The document provides information about various measures of central tendency including arithmetic mean, median, mode, geometric mean, and harmonic mean. It defines each measure and provides examples of calculating them using data from frequency distributions. The arithmetic mean is the most common average and is calculated by summing all values and dividing by the total number of values. The median is the middle value when values are arranged in order. The mode is the most frequent value. The geometric mean is calculated by taking the nth root of the product of n values. The harmonic mean gives the greatest weight to the smallest values and is used to average rates.

1. Measures of Central
Tendency
Mod 5

2. Average
› A single significant figure, which sums up the
characteristics of a group of figures.
› It conveys the general idea of the whole group
› It is generally located at the center or middle of the
distribution

3. Various Measures of Central tendency are;
› Arithmetic Mean
› Median
› Mode
› Geometric Mean
› Harmonic Mean

4. 1. Arithmetic Mean
› It is a Mathematical average
› It represents the whole data using a single figure
› It is the simplest and most widely used measure
›
Mean in Individual Series
Mean of a Discrete freq. distribution
Mean of a continuous freq. distribution

5. Mean in an Individual Series
› If the individual values in a set of variables are,
x1, x2, x3……..xn, then the Arithmetic mean will be;

6. Mean of a discrete frequency distribution
› Let the discrete variables be x1, x2, x3…….xn
› Let the corresponding frequencies be f1, f2, f3……fn
› Then the arithmetic mean = Σfx
N
(N= Σf)

7. Mean in Continuous Frequency distribution
› Here, you have to consider the mid values of each class
as ‘x’
› Then the same formula of discrete frequency distribution
can be applied.
Class Frequency (f) Mid point of class
(x)
fx
0-10 14 5 14×5 = 70
10-20 18 15 18 ×15 = 270
Σf = N Σfx

8. Find the arithmetic mean.
Temperature Number of days
-40 to -30 10
-30 to -20 28
-20 to -10 30
-10 to 0 42
0 to 10 65
10 to 20 180
20 to 30 10

9. › Arithmetic Mean = 1565/365 = 4.28 degree Celsius
Temperature Number of days (f) Mid Point (x) fx
-40 to -30 10 -35 -350
-30 to -20 28 -25 -700
-20 to -10 30 -15 -450
-10 to 0 42 -5 -210
0 to 10 65 5 325
10 to 20 180 15 2700
20 to 30 10 25 250
N = 365 Σfx = 1565

10. Find the AM
Marks Number of Students
20-29 10
30-39 8
40-49 6
50-59 4
60-69 2

11. › Arithmetic Mean = 1135/30 = 37.83
Marks Number of
Students (f)
Midpoint (x) fx
20-29 10 24.5 245
30-39 8 34.5 276
40-49 6 44.5 267
50-59 4 54.5 218
60-69 2 64.5 129
N = 30 Σfx = 1135

12. Find the AM
Marks Number of Students
Less than 10 5
Less than 20 17
Less than 30 31
Less than 40 41
Less than 50 49

13. › Am = 1265/49 = 25.81
Marks Number of
Students
Class New
frequency (f)
Midpoint (x) fx
Less than 10 5 0-10 5 5 25
Less than 20 17 10-20 12 15 180
Less than 30 31 20-30 14 25 350
Less than 40 41 30-40 10 35 350
Less than 50 49 40-50 8 45 360
N = 49 Σfx = 1265

14. 2. Median
› Median is the value of that item which occupies the
central position, when the items are arranged in
ascending or descending order of their magnitude.
› Therefore, Median is the value of that item, which has
equal number of items above and below.
› Hence Median is a Positional average.

15. Median of Individual Series
› Arrange the items in ascending or descending order
› Then take the (n+1)/2th item.
› When ‘n’ is even, there will be two middle terms, then the
median will be the average of them.

16. Median of Discrete frequency distribution
› Here, median will be the size of (N+1)/2th item.
› Here N = Σf
› Mode = N+1/2th item = 56/2 = 28th Item.
Age Frequency Cumulative Frequency
2 14 14
6 22 36
10 19 55
N (Σf) = 55

17. Find the Median
Marks Frequency CF
5 3
8 12
10 8
15 7
20 5
25 4

18. Median of a Continuous Frequency
Distribution
› The first step is to find the Median class
› Median class corresponds to the cumulative frequency
which includes N/2
› Here also, N = Σf
› Once the Median class is found, the Median has to be
calculated using the interpolation formula.

