©. S. Bhattacharjee Ch 5_1

©. S. Bhattacharjee Ch 5_2

Variability refers to the extent to which the observations vary from one another from some average. A measure of variation is designed to state the extent to which the individual measures differ on an average from the mean.
©. S. Bhattacharjee
Continued…..
Ch 5_3

Measures of variation are needed for four basic purposes:
To determine the reliability of an average;
To serve as a basis for the control of the variability;
To compare two or more series with regard to their variability;
To facilitate the use of other statistical measures
©. S. Bhattacharjee Ch 5_4

A good measure of variation should possess the following properties:
It should be simple to understand.
It should be easy to compute.
It should be rigidly defined.
It should be based on each and every observation of the distribution.
It should be amenable to further algebraic treatment.
It should have sampling stability.
It should not be unduly affected by extreme observations.
©. S. Bhattacharjee Ch 5_5

The following are the important methods of studying variation:
The Range
The Interquartile Range or Quartile Deviation.
The Average Deviation
The Standard Deviation
The Lorenz Curve.
Of these, the first four are mathematical and the last is a graphical one.
©. S. Bhattacharjee Ch 5_6

The range is defined as the distance between the highest and lowest scores in a distribution.
It may also be defined as the difference between the value of the smallest observation and the value of the largest observation included in the distribution.
©. S. Bhattacharjee Ch 5_7

Despite serious limitations range is useful in the following cases:
Quality control: Range helps check quality of a product.
The object of quality control is to keep a check on the quality of the product without 100% inspection.
Fluctuation in the share prices: Range is useful in studying the variations in the prices of stocks and shares and other commodities etc.
Weather forecasts: The meteorological department does make use of the range in determining the difference between the minimum temperature and maximum temperature.
©. S. Bhattacharjee Ch 5_8

Merits:
Among all the methods of studying variation, range is the simplest to understand and the easiest to compute.
It takes minimum time to calculate the value of range. Hence, if one is interested in getting a quick rather than a very accurate picture of variability, one may compute range.
©. S. Bhattacharjee Ch 5_9

Limitations:
Range is not based on each and every observation of the distribution.
It is subject to fluctuations of considerable magnitude from sample.
Range cannot be computed in case of open-end distributions.
Range cannot tell anything about the character of the distribution within two extreme observations.
©. S. Bhattacharjee Ch 5_10

Example: O bserve the following three series
Series A: 6,
Series B: 6
Series C: 6
46
6
10
46
6
15
46
6
25
46
46
30
46
46
32
46
46
40
46
46
46
In all the three series range is the same (i.e., 46-6=40), but it does not mean that the distributions are alike. The range takes no account on the form of the distribution within the range. Range is, therefore, most unreliable as a guide to the variation of the values within a distribution.
©. S. Bhattacharjee Ch 5_11

Inter-quartile range represents the difference between the third quartile and the first quartile. In measuring inter-quartile range the variation of extreme observations is discarded.
Continued…..
©. S. Bhattacharjee Ch 5_12

One quartile of the observations at the lower end and another quartile of the observations at the upper end of the distribution are excluded in computing the interquartile range. In other words, inter-quartile range represents the difference between the third quartile and the first quartile. Symbolically,
Inerquartile range = Q
3
– Q
1
Very often the interquartile range is reduced to the form of the semi-interquartile range or quartile deviation by dividing it by 2.
©. S. Bhattacharjee Ch 5_13

The formula for computing inter-quartile deviation is stated as under:
Q .
D .

