Mean Deviation for a Grouped Data
 Discrete frequency distribution:
Tabulating the results, we get.
Values:  
Frequency: 
 Mean deviation about mean.
Then we find the absolute values of the deviations of each reading from ie for .
Then M.D
 Mean deviation about Median.
 Observations are arranged in ascending order.
 Cumulative frequencies are obtained.
 Identify the observation corresponding to the cumulative frequency either equal to or just greater than (where ) or the last cumulative frequency in the table)
Example1:
Find the mean deviation about the mean for the following data:
Value  10  11  12  13  14 
Frequency  3  12  18  12  3 
Solution:


10 11 12 13 14 
3 12 18 12 3 
30 132 216 156 42 
2 1 0 1 2 
6 12 0 12 6 
48 
576 
36 
M.D = = 0.75
Example 2:
Calculate M.D from the median for the following (discrete) frequency distribution.
Value of the item  12  13  14  25  26  27  38  40 
Frequency  2  3  5  8  7  3  2  1 
Solution:
N = 31; 15.5, Median M = 25
Value 

12 13 14 25 26 27 38 40 
2 3 5 8 7 3 2 1 
2 5 10 18 25 28 30 31 
13 12 11 0 1 2 13 15 
26 36 55 0 7 6 26 15 
31 
171 
 M.D for a continuous frequency distribution.
Example 3:
Calculate the mean deviation from the mean.
Class interval : 
2  4 
4  6 
6  8 
8  10 
Frequency: 
5 
6 
3 
1 
Solution:
Class  Frequency  Midvalue  
2  4 4  6 6  8 8  10 
5 6 3 1 
3 5 7 9 
15 30 21 9 
1 0 6 4 
15 
75 
20 
M.D
Note: It is not always easy to calculate the mean of a continuous distribution by using the formula as the numbers may be quite big. In such cases, the short cut method or step  deviation method can be accepted. Recall the steps involved in the short  cut method.
 We take an assured mean, Say A, which is in the middle or just close to it in the given data.
 The deviations of the observations (mid  values) are taken from the assumed mean
 If there is a common factor for all deviations, we divide them by this common factor .
 is the formula for calculating .
Example 4: Find the mean deviation about the mean for the following data.
Class interval 
0  10 
10  20 
20  30 
30  40 
40  50 
50  60 
60  70 
Frequency 
8 
12 
10 
8 
3 
2 
7 
Solution:
Assumed mean A =35
Class 
MidValue 
Frequency 

0  10
10  20 20  30 30  40 40  50 50  60 60  70 
5 15 25 35 45 55 65 
8 12 10 8 3 2 7 
âˆ’ 3 âˆ’ 2 âˆ’ 1 0 1 2 3 
âˆ’ 24 âˆ’ 24 âˆ’ 10 0 3 4 21 
24 14 4 6 16 26 36 
192 168 40 48 48 52 252 
800 
A = 35
h = 10
= 35  6 = 29.
M.D
Note: To calculate the mean deviation about the median in a continues distribution we have to calculate median using the formula :
where is the lower limit of the median class
N is the total frequency.
C is the cumulative frequency of pre  median class
is the frequency of the median class.
is the interval of the uniform class interval.
Let us taken an example.
Example 5: Calculate the mean deviation about the median for the following data.
Marks 
0  10 
10  20 
20  30 
30  40 
40  50 
50  60 
60  70 
70  80 
No. of students 
18 
16 
15 
12 
10 
5 
2 
2 
Solution:
Class 
Mid  mark 

0  10
10  20 20  30 30  40 40  50 50  60 60  70 70  80 
18 16 15 12 10 5 2 2 
18 34 49 61 71 76 78 80 
5 15 25 35 45 55 65 75 
19 9 1 11 21 31 41 51 
342 144 15 132 210 155 82 102 
80 
1182 
Median class is 20  30
.
Mean deviation about M = = = 14.775
Example 6: Calculate the mean deviation fun the median from the following data
Sales (in 1000'sRs)  Number of shops 
4  5 6  7 8  9 10  11 12  13 14  15 
4 10 20 15 8 3 
60 
Solution:
The class intervals given are not continuous. Convert it to a continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
Class 

3.5  5.5 5.5  7.5 7.5  9.5 9.5  11.5 11.5  13.5 13.5  15.5 
4 10 20 15 8 3 
4 14 34 49 57 60 
4.5 6.5 8.5 10.5 12.5 14.5 
4.6 2.6 0.6 1.4 3.4 5.4 
18.4 26.0 12.0 21.0 27.2 16.2 
60 
120.8 
M = = 7.5 + 1.6 = 9.1
Mean deviation about M = =
Some Observations about mean deviation
Even though mean deviation is based on all observations, it gives up the sign of the deviations and takes only the numerical deviations into consideration, therefore it is not suited for algebraic treatment. It lacks a scientific approach. This implies that we have to look for a more accurate and scientific measure of dispersion. Standard deviation is such a measure.