1. INTRODUCTION
Dispersion refers to the extent to which the items vary from one another and from the central value.It may be noted that the measures of dispersion or variation measure only the degree but not the direction of the variation. The measures of dispersion are also called averages of the second order because they are based on the deviations of the different values from the mean or other measures of central tendency which are called averages of the first order.

2. DEFINITION
In the words of Bowley “Dispersion is the measure of the variation of the items”
According to Conar “Dispersion is a measure of the extent to which the individual items vary”

3. METHODS OF MEASURING DISPERSION
Range
Inter Quartile Range and Quartile Deviation
Mean Deviation
Standard Deviation
Coefficient of variation
Lorenz curve

4. RANGE Range is the simplest method of measuring dispersion
It is defined as the difference between the smallest and the largest observations in a given set of data.
Its formula is:
R=L-S
Where, R= Range
L= Largest value
S= Smallest value
The relative measure of Range, also called coefficient of range is defined as:
Coefficient of Range= L-S/L+S

5. CALCULATION OF RANGE
Range is calculated with the help of following series :
Individual series
Discrete series
Continuous series
Let us explain these series with the help of examples

6. 1. INDIVIDUAL SERIES
Ex 1. Five students obtained the following marks in statistics: 20, 35, 25, 30, 15 Find the range and coefficient of range. Sol. Here L=35, S=15 range= L-S range= 35-15=20 Coefficient of range= L-S/L+S =35-15/35+15 = 20/15 = 0.40

7. Ex 2.Find the range and coefficient of range from the data:
2. DISCRETE SERIES
Ex 2.Find the range and coefficient of range from the data:
marks 10 20 30 40 50 60 70
No of students 15 18 25 16 9
Sol. Here L=70, S=10
range= L-S
range= 70-10=60
Coefficient of range= L-S/L+S
= 70-10/70+10
= 0.75

8. Ex 3. find range and coefficient of range from the series:
3. CONTINUOUS SERIES
Ex 3. find range and coefficient of range from the series:
size
5-10
10-15
15-20
20-25
25-30
frequency 4 9 15 30 40
Sol. Range= L-S
here, L= upper limit of the largest class=30
S= lower limit of the smallest class=5
range= 30-5=25
Coefficient of range= L-S/L+S
=30-5/30+5
=5/7

9. INTERQUARTILE RANGE & QUARTILE DEVIATION
Interquartile range is the difference b/w the upper quartile (Q3) and lower quartile (Q1)
Its formula is:
interquartile range= Q3- Q1
Quartile deviation is the half of the difference b/w the upper quartile (Q3) and lower quartile (Q1)
Quartile deviation =Q3 - Q1/2
the relative measure of Quartile deviation also called coefficient of Quartile deviation is defined as:
coefficient of Quartile deviation= Q3 - Q1/Q3+Q1

10. CALCULATION OF INTERQUARTILE RANGE & QUARTILE DEVIATION
It is calculated with the help of following series :
Individual series
Discrete series
Continuous series
Let us explain these series with the help of examples

11. 1. INDIVIDUAL SERIES
Ex 1. Find the interquartile range, quartile deviation and coefficient of quartile deviation from the data
28,18,20,24,27,30,15
Sol. Arrange the data in ascending order:
15,18,2024,27,28,30
Q1= size of (n+1/4)th item Q3= size of 3(n+1/4)th item
= size of (7+1/4)th item = size of 3(7+1/4)th item
= size of (2)nd item = size of 6th item
= 18 marks = 28 marks
interquartile range=Q3-Q1
= 28-18=10
quartile deviation= Q3 - Q1/2
= 10/2=5
coefficient of quartile deviation= Q3 - Q1/Q3+Q1
= 28-18/28+18
= 0.217

12. 2. DISCRETE SERIES
Wages 10 20 30 40 50 60
No of workers 2 8 35 42
Q1= size of (n+1/4)th item Q3= size of 3(n+1/4)th item
= size of (127+1/4)th item = size of 3(127+1/4)th item
= size of 32nd item= = size of 96th item=50
interquartile range=Q3-Q1
= 50-40=10
quartile deviation= Q3 - Q1/2
= 50-40/2=5
coefficient of quartile deviation= Q3 - Q1/Q3+Q1
= 50-40/50+40
= 0.11

