1 of 62

Measures of Dispersion

Dr Anshul Singh Thapa

2 of 62

MEASURES OF DISPERSION FROM�AVERAGE

  • Recall that dispersion was defined as the extent to which values differ from their average. Range and Quartile Deviation are not useful in measuring, how far the values are, from their average. Yet, by calculating the spread of values, they do give a good idea about the dispersion. Two measures which are based upon deviation of the values from their average are Mean Deviation and Standard Deviation.
  • Since the average is a central value, some deviations are positive and some are negative. If these are added as they are, the sum will not reveal anything. In fact, the sum of deviations from Arithmetic Mean is always zero.
  • Mean Deviation tries to overcome this problem by ignoring the signs of deviations, i.e., it considers all deviations positive. For standard deviation, the deviations are first squared and averaged and then square root of the average is found. We shall now discuss them separately in detail.

3 of 62

Mean Deviation

  • Mean Deviation is the arithmetic average of the deviations of all the values taken from some average value (mean, median, mode) of the series, ignoring signs (+ or -) of the deviations.
  • Clark and Schakde “Mean deviation is the arithmetic average of deviations of all the values taken from a statistical average (mean, median or mode) of series. In taking deviation of values, Algebraic signs + and - are not taken into consideration, that is negative deviations are also treated as positive deviation.

4 of 62

MEAN DEVIATION

UNGROUPED DATA

DISCRETE SERIES

DIRECT METHOD

ASSUMED MEAN METHOD

FROM MEDIAN

GROUPED DATA

FROM ARITHMETIC MEAN

CONTINUOUS SERIES

DIRECT METHOD

SHORT CUT METHOD

FROM ARITHMETIC MEAN

SHORT CUT METHOD

SHORT CUT METHOD

FROM MEDIAN

FROM MEDIAN

FROM ARITHMETIC MEAN

DIRECT METHOD

DIRECT METHOD

5 of 62

MEAN DEVIATION

UNGROUPED DATA

DISCRETE SERIES

FROM MEDIAN

GROUPED DATA

FROM ARITHMETIC MEAN

CONTINUOUS SERIES

FROM ARITHMETIC MEAN

FROM MEDIAN

FROM MEDIAN

FROM ARITHMETIC MEAN

ΣI dx I

n

ΣI dx I

n

ΣI dx I

n

ΣI dx I

n

ΣI dx I

n

ΣI dx I

n

6 of 62

CALCULATION OF MEAN DEVIATION FOR UNGROUPED DATA

STEPS (FROM MEDIAN)

STEPS (FROM MEAN)

I. Arrange the data in ascending order.

II. Calculate the median

Me = Size of (N + 1/ 2)th item

III. Find out the deviation of the items from median. Ignore (-) and (+) signs. Express it by I dm I sign.

IV. Sum up the deviations, and express the same by using the relevant formula to find out the required answer.

MDm = ΣI dm I

N

V. After that solve the question with the help of formula.

  1. Calculate Arithmetic mean by adding up the data.
  2. Find out the deviation of the items from mean. Ignore (-) and (+) signs. Express it by I dx I sign.
  3. Sum up the deviations.
  4. Use the relevant formula to find the required answer:

MDx = ΣI dx I

N

7 of 62

Calculation of Mean Deviation from Arithmetic Mean for ungrouped data – Direct Method

Steps:

  • The Arithmetic Mean of the values is calculated
  • Difference between each value and the Arithmetic Mean is calculated. All differences are considered positive. These are denoted as |d|
  • The Arithmetic Mean of these differences (called deviations) is the Mean Deviation.

8 of 62

Example

X

2

4

7

8

9

9 of 62

Example

X

2

4

7

8

9

30

10 of 62

Example

X

|dx| = X – X = 6

2

4

7

8

9

11 of 62

Example

X

|dx| = X – X = 6

2

4

4

2

7

1

8

2

9

3

12 of 62

Example

X

|dx| = X – X = 6

2

4

4

2

7

1

8

2

9

3

Σ|dx| = 12

13 of 62

Example

X

|dx| = X – X = 6

2

4

4

2

7

1

8

2

9

3

Σ|dx| = 12

  • The M Dx = Σ|dx| = 12 = 2.4

N 5

14 of 62

Calculation of Mean Deviation from Arithmetic Mean for ungrouped data – Assumed Mean Method

  • Mean Deviation can also be calculated by calculating deviations from an assumed mean. This method is adopted especially when the actual mean is a fractional number. (Take care that the assumed mean is close to the true mean). For the values in example 3, suppose value 7 is taken as assumed mean, M.D. can be calculated as under:

15 of 62

Example

  • In such cases, the following formula is used:

