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Mean Deviation for a Grouped Data

  1. Discrete frequency distribution:
Let the given data consists of n discrete values with frequencies respectively.
Tabulating the results, we get.
Values:
Frequency:
  1. Mean deviation about mean.
Let be the mean of the given data.

Then we find the absolute values of the deviations of each reading from ie for .
Then M.D
  1. Mean deviation about Median.
To find the median, follow the steps given below.
  1. Observations are arranged in ascending order.
  2. Cumulative frequencies are obtained.
  3. Identify the observation corresponding to the cumulative frequency either equal to or just greater than (where ) or the last cumulative frequency in the table)
If M is the median, M.D (M) =

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(M) =
  1. M.D for a continuous frequency distribution.
Here the values will be given in continuous class intervals and we can reasonably assume that the frequency of each class is centred at its mid-point. Having found the mid-value, we can precede as for discrete frequency distribution this with an example.

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 Mid-value

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.
  1. We take an assured mean, Say A, which is in the middle or just close to it in the given data.
  2. The deviations of the observations (mid - values) are taken from the assumed mean
  3. If there is a common factor for all deviations, we divide them by this common factor .
  4. is the formula for calculating .
Let us see an example using this.

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

Mid-Value

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

3

2

1

0

1

2

3

24

24

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.




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