Q
3

Q
1
2
Q.D. = Quartile deviation
Quartile deviation gives the average amount by which the two Quartiles differ from the median. In asymmetrical distribution, the two quartiles (Q
1 and Q
3
) are equidistant from the median, i.e., Median ± Q.D. covers exactly 50 per cent of the observations.
©. S. Bhattacharjee Ch 5_14

When quartile deviation is very small it describes high uniformity or small variation of the central 50% observations, and a high quartile deviation means that the variation among the central observations is large.
Quartile deviation is an absolute measure of variation.
The relative measure corresponding to this measure, called the coefficient of quartile deviation, is calculated as follows:

Q
1
Co-efficient of Quartile deviation

Q
3
Q
3
 Q
1
Coefficient of quartile deviation can be used to compare the degree of variation in different distributions.
©. S. Bhattacharjee Ch 5_15

The process of computing quartile deviation is very simple. It is computed based on the values of the upper and lower quartiles. The following illustration would clarify the procedure.
Example:
You are given the frequency distribution of 292 workers of a factory according to their average weekly income.
Calculate quartile deviation and its coefficient from the following data:
Continued…………
©. S. Bhattacharjee Ch 5_16

Example:
Weekly Income
Below 1350
1350-1370
1370-1390
1390-1410
1410-1430
1430-1450
1450-1470
1470-1490
1490-1510
1510-1530
1530 & above
No. of workers
8
16
39
58
60
40
22
15
15
9
10
Continued…………
©. S. Bhattacharjee Ch 5_17

Example:
Calculation of Quartile deviation
Weekly Income
Below 1350
1350-1370
1370-1390
1390-1410
1410-1430
1430-1450
1450-1470
1470-1490
1490-1510
1510-1530
1530 & above
No. of workers
8
16
39
58
60
40
22
15
15
9
10
N = 292
©. S. Bhattacharjee
Continued………… c.f.
8
24
63
121
181
221
243
258
273
282
292
Ch 5_18

Example:
Median = Size of
N th observatio n

2
Median lies in the class 1410 - 1430
292
2

146 th observatio n
N
 p .
c .
f
Medain

L
 2  i f


1410

1410

146
8

121

60

333

20
1418 .
333
Q
1
Q
1

Size lies in of
N th observatio n

4 the class 1390

1410 .
292

73 rd
4 observatio n
©. S. Bhattacharjee
Continued…………
Ch 5_19

Example:
Q
1

L

N
4
 f p .
c .
f .
 i

1390

73

58
63

20

1390

3

448

1393

448
Q
3

Size of
3 N th observatio n
4

3

292
4

219 th observatio n Q
3 lies in the class 1430

1449
Q
3

L

3 N
4
Coeffiecnt of
 f p .
c .
f .
 i

1430

Q .
D .

Q
3
Q
3

Q
1
 Q
1

219

40
181

20
1449

1393

448
1449

1393

448


1430
55


19
552
2842


1449
448

55552
2842

448

0

020 .
©. S. Bhattacharjee Ch 5_20

Merits:
In certain respects it is superior to range as a measure of variation
It has a special utility in measuring variation in case of open-end distributions or one in which the data may be ranked but measured quantitatively.
It is also useful in erratic or highly skewed distributions, where the other measures of variation would be warped by extreme value.
The quartile deviation is not affected by the presence of extreme values.
©. S. Bhattacharjee Ch 5_21

Limitations:
Quartile deviation ignores 50% items, i.e., the first 25% and the last 25%. As the value of quartile deviation does not depend upon every observation it cannot be regarded as a good method of measuring variation.
It is not capable of mathematical manipulation.
Its value is very much affected by sampling fluctuations.
It is in fact not a measure of variation as it really does not show the scatter around an average but rather a distance on a scale, i.e., quartile deviation is not itself measured from an average, but it is a positional average.
©. S. Bhattacharjee Ch 5_22

Average deviation refers to the average of the absolute deviations of the scores around the mean.
It is obtained by calculating the absolute deviations of each observation from median ( or mean), and then averaging these deviations by taking their arithmetic mean.
Continued…….
©. S. Bhattacharjee Ch 5_23

Ungrouped data
The formula for average deviation may be written as:
A .
D .
( Med .)