MDx = Σ|dx| + (X – Ax) (ΣfB – ΣfA)

n

X

|dx| = Ax – X

2

5

4

3

7

0

8

1

9

2

Σ|dx| = 11

16 of 62

  • Where Σ|dx| is the sum of absolute deviations taken from the assumed mean.
  • x is the actual mean.
  • Ax is the assumed mean used to calculate deviations.
  • ΣfB is the number of values below the actual mean including the actual mean.
  • ΣfA is the number of values above the actual mean.
  • Substituting the values in the above formula:

MDx = 11 + (6 – 7) (3 – 2) = 12 = 2.4

5 5

17 of 62

Mean Deviation from median for ungrouped data

Direct Method

  • M.D. from the Median can be calculated as follows:
    • Calculate the median which is 4400 in this case.
    • Calculate the absolute deviations from median, denote them as |d|.
    • Find the average of these absolute deviations. It is the Mean Deviation.

18 of 62

Mean Deviation from median for ungrouped data

Median = Size of N + 1 th item = 5+ 1 = 3rd item

2 2

Size of the third item is 4400

M.D.median = |dx| = 1200 = 240

n 5

Income

(X)

4000

4200

4400

4600

4800

19 of 62

Mean Deviation from median for ungrouped data

Median = Size of N + 1 th item = 5+ 1 = 3rd item

2 2

Size of the third item is 4400

Income

(X)

4000

4200

4400

4600

4800

20 of 62

Mean Deviation from median for ungrouped data

Median = Size of N + 1 th item = 5+ 1 = 3rd item

2 2

Size of the third item is 4400

Income

(X)

Deviation from Median 4400

|dx|

4000

400

4200

200

4400

0

4600

200

4800

400

21 of 62

Mean Deviation from median for ungrouped data

Median = Size of N + 1 th item = 5+ 1 = 3rd item

2 2

Size of the third item is 4400

Income

(X)

Deviation from Median 4400

|dx|

4000

400

4200

200

4400

0

4600

200

4800

400

Σ|dx| = 1200

22 of 62

Mean Deviation from median for ungrouped data

Median = Size of N + 1 th item = 5+ 1 = 3rd item

2 2

Size of the third item is 4400

M.D.median = |dx| = 1200 = 240

n 5

Income

(X)

Deviation from Median 4400

|dx|

4000

400

4200

200

4400

0

4600

200

4800

400

Σ|dx| = 1200

23 of 62

Mean Deviation from median for ungrouped data

Short-cut method

  • To calculate Mean Deviation by short cut method, a value (A) is used to calculate the deviations and the following formula is applied:

MDmedian = Σ|dx| + (Median – A) (ΣfB – ΣfA)

n

  • Where, A = the constant from which deviations are calculated. (Other notations are the same as given in the assumed mean method).