13. 3. CONTINUOUS SERIES Age(years) 0-20 20-40 40-60 60-80 80-100 persons15 20 11
Age(years) f cf
0-20 4
20-40 10 14
40-60 15 29
60-80 20 49
80-100 11 60
N=60

14. Q1= size of (n/4)th item Q3 = size of 3(n/4)th item =(60/4)th item = 3(60/4)th item =15th item = 45th item Q1 lies in the class Q3 lies in the class Q1= l1 + n/4-cf * i Q3= l1 + n/4-cf * i f f = * 20 = * = = 76 interquartile range=Q3-Q1 = =34.67 quartile deviation= Q3 - Q1/2 = /2 coefficient of quartile deviation= Q3 - Q1/Q3+Q1 = / = 0.29

15. MEAN DEVIATION
Mean Deviation is also known as average deviation. In this case deviation taken from any average especially Mean, Median or Mode. While taking deviation we have to ignore negative items and consider all of them as positive.

16. Formulae to calculate mean deviation
M.D. from Mean=
M.D. from Median=

17. Coefficient of Mean Deviation= =
Coefficient of

18. Calculation of mean deviation
Individual series = =

19. Example: Calculate mean deviation from mean as well as from median and coefficient of mean deviation from the following data: Marks: 20 , 22 , 25 , 38 , 40 , 50 , 65 , 70 , 75
Solution:
Median= Size of
th item
=Size of
th item = 40

20. Deviations from Median 40
Marks
(X)
Deviations from mean 45
Deviations from Median 40 20 25 22 23 28 15 38 7 2 40 5 50 10 65 70 30 75 35
N=9, ΣХ=405

21. M.D. from Mean= =
Coeff. Of
Median= 40
M.D. from Median=
Coeff of

22. Discrete Series

23. Example: Calculate the mean deviation from median and mean and their coefficient from the following table: X: 20 30 40 50 60 70
Frequency : 8 12 10 6 4
Solution:
M = Size of
th item
= Size of
th item
= Size of 30.5th item
M =40

24. Calculation of mean deviation from medianX f
c.f.
M=40 20 8
160 30 12 10
120 40 50
100 60 6 56 70 4
N=60
M=40
Coeff of

25. Calculation of mean deviation from meanX f fX 20 8
160 21
168 30 12
360 11
132 40
800 1 50 10
500 9 90 60 6 19
114 70 4
280 29
116
N=60
ΣfX=2460
Coeff of

26. Continuous Series
,where
,where

27. Example: Calculate the mean deviation from mean and its coefficient from the following data:
Marks
0-10
10-20
20-30
30-40
40-50
No. of students 5 8 15 16 6
Solution:

28. Calculation of mean deviation from mean
Marks f m fm
0-10 5 25 22
110
10-20 8 15
120 12 96
20-30
375 2 30
30-40 16 35
560
128
40-50 6 45
270 18
108
N=50
Σfm=1350
Coeff. of

29. Example: Calculate means deviation from median and its coefficient from the following data
Size
frequency 4 6 10 8 5
Solution:
Item Median lies in

30. Calculation of mean deviation from median
Size f cf m
M=153 4
110 43
172 6 10
130 23
138 20
150 3 30 8 28
170 17
136 5 33
190 37
185
N=33
M=153
Coeff of

31. Short-cut method for mean deviation
If value of the average comes out to be in fractions, M.D. is calculated using the following formula:
Individual Series-
Discrete and Continuous Series-

32. Individual Series
Example: Using shortcut method , calculate the mean deviations from mean and median from the following data: 7 , 9 , 13 , 13 , 15 , 17 , 19 , 21 , 23

33. X
Taking
Taking M 7 9 13 15
M=15 17 19 21 23
ΣX=137
N=9

34. M = Size of
th item = 15
From mean:

35. From Median:

36. Continuous Series
Example: Using shortcut method , calculate the mean deviations from mean and median from the following data:
Marks
0-10
10-20
20-30
30-40
40-50
No. of Students 6 28 51 11 4

37. Calculation of mean deviation from mean
MarksX f X fX
0-10 6 5 30
10-20 28 15
420
20-30 51 25
1275
30-40 11 35
385
40-50 4 45
180
ΣfХ=2290

39. Calculation of mean deviation from median
MarksX f cf X fX
0-10 6 5 30
10-20 28 34 15
420
M = 23.14
20-30 51 85 25
1275
30-40 11 96 35
385
40-50 4
100 45
180
ΣfХ=2290
Median = 23.14

41. STANDARD DEVIATION
The concept of standard deviation was first introduced by Karl Pearson in The standard deviation is the most useful and the most popular measure of dispersion. Just as the arithmetic mean is the most of all the averages, the standard deviation is the best of all measures of dispersion.