X

Med
N
If the distribution is symmetrical the average (mean or median) ± average deviation is the range that will include 57.5 per cent of the observation in the series. If it is moderately skewed, then we may expect approximately 57.5 per cent of the observations to fall within this range. Hence if average deviation is small, the distribution is highly compact or uniform, since more than half of the cases are concentrated within a small range around the mean.
©. S. Bhattacharjee Ch 5_24

Ungrouped data
The relative measure corresponding to the average deviation, called the coefficient of average deviation, is obtained, by dividing average deviation by the particular average used in computing average deviation.
Thus, if average deviation has been computed from median, the coefficient of average deviation shall be obtained by dividing average deviation by the median.
Coefficien t of A .
D .

Med .


A .
D .
Median
If mean has been used while calculating the value of average deviation, in such a case coefficient of average deviation is obtained by dividing average deviation by the mean.
©. S. Bhattacharjee Ch 5_25

Example:
Calculate the average deviation and coefficient of average deviation of the two income groups of five and seven workers working in two different branches of a firm:
Branch 1
Income (Tk)
4,000
4,200
4,400
4,600
4,800
Branch II
Income (Tk)
3,000
4,000
4,200
4,400
4,600
4,800
5,800
©. S. Bhattacharjee
Continued…..
Ch 5_26

Calculation of Average deviation
Branch 1
│X- Med 
Income (Tk) Med.=4,400
│X- Med 
Income (Tk) Med.= 4,400
4,000 400 3,000 1,400
4,200 200
4,400 0
4,600 200
4,800 400
N= 5
 │X- Med 
=1,200
4,000 400
4,200 200
4,400 0
4,600 200
4,800 400
5,800 1,400
N = 7
 │X- Med 
= 4000
Continued…..
©. S. Bhattacharjee Ch 5_27

Brach I:
Brach II:
Coeff .
of coeff .
of
A .
D .


X
N

Med .

1200
5

240
A .
D .
A .
D .
A .
D .


Median

X

N

240
4 , 400
Med .


0

054
4 , 000
7

571

43
A .
D .

57143

0

13
4 , 400
©. S. Bhattacharjee Ch 5_28

Grouped data
In case of grouped data, the formula for calculating average deviation is :
A .
D .
( Med .)

 f X

Med
N
©. S. Bhattacharjee
Continued………..
Ch 5_29

Example:
Calculation of Average Deviation from mean from the following data:
Sales
(in thousand Tk)
No. of days
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
3
6
11
3
2
Continued……..
©. S. Bhattacharjee Ch 5_30

Sales
(in thousand
Tk)
10 – 20
20 – 30
Calculation of Average deviation m.p
X
15
25 f
3
6

X

35

10
(=d)
–2
–1 fd
– 6
– 6
X

X
18
8
30 – 40 35 11 0 0 2
40 – 50
50 – 60
45
55
3
2
N = 25
+ 1
+ 2
+ 3
+ 4
 fd = –5
12
22 f
X

X
54
48
22
36
 f
44
X

X
= 204
Continued……
©. S. Bhattacharjee Ch 5_31

A .
D .


X

A
 f
N
X

N fd

X
 i

35

5
25

10

35

2

33
A .
D .

204
25

8

16
Thus the average sales are Tk. 33 thousand per day and the average deviation of sales is Tk. 8.16
thousand.
©. S. Bhattacharjee Ch 5_32

It is especially effective in reports presented to the general public or to groups not familiar with statistical methods.
This measure is useful for small samples with no elaborate analysis required.
Research has found in its work on forecasting business cycles, that the average deviation is the most practical measure of variation to use for this purpose.
©. S. Bhattacharjee Ch 5_33

Merits:
The outstanding advantage of the average deviation is its relative simplicity. It is simple to understand and easy to compute.
Any one familiar with the concept of the average can readily appreciate the meaning of the average deviation.
It is based on each and every observation of the data.
Consequently change in the value of any observation would change the value of average deviation.
©. S. Bhattacharjee Ch 5_34