19. Class Frequency C.F
10-30 8 8
30-50 6 14
50-70 12 26
N=26
› Median = 26/2th item, that is 13th item.
› 13 is included in the CF 14, which means Median class is
30-50
› Median =

20. › L1 = Lower class limit of the median class
› Cf = Cumulative frequency of the class just proceeding
the median class
› f = Frequency of the median class
› C = Interval of the Median class

21. Calculate the Median
Class Frequency CF
0-10 8
10-20 12
20-30 20
30-40 23
40-50 18
50-60 7
60-70 2

22. Calculate the Median
Class Frequency
10-20 4
20-40 10
40-70 26
70-120 8
120-140 2

23. 3. Mode
› Mode is the value of the item of a series, which occurs
most frequently.
› When no item appears to have repeating more than
others, the mode is considered as ‘ill defined’
› In such cases, mode is found using the formula,
Mode = 3 Median - 2 Mean

24. Mode in Individual Series
› Consider the most repeated item in the given series.

25. Mode in Discrete frequency distribution
› Mode will be that particular item with the maximum
frequency.

26. Mode in Continuous Frequency Distribution
› First step is to find the class with the maximum
frequency.
› The class with the maximum frequency is then termed as
Model Class
› From the Model class, you have to calculate the Mode
using interpolation formula.

27. › L1 = Lower limit of the Model Class
› F1 = Frequency of the Model class
› Fo = Frequency of the class just preceding the Model
class
› F2 = Frequency of the class just succeeding the Model
class

28. Calculate Mode
Wages No: of Workers
0-10 15
10-20 20
20-30 25
30-40 24
40-50 12
50-60 31
60-70 71
70-80 52

29. Calculate Mean, Median & Mode
Class Frequency
10-15 4
15-20 8
20-25 18
25-30 30
30-35 20
35-40 10
40-45 5
45-50 2

30. Class Frequency Mid Point (X) fx CF
10-15 4 12.5 50 4
15-20 8 17.5 140 12
20-25 18 22.5 405 30
25-30 30 27.5 825 60
30-35 20 27.5 650 80
35-40 10 32.5 375 90
40-45 5 37.5 212.5 95
45-50 2 42.5 95 97
N = 97 Σfx = 2752.5

31. Answers
› AM = 28.37
› Median Class = 25-30
› Median = 28.08
› Model Class = 25-30
› Mode = 27.72

32. Find
› The mean marks obtained by all 50 boys in a class is 40
and mean marks of all 30 girls of the same class is 46.
Find the mean marks of the class.

33. Find
› Average monthly production of a certain factory for the
first 9 months is 2584 units and for the remaining 3
months is 2416 units. Calculate the average monthly
production for the year.

34. Find
› The mean weight of 150 students in a certain class is 60
Kgs. The mean weight of boys and girls in the class is 70
and 55 kgs respectively. Find the number of boys and
girls in the class.

35. Finding Log Values
› Step 1
› Consider log (12.69)
Finding the number of digits on the left hand side of the
decimal point.
Eg: 12.69 (Number of digits to the left of the decimal point is 2)
So here, n = 2

36. Step 2
› Find n-1
› Here, in case of 12.69, n=2 and n-1 = 1
› This means that the value of log(12.69) would start with 1

37. 00

38. Step 3
› Consider the first two digits of the number (12.69) and
plot the same in the vertical bar of the Log chart
› Here, the first two numbers is 12

40. Step 4
› Consider the 3rd digit of the number and plot the
corresponding value in the horizontal bar of the chart
› Here the number is 12.69, so 3rd number is 6
› Find the value in the chart which corresponds to the
intersection of both the vertical and horizontal figures.
› That is the number corresponds to 12 and 6

43. › The intersection value = 1004

44. Step 5
› Consider the fourth digit of the number and find its
corresponding value in the mean difference column of the
chart. (Number = 12.69, 4th digit is 9)
› Find the intersection point of the vertical column and the
mean difference value.
› Add the corresponding mean difference value to the
previous intersection figure (Here 1004).
› Write the resulting number after the decimal point

46. › So the value now becomes 1004 + 31 = 1035
› Therefore log (12.69) = 1.1035

47. Find the log values for the following;
› 57.5
› 87.75
› 53.5
› 73.5
› 81.75

48. Calculating the Antilog Values.
› Consider Antilog (2.8675)
› Step one is to calculate the Antilog value corresponds to
.8675 using Antilog table (same procedures apply as in
the case of Log)
› The resultant value is 7362 + 8 = 7370

49. Step 2
› Step two is to find where to put the decimal point
› Consider the number to the left of decimal point, which is
‘2’ in this case (2.8675).
› Add 1 to that number. (2+1 = 3)
› This mean that you have to put the decimal point after
the first 3 digits.
› So AL (2.8675) = 737.0

50. Geometric Mean
› If there are ‘n’ values in a series, then their GM is defined
as the n’th root of the product of those n values
› It is a mathematical average