24 of 62

Mean Deviation from mean for discrete series

X

10

11

12

13

14

f

3

12

18

12

3

25 of 62

x

f

10

3

11

12

12

18

13

12

14

3

26 of 62

x

f

10

3

11

12

12

18

13

12

14

3

Σf = 48

27 of 62

x

f

fx

10

3

30

11

12

132

12

18

216

13

12

156

14

3

42

Σf = 48

28 of 62

x

f

fx

10

3

30

11

12

132

12

18

216

13

12

156

14

3

42

Σf = 48

Σfx = 576

29 of 62

  • Actual Mean = Σfm = 576 = 12

Σf 48

x

f

fx

10

3

30

11

12

132

12

18

216

13

12

156

14

3

42

Σf = 48

Σfx = 576

30 of 62

  • Actual Mean = Σfm = 576 = 12

Σf 48

x

f

fx

|d|

10

3

30

2

11

12

132

1

12

18

216

0

13

12

156

1

14

3

42

2

Σf = 48

Σfx = 576

31 of 62

  • Actual Mean = Σfm = 576 = 12

Σf 48

x

f

fx

|d|

f|d|

10

3

30

2

6

11

12

132

1

12

12

18

216

0

0

13

12

156

1

12

14

3

42

2

6

Σf = 48

Σfx = 576

32 of 62

  • Actual Mean = Σfm = 576 = 12

Σf 48

x

f

fx

|d|

f|d|

10

3

30

2

6

11

12

132

1

12

12

18

216

0

0

13

12

156

1

12

14

3

42

2

6

Σf = 48

Σfx = 576

36

33 of 62

  • Actual Mean = Σfm = 576 = 12

Σf 48

  • M.D.x = Σf|d| = 36 = 0.75

Σf 48

x

f

fx

|d|

f|d|

10

3

30

2

6

11

12

132

1

12

12

18

216

0

0

13

12

156

1

12

14

3

42

2

6

Σf = 48

Σfx = 576

36

34 of 62

Mean Deviation from mean for discrete series

Height

158

159

160

161

162

163

164

165

166

Students

15

20

32

35

33

22

20

10

8

Mean Deviation Mean = Σf|d| = 338.59 = 1.74

Σf 195

35 of 62

Mean Deviation from median for discrete series

X

10

11

12

13

14

f

3

12

18

12

3

36 of 62

x

f

10

3

11

12

12

18

13

12

14

3

Σf = 48

37 of 62

x

f

c.f

10

3

3

11

12

15

12

18

33

13

12

45

14

3

48

Σf = 48

38 of 62

  • Median = size of (N+1)/2 th item
  • Median = (48 + 1) = 49 = 24.5th item

2

Size of 5th item lies in 33 cumulative frequecy therefore the median is 12

x

f

c.f

10

3

3

11

12

15

12

18

33

13

12

45

14

3

48

Σf = 48

39 of 62

  • Median = size of (N+1)/2 th item
  • Median = (48 + 1) = 49 = 24.5th item

2

Size of 5th item lies in 33cumulative frequecy therefore the median is 12

x

f

|d|

c.f

10

3

2

3

11

12

1

15

12

18

0

33

13

12

1

45

14

3

2

48

Σf = 48

40 of 62

  • Median = size of (N+1)/2 th item
  • Median = (48 + 1) = 49 = 24.5th item

2

Size of 5th item lies in 33cmlativefrequecy therefore the median is 12

x

f

|d|

f|d|

c.f

10

3

2

6

3

11

12

1

12

15

12

18

0

0

33

13

12

1

12

45

14

3

2

6

48

Σf = 48

Σfd = 36

41 of 62

  • Median = size of (N+1)/2 th item
  • Median = (48 + 1) = 49 = 24.5th item

2

Size of 5th item lies in 33cmlativefrequecy therefore the median is 12

  • Mean Deviation median = Σf|d| = 36 = 0.75

Σf 48

x

f

|d|

f|d|

c.f

10

3

2

6

3

11

12

1

12

15

12

18

0

0

33

13

12

1

12

45

14

3

2

6

48

Σf = 48

Σfd = 36

42 of 62

Mean Deviation from Mean for�Continuous distribution

Steps:

  • Calculate the mean of the distribution.
  • Calculate the absolute deviations |d| of the class midpoints from the mean.
  • Multiply each |d| value with its corresponding frequency to get f|d| values. Sum them up to get Σf|d|.
  • Apply the formula

43 of 62

Example

Class – interval

Frequency

10 – 20

5

20 – 30

8

30 – 50

16

50 – 70

8

70 – 80

3

40

44 of 62

Class Interval

f

10 – 20

5

20 – 30

8

30 – 50

16

50 – 70

8

70 – 80

3

45 of 62

Class Interval

f

10 – 20

5

20 – 30

8

30 – 50

16

50 – 70

8

70 – 80

3

Σf = 40

46 of 62

Class Interval

f

Mid point

m

10 – 20

5

15

20 – 30

8

25

30 – 50

16

40

50 – 70

8

60

70 – 80

3

75

Σf = 40

47 of 62

Class Interval

f

Mid point

m

fm

10 – 20

5

15

75

20 – 30

8

25

200

30 – 50

16

40

640

50 – 70

8

60

480

70 – 80

3

75

225

Σf = 40

48 of 62

Class Interval

f

Mid point

m

fm

10 – 20

5

15

75

20 – 30

8

25

200

30 – 50

16

40

640

50 – 70

8

60

480

70 – 80

3

75

225

Σf = 40

Σfm = 1620

49 of 62

Class Interval

f

Mid point

m

fm

10 – 20

5

15

75

20 – 30

8

25

200

30 – 50

16

40

640

50 – 70

8

60

480

70 – 80

3

75

225

Σf = 40

Σfm = 1620

  • Actual Mean = Σfm = 1620 = 40.5

Σf 40

  • M.D.x = Σf|d| = 519 = 12.97

Σf 40

50 of 62

Class Interval

f

Mid point

m

fm

|d|

10 – 20

5

15

75

25.5

20 – 30

8

25

200

15.5

30 – 50

16

40

640

0.5

50 – 70

8

60

480

19.5

70 – 80

3

75

225

34.5

Σf = 40

Σfm = 1620

  • Actual Mean = Σfm = 1620 = 40.5

Σf 40

  • M.D.x = Σf|d| = 519 = 12.97

Σf 40

51 of 62

Class Interval

f

Mid point

m

fm

|d|

f|d|

10 – 20

5

15

75

25.5

127.5

20 – 30

8

25

200

15.5

124.0

30 – 50

16

40

640

0.5

8.0

50 – 70

8

60

480

19.5

156.0

70 – 80

3

75

225

34.5

103.5

Σf = 40

Σfm = 1620

  • Actual Mean = Σfm = 1620 = 40.5

Σf 40

  • M.D.x = Σf|d| = 519 = 12.97

Σf 40

52 of 62

Class Interval

f

Mid point

m

fm

|d|

f|d|

10 – 20

5

15

75

25.5

127.5

20 – 30

8

25

200

15.5

124.0

30 – 50

16

40

640

0.5

8.0

50 – 70

8

60

480

19.5

156.0

70 – 80

3

75

225

34.5

103.5

Σf = 40

Σfm = 1620

Σf|d| = 519.0

  • Actual Mean = Σfm = 1620 = 40.5

Σf 40

  • M.D.x = Σf|d| = 519 = 12.97

Σf 40

53 of 62

Class Interval

f

Mid point

m

fm

|d|

f|d|

10 – 20

5

15

75

25.5

127.5

20 – 30

8

25

200

15.5

124.0

30 – 50

16

40

640

0.5

8.0

50 – 70

8

60

480

19.5

156.0

70 – 80

3

75

225

34.5

103.5

Σf = 40

Σfm = 1620

Σf|d| = 519.0

  • Actual Mean = Σfm = 1620 = 40.5

Σf 40

  • M.D.x = Σf|d| = 519 = 12.97

Σf 40

54 of 62

Mean Deviation from median for�Continuous distribution

  • The procedure to calculate Mean Deviation from the median is the same as it is in case of M.D. from Mean, except that deviations are to be taken from the median as given below:

Class Interval

f

20 – 30

5

30 – 40

10

40 – 60

20

60 – 80

9

80 – 90

6

Σf = 50

55 of 62

Mean Deviation from median for�Continuous distribution

  • The procedure to calculate Mean Deviation from the median is the same as it is in case of M.D. from Mean, except that deviations are to be taken from the median as given below:

Class Interval

Mid Value

f

20 – 30

25

5

30 – 40

35

10

40 – 60

50

20

60 – 80

70

9

80 – 90

85

6

Σf = 50

56 of 62

Mean Deviation from median for�Continuous distribution

  • The procedure to calculate Mean Deviation from the median is the same as it is in case of M.D. from Mean, except that deviations are to be taken from the median as given below:

Class Interval

Mid Value

f

c.f

20 – 30

25

5

5

30 – 40

35

10

15

40 – 60

50

20

35

60 – 80

70

9

44

80 – 90

85

6

50

Σf = 50

57 of 62

Median Class: N/2 = 50/2 = 25

Median = L + (N/2 - c.f.) × i

f

Median = 40 + (25 - 15) x 20 = 50

20

58 of 62

Mean Deviation from median for�Continuous distribution

  • The procedure to calculate Mean Deviation from the median is the same as it is in case of M.D. from Mean, except that deviations are to be taken from the median as given below:

Class Interval

Mid Value

f

c.f

20 – 30

25

5

5

30 – 40

35

10

15

40 – 60

50

20

35

60 – 80

70

9

44

80 – 90

85

6

50

Σf = 50

59 of 62

Mean Deviation from median for�Continuous distribution

  • The procedure to calculate Mean Deviation from the median is the same as it is in case of M.D. from Mean, except that deviations are to be taken from the median as given below:

Class Interval

Mid Value

f

c.f

|d|

20 – 30

25

5

5

25

30 – 40

35

10

15

15

40 – 60

50

20

35

0

60 – 80

70

9

44

20

80 – 90

85

6

50

35

Σf = 50

60 of 62

Mean Deviation from median for�Continuous distribution

  • The procedure to calculate Mean Deviation from the median is the same as it is in case of M.D. from Mean, except that deviations are to be taken from the median as given below:

Class Interval

f

c.f

|d|

f|d|

20 – 30

5

5

25

125

30 – 40

10

15

15

150

40 – 60

20

35

0

0

60 – 80

9

44

20

180

80 – 90

6

50

35

210

Σf = 50

61 of 62

Mean Deviation from median for�Continuous distribution

  • The procedure to calculate Mean Deviation from the median is the same as it is in case of M.D. from Mean, except that deviations are to be taken from the median as given below:

Class Interval

f

c.f

|d|

f|d|

20 – 30

5

5

25

125

30 – 40

10

15

15

150

40 – 60

20

35

0

0

60 – 80

9

44

20

180

80 – 90

6

50

35

210

Σf = 50

Σf|d| = 665

62 of 62

Median = L + (N/2 - c.f.) × i

f

Median = 40 + (25 - 15) x 20 = 50

20

Mean Deviation median = Σf|d| = 665

Σf 50