Merits:
Average deviation is less affected by the values of extremes observation.
Since deviations are taken from a central value, comparison about formation of different distributions can easily be made.
©. S. Bhattacharjee Ch 5_35

Limitations:
The greatest drawback of this method is that algebraic signs are ignored while taking the deviations of the items. If the signs of the deviations are not ignored, the net sum of the deviations will be zero if the reference point is the mean, or approximately zero if the reference point is median.
The method may not give us very accurate results.
The reason is that average deviation gives us best results when deviations are taken from median. But median is not a satisfactory measure when the degree of variability in a series is very high.
©. S. Bhattacharjee
Continued…….
Ch 5_36

Limitations:
Compute average deviation from mean is also not desirable because the sum of the deviations from mean ( ignoring signs) is greater than the sum of the deviations from median (ignoring signs).
If average deviation is computed from mode that also does not solve the problem because the value of mode cannot always be determined.
It is not capable of further algebraic treatment.
It is rarely used in sociological and business studies.
©. S. Bhattacharjee
Continued…….
Ch 5_37

Standard deviation is the square root of the squared deviations of the scores around the mean divided by
N. S represents standard deviation of a sample; ∂, the standard deviation of a population.
Standard deviation is also known as root mean square deviation for the reason that it is the square root of the means of square deviations from the arithmetic mean.
The formula for measuring standard deviation is as follows :
 
 
X

X

2
N
©. S. Bhattacharjee Ch 5_38

This refers to the squared deviations of the scores around the mean divided by N.
A measure of dispersion is used primarily in inferential statistics and also in correlation and regression techniques; S 2 represents the variance of a sample ; ∂ 2 , the variance of a population.
If we square standard deviation, we get what is called
Variance.
  2
 
©. S. Bhattacharjee Ch 5_39

Ungrouped data
Standard deviation may be computed by applying any of the following two methods:
By taking deviations from the actual mean
By taking deviations from an assumed mean
Continued……..
©. S. Bhattacharjee Ch 5_40

Ungrouped data
By taking deviations from the actual mean:
When deviations are taken from the actual mean, the following formula is applied:
 
 
X

X

2
N
If we calculate standard deviation without taking deviations, the above formula after simplification
(opening the brackets) can be used and is given by:
Continued……..
©. S. Bhattacharjee Ch 5_41

 

X
2
N
 


 
X
N



 2 or
 

X
2
N

 
2
By taking deviations from an assumed mean: When the actual mean is in fractions, say 87.297, it would be too cumbersome to take deviations from it and then find squares of these deviations. In such a case either the mean may be approximated or else the deviations be taken from an assumed mean and the necessary adjustment be made in the value of standard deviation.
©. S. Bhattacharjee Ch 5_42

The former method of approximation is less accurate and therefore, invariably in such a case deviations are taken from assumed mean.
When deviations are taken from assumed mean the following formula is applied:
 
 d
2
N
 


  d
N



 2
Where d


X

A

©. S. Bhattacharjee Ch 5_43

Example:
Find the standard deviation from the weekly wages of ten workers working in a factory:
D
E j
I
F
G
H
Workers
A
B
C
©. S. Bhattacharjee
Weekly wages (Tk)
1320
1310
1315
1322
1326
1340
1325
1321
1320
1331
Ch 5_44

Workers
A
B
C
D
E
F
G
H
I
J
N= 10
Calculations of Standard Deviation
Weekly wages
(Tk)
1320
1310
1315
1322
1326
1340
1325
1321
1320
1331
 x=13230

X

X

+17
+2
- 2
- 3
- 3
- 13
- 8
- 1
+3
+8


X

X


X

X

2
1
9
9
169
64
289
4
4
9
64

X

X

2
= 622
©. S. Bhattacharjee Ch 5_45
Continued…….

We know
Since
 
 
X

X

2
..........
.....( i )
N
 d
X


X

A
 d
A
X

A
 d
Subtractin g X from X , we get

X

X


 d
 d

Continued…….
©. S. Bhattacharjee Ch 5_46

 

X


  d
 d

2
N
X


X
N

 d
2
N

 d
N

13230
10

Tk .
1323
2
 
 
X

X

2
N

622

10
62

7

89
.
2
If, in the above question, deviations are taken from 1320 instead of the actual mean 1323, the assumed mean method will be applied and the calculations would be as follows:
Continued…….
©. S. Bhattacharjee Ch 5_47

Calculation of standard deviation (assumed mean method )
Workers Weekly wages
(Tk)

X

A

 d
A = 1320 d 2
C
D
E
A
B
F
G
H j
I
N= 10
©. S. Bhattacharjee
1320
1310
1315
1322
1326
1340
1325
1321 +1
1320
1331
0
+11
 d=30
Continued…….
0
-10
-5
+2
+6
+20
+5
0
100
25
4
36
400
25
1
0
121
 d 2 =712
Ch 5_48

 
 d
2
N
 


  d
N



 2

712
10




30
10 

 2

71

2

9

62

2

7

89
Thus the answer remains the same by both the methods. It should be noted that when actual mean is not a whole number, the assumed mean method should be preferred because it simplifies calculations.
©. S. Bhattacharjee Ch 5_49

Grouped data
In grouped frequency distribution, standard deviation can be calculated by applying any of the following two methods:
By taking deviations from actual mean.
By taking deviations from assumed mean.
©. S. Bhattacharjee
Continued……..
Ch 5_50

Grouped data
Deviations taken from actual mean: When deviations are taken from actual mean, the following formula is used:
 
 f

X

X
2

N
If we calculate standard deviation without taking deviations, then this formula after simplification
(opening the brackets ) can be used and is given by
 
 fX
2
N


N fX
2 or
 
 fX
2
N

 
2
Continued……..
©. S. Bhattacharjee Ch 5_51

Grouped data
Deviations taken from assumed mean: When deviations are taken from assumed mean, the following formula is applied :
 
 fd
2
N
 


 
N fd



 2
 i , d
 x

A i where
Continued……..
©. S. Bhattacharjee Ch 5_52

Example:
A purchasing agent obtained samples of 60 watt bulbs from two companies. He had the samples tested in his own laboratory for length of life with the following results:
Length of life (in hours)
1,700 and under 1,900
1,900 and under 2,100
2,100 and under 2,300
2,300 and under 2,500
2,500 and under 2,700
Samples from
Company A Company B
10 3
16 40
20 12
8 3
6 2
Continued……..
©. S. Bhattacharjee Ch 5_53

Example:
1. Which Company’s bulbs do you think are better in terms of average life?
2. If prices of both types are the same, which company’s bulbs would you buy and why?
©. S. Bhattacharjee
Continued……..
Ch 5_54

Example:
Length of life
(in hours)
Midpoint Sample from Co. A
1,700– 1,900
1,900–2,100
2,100–2,300
2,300–2,500
1800
2000
2200
2400 f
10
16
20
8 d
–2
–1
0
1 fd
–20
–16
0
+8
2,500–2,700 d

X
2600 6 2 +12
N=60 d=0
 fd
= –16 i

A
, where

Assumed
Here, A = 2200 i = 200 mean fd 2
40
16
0
8
24
 fd 2
=88
Samples from Co. B f
3
40
12 d
–2
–1
0 fd
– 6
– 40
0
3 1 +3
2 2 +4
N=60 d=0
 fd
= –39 fd 2
12
40
0
3
8
 fd 2
= 63
Continued……..
©. S. Bhattacharjee Ch 5_55

Example:
Here, N = 60
For Company A :
A = 2,200
 fd = - 16
 fd 2 = 88
X
 

A


N fd
 i

2 , 200

16
60

200

2 , 146

67

N fd
2


N fd
2
 i

88
60

16
60
2

200

C .
V .