51. Geometric Mean in Individual Series
This can be expressed in another form as;

52. Calculate the geometric mean for the
following figures
› 57.5, 87.75, 53.5, 73.5, 81.75

53. Geometric Mean in Discrete/Continuous
Series

54. Find GM
Size Frequency
5 2
8 3
10 4
12 1

55. Find GM
Weight Number
100-104 24
105-109 30
110-114 45
115-119 65
120-124 72
125-129 84
130-134 124
135-139 58

56. Find GM
Marks Number of Students
0-10 5
10-20 7
20-30 15
30-40 25
40-50 8

57. Marks f Mid Point (X) Log (x) F × log
(x)
0-10 5 5
10-20 7 15
20-30 15 25
30-40 25 35
40-50 8 45
60

58. Marks f Mid Point (X) Log (x) F × log
(x)
0-10 5 5 0.6990 3.495
10-20 7 15 1.1761 8.2327
20-30 15 25 1.3979 20.9685
30-40 25 35 1.5441 38.6025
40-50 8 45 1.6532 13.2256
60 84.5243

59. Merits and Limitations of GM
› Helps in averaging ratios and percentages
› Not affected by extreme values
› Difficult to understand
› Not applicable on negative values

60. Harmonic Mean
› Harmonic Mean of a set of n Values is defined as the
reciprocal of the mean of the reciprocals of those values
› Harmonic Mean is used when it is desired to give the
greatest weight to the smallest items.
› It is predominantly used in averaging rates.

61. Harmonic Mean – Individual Series

62. Harmonic Mean in Individual Series

63. Find Harmonic Mean for the following
› 2,3,4,5

64. Find HM
X 1/X
2 0.50
3 0.33
4 0.25
5 0.20
Σ 1.28

65. Find Harmonic Mean for the following
› 8,14,16,10,24,18

66. X 1/X
8 0.125
14 0.071
16 0.062
10 0.1
24 0.042
18 0.055
0.455
n/0.455 = 6/0.455
= 13.18

67. HM in a Frequency Distribution

68. Size Frequency
6 20
10 40
14 30
18 10
Find HM

69. Size (X) Frequency (f) 1/X f (1/x)
6 20 0.1667 3.334
10 40 0.1000 4.000
14 30 0.0714 2.142
18 10 0.0556 0.556
N = 100 Σ = 10.032
HM = 100/10.032
= 9.97

70. Class Frequency
10-20 4
20-30 6
30-40 10
40-50 7
50-60 3
Find HM

71. Class Frequency Mid Point
(x)
1
x
f× 1
x
10-20 4 15 0.067 0.268
20-30 6 25 0.040 0.240
30-40 10 35 0.029 0.290
40-50 7 45 0.022 0.154
50-60 3 55 0.018 0.054
30 1.006
HM = 30/1.003
= 29.82

72. Find GM & HM
Weight Frequency
10-20 11
20-30 19
30-40 9
40-50 10
50-60 17
60-70 16
70-80 15
80-90 23

73. Weight Frequency Mid Point (X) Log(x) F × log (x)
10-20 11 15 1.1761 12.94
20-30 19 25 1.3971 26.54
30-40 9 35 1.5441 13.90
40-50 10 45 1.6532 16.53
50-60 17 55 1.7404 29.59
60-70 16 65 1.8129 29.00
70-80 15 75 1.8751 28.13
80-90 23 85 1.9294 44.38
120 201.01
GM = AL (201.01/120)
= AL (1.6751) = 47.33

74. Measures of Dispersion
(Measures of Variability)

75. Dispersion
 Dispersion refers to the variability in the size of items.
 It represents the spread or scatter of the value in a
series.
 Measures of dispersion measure the variability in a
series
 It tell us the extent to which the values of a series differ
from their average or among themselves.

76. Measures of
Dispersion
Absolute
Measures
Relative
Measures

77. Absolute Measures of Dispersion
1. Range
2. Quartile Deviation
3. Mean Deviation
4. Standard Deviation

78. 1. Range
› Range is the simplest possible measure of dispersion
› It is the difference between the highest and lowest values
in a series
› Range of an individual series ;
= Highest Value- Lowest Value (H-L)
› Range measures the maximum variation in the values of a
series.

79. Coefficient of Range
› Coefficient of Range is the relative measure based on
range.
› Coefficient of Range =
𝑯−𝑳
𝑯+𝑳

80. Uses of Range
› Used to study Data, for which variations are low.
› Application in Diagnosing
› Applications in Quality Control
› Applications in Weather forecasting etc

81. 2. Quartile Deviation (Semi Interquartile range)
› Defined as half the distance between the third and first
Quartiles
› Quartile Deviation =
𝑸𝟑−𝑸𝟏
𝟐
› Coefficient of Quartile Deviation =
𝑸𝟑−𝑸𝟏
𝑸𝟑+𝑸𝟏

82. QD in Individual Series
› First, arrange the variables in the ascending order
› Q1= 𝑺𝒊𝒛𝒆 𝒐𝒇
𝒏+𝟏
𝟒
th item
› Q3= 𝑺𝒊𝒛𝒆 𝒐𝒇
𝒏+𝟏
𝟒
×3th item
› