# Mean Deviation

#### Topics

• Mean deviation for grouped data
• Mean deviation for ungrouped data

## Notes

An absolute measure of dispersion is the mean of these deviations.  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 ( x ) is zero.
M.D(a) = ("Sum of absolute values of deviations from a")/ ("Number of observations")
Remark :  Mean deviation may be obtained from any measure of central tendency. However, mean deviation from mean and median are commonly used in statistical studies.

1)  Mean deviation for ungrouped data :

Let  n observations be x_1, x_2, x_3, ...., 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 xi from a, i.e., x_1 – a, x_2 – a, x_3 – a,. . . , x_n– a

Step 3- Find the absolute values of the deviations, i.e., drop the minus sign (–), if it is
there, i.e., |x_1 - a| , |x_2 - a| , | x_3 - a |, ..., |x_n-a|

Step 4- Find the mean of the absolute values of the deviations. This mean is the mean deviation about a, i.e.,
M.D.(a) = $\displaystyle\sum_{i=1}^{n} |x_i - a|$
Thus M.D.(bar x) = 1/n $\displaystyle\sum_{i=1}^{n} |x_i - \bar{x} |$ , where bar x = Mean

and M.D.(M) = 1/n $\displaystyle\sum_{i=1}^{n} |x_i - M|$, where M = Median

For example : 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
find the mean ( x ) of the given data
bar x = 1/20 sum _(i=1)^20 x_i = 200 / 20 = 10
The absolute values of the deviations from mean, i.e., |x_i - bar x| are
2,7,8,7,6,1,7,9,10,5,2,7,8,7,6,1,7,9,10,5
Therefore sum_(i=1)^20 |x_i - bar x| = 124
and M.D.(bar x) = 124/20 = 6.2

2)  Mean deviation  for grouped data :
Data can be grouped into two ways :
(a)  Discrete frequency distribution,
(b) Continuous frequency distribution.
Let us discuss the method of finding mean deviation for both types of the data.

Discrete frequency distribution:  Let the given data consist of n distinct values x_1, x_2, ..., x_n occurring with frequencies f_1, f_2 , ..., f_n respectively. This data can be represented in the tabular form as given below, and is called discrete frequency distribution:
x : x_1    x_2 x_3 ... x_n
f : f_1     f_2 f_3 ... f_n

First of all we find the mean
bar x of the given data by using the formula$\bar{x} =\frac{\displaystyle\sum_{i=1}^{n} x_i f_i }{ \displaystyle\sum_{i=1}^{n} f_i\ }= \frac {1}{N} \displaystyle\sum_{i=1}^{n} x_i f_i$ ,
where summation x_i f_i denotes the sum of the products of observations xi with their respective frequenciesf_i and  N =$\displaystyle\sum_{i=1}^{n} f_i$  is the sum of the frequencies.The mean of the absolute values of the deviations, which is the required mean deviation about the mean.

M.D.$\bar{x} =\frac{\displaystyle\sum_{i=1}^{n} f_i |x_i - \bar{x}| }{ \displaystyle\sum_{i=1}^{n} f_i\ }= \frac {1}{N} \displaystyle\sum_{i=1}^{n} f_i |x_i - \bar{x}|$ ,
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 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
M.D.(M) =$\frac {1}{N} \displaystyle\sum_{i=1}^{n} f_i |x_i -M|$

(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.
The assumption that the frequency in each class is centred at its mid-point.
The mid-point of each given class and proceed further as for a discrete frequency distribution to find the mean deviation.

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.
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

Median = l + (N/2 - C)/f xx h
where median class is the class interval whose cumulative frequency is just greater than or equal to  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 xi of each class from the median i.e. |x_i - M|are obtained.
M.D.(M) = $\frac {1}{N} \displaystyle\sum_{i=1}^{n} f_i |x_i -M|$

3) Limitations of mean deviation:
In a series, where the degree of variability is very high, the median is not a representative central tendency. 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.

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