1

467

0

071

200

X

100


1

182

200
236

4
2146

67

100

11 per

236

4 cent .
©. S. Bhattacharjee
Continued……..
Ch 5_56

For Company B :

 fd
 
39 fd
2 
63
A

2200
X

2200

39
60

200

2200

130

2070
 
63
60

39
60
2

200

1

05
 
42

200
 
794

200

158

8
C .
V .

158

8
2070

100

7

67 per cent .
Continued……..
©. S. Bhattacharjee Ch 5_57

Illustration :18
You are given the data pertaining to kilowatt hours electricity consumed by 100 persons in Deli.
Consumption (K. Wait hours)
0 but less than 10
10 but less than 20
20 but less than 30
30 but less than 40
40 but less than 50
No. of users
6
25
36
20
13
Calculate the mean and the standard deviation.
©. S. Bhattacharjee
Continued……..
Ch 5_58

Solution:
Calculation of Mean and standard Deviation (Taking deviation from assumed mean)
Consumption
K. wait hours
0–10
10–20
20–30
30–40
40–50 m.p
(X)
5
15
25
35
45
No. of Users
(f)
6
25
36
20
13
N=100

X

25

10 d
–2
–1
0
+1
+2 fd fd 2
–12
–25
0
+20
24
25
0
20
+26 52
 fd =9
 fd 2 =121 c.f.
6
31
67
87
100
©. S. Bhattacharjee
Continued……..
Ch 5_59

(i)
X

A


N fd
 i

25

9
100

10

25 .
9 k .
wait hours
(ii)
 

 fd
2
N
 


 
N fd



 2
 i

121
100




9
100


 2

10
1 .
21

.
008

10

1 .
096

10

10 .
96
©. S. Bhattacharjee Ch 5_60

Calculation of Mean and the Standard Deviation
Consumption
K. wait hours
0–10
10–20
20–30
30–40
40–50 nt
X
(Taking deviation from assumed mean)
Midpoi No. of
Users f fX

X

X

2

X

X

5
15
25
6
25
36
30 –2 436.81
375 –10.9
118.81
900
–0.9
0.81
35
45
20 700
13 585
N=100
 fX=
2590
9.1
82.81
19.1
364.81
f

X

X

2
2620.6
2970.25
29.16
1656.20
4742.53
 f

X

X

2 
12019
Continued……..
©. S. Bhattacharjee Ch 5_61

X


N fX

2590
100

25 .
9
   f

X

X

2
N

12019
100

120 .
19

10 .
96
©. S. Bhattacharjee Ch 5_62

1. Since average length of life is greater in case of company A, hence bulbs of company A are better.
2. Coefficient of variation is less for company B.
Hence if prices are same, we will prefer to buy company B’s bulbs because their burning hours are more uniform.
©. S. Bhattacharjee Ch 5_63

Example:
For two firms A and B belonging to same industry, the following details are available:
Firm A
Firm B
Number of Employees
100
200
Average monthly wage:
Tk. 4,800
Tk. 5,100
Standard deviation:
Tk. 600
Tk. 540
Find i. Which firm pays out larger amount as wages?
ii. Which firm shows greater variability in the distribution of wages?
iii. Find average monthly wage and the standard deviation of the wages of all employees in both the firms.
©. S. Bhattacharjee Ch 5_64

i. For finding out which firm pays larger amount, we have to find out

X.
X


N
X or

X

N X
Firm A : N = 100, X = 4800,
 
X=100 ×4800 =4,80,000
Firm B : N = 200, X
= 5100,
 
X=200 ×5100 =10,20,000
Hence firm B pays larger amount as monthly wages.
©. S. Bhattacharjee Ch 5_65

ii. For finding out which firm pays greater variability in the distribution of wages, we have to calculate coefficient of variation.
Firm A :
Firm B :
C .
V .


