & 508 \\
\hline
\end{array}
\)
The class interval containing \(\frac{ N ^{\text {th }}}{2}\) or \(25^{\text {th }}\) item is \(20-30\). Therefore, \(20-30\) is the median class. We know that
\(
\text { Median }=l+\frac{\frac{ N }{2}- C }{f} \times h
\)
Here \(l=20, C =13, f=15, h=10\) and \(N =50\)
Therefore, \(\quad\) Median \(=20+\frac{25-13}{15} \times 10=20+8=28\)
Thus, Mean deviation about median is given by
M.D. \(( M )=\frac{1}{ N } \sum_{i=1}^6 f_i\left|x_i- M \right|=\frac{1}{50} \times 508=10.16\)
Limitations of mean deviation
In a series, where the degree of variability is very high, the median is not a representative central tendency. Thus, the mean deviation about median calculated for such series can not be fully relied.
The sum of the deviations from the mean (minus signs ignored) is more than the sum of the deviations from median. Therefore, the mean deviation about the mean is not very scientific. Thus, in many cases, mean deviation may give unsatisfactory results. Also mean deviation is calculated on the basis of absolute values of the deviations and therefore, cannot be subjected to further algebraic treatment. This implies that we must have some other measure of dispersion. Standard deviation is such a measure of dispersion.