42. Calculation of Standard Deviation
Individual Series
In case of the individual series, standard deviation can be compared by applying any of the three methods:
1) Actual Mean Method
When deviations are taken from the actual mean, the following formula is used :
σ = ∑ ( X – X)² or ∑ x ² where , x = X – X
N N

43. Steps for Calculation
Calculate the actual mean (x) of the series
Then take the deviation of the items from the mean i.e. find X – X and denote these deviations by x.
Square these deviations and obtain the total i.e. ∑ x2
Divide ∑ x2 by the total number of items i.e. N and take the square root of it. The result will give the value of the standard deviation.

44. 2) Assumed Mean Method When the actual mean is not a whole number but in fraction then it becomes difficult to take deviations from mean and obtain the square of these deviations. To save time and labor we use assumed mean method or called short-cut method. When deviations are taken fro assumed mean , the following formula is used :

45. Steps for Calculation
Any of the items in the series is taken as assumed mean.
Take the deviations of the items from the assumed mean, i.e. X – A and denote these deviations by ‘d’. Sum up these deviations to obtain ∑ d.
Then square these deviations taken from assumed mean and obtain the total , i.e. ∑d²
Substitute the values of ∑d² , ∑ d and N in the above formula. The result will give the value of standard deviation.

46. X
X= 20
x= X – X x² 16 -4 20 18 -2 4 19 -1 1 28 8 64 17 -3 9 22 2
N = 10, ∑x= 200
∑x= 0
∑x² = 98
Example –Calculate the standard deviation from the following : X = 16, 20, 18, 19, 20, 20, 28, 17, 22, 20 Solution : Calculation of Standard Deviation

47. X
A = 20
d= X – A d² 7
-13
169 10
-10
100 12 -8 64 13 -7 49 15 -5 25
20 = A 21 1 28 8 29 9 81 35
225
N = 10
∑d= -10
∑d² = 778
Example –Calculate standard deviation of the following series: 7, 10, 12, 13, 15, 20, 21, 28, 29, 35 Use assumed method Solution : Calculation of Standard Deviation

48. 2) Method based on Use of Actual data When the number of observations are few, standard deviations can be calculated by using the actual data. When this method is used, following formula is used :

49. Steps for Calculation
Firs we find the sum of the items, i.e. ∑ x
Then the values of x are squared up and added to get ∑ x²
Substitute the values in the above formula. The result will give the value of standard deviation.

50. X X² 16
256 20
400 18
324 19
361 28
784 17
289 22
484
N = 10, ∑x= 200
∑x²= 4098
Example –Calculate the standard deviation from the following : X = 16, 20, 18, 19, 20, 20, 28, 17, 22, 20 Solution : Calculation of Standard Deviation

51. DISCRETE SERIES
There are 3 methods to find standard deviation standard deviation is denoted by symbol= σ
Actual mean method:√∑fx²⁄N
Where, x=X-mean
2) Assumed mean method: √∑fd²/N-(∑fd/N)²
Where, d=X-A
3) Step deviation method: √∑fd’²/N-(fd’/N)²*i
Where d’=X-A/i and i= common factor

52. Step deviation method X:10,20,30,40,50,60,70 F: 3,5,7,9,8,5,3
SD= √fd’²/N-(fd’/N)²*i= √109/40-(1/40)²*10 = 16.5 X f
A=40
d=X-40 d’
fd’
fd’² 10 3
-30 -3 -9 27 20 5
-20 -2
-10 30 7 -1 -7 40 9 50 8 1 60 2 70
N=40
∑fd=1
∑fd’²=109