X

100

600
4800

100

12

50
C .
V .


X

100

540
5100

100

10

59
Since coefficient of variation is greater in case of firm A, hence it shows greater variability in the distribution of wages.
©. S. Bhattacharjee Ch 5_66

iii. Combined average weekly wage:
X
12

N
1
X
1
N
1


N
2
X
2
N
2
N
1
= 100, X
1
= 4800,
N
2
= 200, X
2
= 5100,
X
12


 
4800
  
5100

480000
100


200
1020000

300
TK .
5 , 000
©. S. Bhattacharjee Ch 5_67

Combined Standard Deviation

12

N
1

1
2

N
2

2
2
N
1


N
1
N
2 d
1
2

N
2 d
2
2
N
1

100 ,

1

600 , N
2

200 ,

2

540
X
12
= 4800 - 5000 =200 d
2
= X
2
 X
12
= 5100 - 5000 =100
Continued…….
©. S. Bhattacharjee Ch 5_68


12



100
 
2 
200
2 
100
100

200
2 
200
36000000

5832000

4000000

2000000
300
100

320 , 000

578 .
25
300
2
Hence the combined standard deviation is Tk. 578.25.
©. S. Bhattacharjee Ch 5_69

The choice of a suitable measure depends on the following two factors:
The type of data available
The purpose of investigation
©. S. Bhattacharjee Ch 5_70

It is cumulative percentage curve in which the percentage of items is combined with the percentage of other things as wealth, profit, turnover, etc. The Lorenz curve is a graphic method of studying variation.
©. S. Bhattacharjee Ch 5_71

While drawing the Lorenz curve the following procedure is used:
The size of items and frequencies are both cumulated and the percentages are obtained for the various commutative values.
On the X-axis, start from 0 to 100 and take the per cent of variable.
On the Y-axis, start from 0 to 100 and take the per cent of variable.
Continued…..
©. S. Bhattacharjee Ch 5_72

Draw a diagonal line joining 0 with 100. This is known as line of equal distribution. Any point on this line shows the same per cent on X as on Y.
Plot the various points corresponding to X and Y and join them. The distribution so obtained, unless it is exactly equal, will always curve below the diagonal line.
©. S. Bhattacharjee Ch 5_73

If two curves of distribution are shown on the Lorenz presentation, the curve that is farthest from the diagonal line represents the greater inequality. Clearly the line of actual distribution can never cross the line of equal distribution.
©. S. Bhattacharjee Ch 5_74

Example:
In the following table is given the number of companies belonging to two areas A and B according to the amount of profits earned by them. Draw in the same diagram their
Lorenz curves and interpret them.
Profits earned in Tk.'000
105
150
170
400
6
25
60
84
©. S. Bhattacharjee
No. of companies
Area A Area B
6
11
13
14
2
38
52
28
15
17
10
14
38
26
12
4
Continued…… Ch 5_75

Solution:
Calculation for drawing the Lorenz curve
6
25
60
84
105
150
170
400
Profits earned in
Tk. ‘000
6
31
91
175
280
430
600
1000
Profit
Cumulativ e
Profits
Cumulativ e
Percentag e
0.6
3.1
9.1
17.5
28.0
43.0
60.0
100.0
No. of
Companies
6
11
13
14
15
17
10
14
Cumulative
Number
6
17
30
44
59
76
86
100
Cumulative
Percentage
6
17
30
44
59
76
86
100
No. of
Companies
2
38
52
28
38
26
12
4
Cumulative
Number
Cumulative
Percentage
2
40
92
120
158
184
196
200
1
20
46
60
79
92
98
100
©. S. Bhattacharjee
Continued……..
Ch 5_76

120.0
100.0
80.0
60.0
40.0
20.0
0.0
Lorenz curve
©. S. Bhattacharjee
Per Cent of Companies
Ch 5_77

©. S. Bhattacharjee Ch 5_78