53. CONTINUOUS SERIES
In continuous series we can use any of three methods discussed above in discrete series. BUT step deviation method is commonly used in continuous series, its formula is:
SD=√fd’²/N-(fd’/N)²*i
Where, d’= m-A/i and i= size of class interval

54. EXAMPLE
MARKS
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
NO. OF ST. 5 10 20 40 30 4
Marks f
M.V
d=m-35 d’
fd’
fd’²
0-10 5
-30 -3
-15 45
10-20 10 15
-20 -2 40
20-30 20 25
-10 -1
30-40 35
40-50 30 1
50-60 55 2 80
60-70 65 3 90
70-80 4 75 16 64
N=139
∑=fd’=61
∑=fd’²=
-369

55. variance SD= √fd’²⁄N-(fd’/N)²*i = √369/139-(61/139)²*10= 15.69
Variance is the sqaure of standard deviation, i.e
variance= (S.D)²= σ²

56. COMBINED STANDARD DEVIATION
The combined standard deviation of two or more groups can be calculated by the formula :
₂ ₂ ₂ ₂
σ₁₂ = N₁σ₁ + N₂σ₂ + N₁d₁ + N₂d₂
N₁ + N₂

57. Find the combined mean and standard deviation of the two series.
Series A
Series B
Mean 50 40
Standard deviation 5 6
No. of items
100
150
Find the combined mean and standard deviation of the two series.
Solution:
X‾12 = N1X1 + N2 X2 / N1 + N2
= (100×50)+ (150×40))/( )=11,000/250=44
  σ12 = √ N1(σ12 + d12) + N2 (σ22 + d22) / N1+ N2
 d1 = (X‾1 - X‾12) = = 6
 d2 = (X‾2 - X‾12) = = 4
 σ12 = √(100 [ (52+ 62) ]+ 150 [62+ (-42 )]) / ( )))
  = √(( )/250)  = √(13,900/250)
 = √55.6=7.46

58. Correcting Mean and Standard Deviation
Quite often while tabulating data some values are wrongly entered. For example sometimes 15 may be misread as 51 or 159 may be misread as 59, and we may have used these wrong values in the calculation of the Mean and Standard Deviation. When the error is found out, there are two courses open to us-—either to make all calculations afresh, which is a very tedious task or to correct the values of Mean and Standard, deviation by adjustment and readjustment of wrong and right figures. This process is easy and takes much less time and saves a lot of botheration. We shall discuss these procedures in the examples given below:
Example : The mean and standard deviation of 20 items was found to be 10 and 2 respectively. Later it was found the item 12 was misread as 8. Calculate the correct mean and standard deviation.

59. Solution:
Given N =20, #=10, a=2 Wrong value used 8, correct value (i) Calculation of Correct Mean:
Given N = 20, X = 10, σ = 2   Wrong value used 8, correct value 12.
 X =  (∑X)/N or    X × N = ∑X   or   X × N = 10 x 20 = 200 = ∑X
This is the wrong value of 2X. The correct value would be
 200— = 204    Correct ∑X=204. As such
(ii) Calculations of correct standard deviation: Variance or σ2= (∑(x2)/N- (X2) The wrong value of ∑ (X2) = N (σ2- X2) The correct value of ∑(X2) =2080 — (8)2 + (12)2 = Now correct variance  = (∑(X2))/N- (X)2     = 2160/20- (10.2)2 The correct standard deviation = √(2160/20- (10.2)2)
 = √( )   = √3.96
=1.989

60. LORENZ CURVE
It is a graphical method of studying dispersion. Lorenz curve was given by famous statistician Max O Lorenz. Lorenz curve has great utility in the study of degree of inequality in the distribution of income and wealth between the countries. It is also useful for comparing the distribution of wages, profit etc over different business groups. Lorenz curve is a cumulative percentage curve in which the percentage of frequency is combined with percentage of other items such as income , profit, wages etc

61. EXAMPLE Income 100 200 400 500 800 No. of persons 80 70 50 30 20
Cumulative Income
Cumulative Percentage
No. of persons
Cumulative total
Cumulative percentage
100
100 *100 = 5
200 80
80*100=32
250
300
300*100=15
2000 70
150
150*100=60
400
700
700*100=35 50
200*100=80
500
1200
1200*100=60 30
230
230*100=92
800
2000*100=100 20
250*100=100

62. THANK YOU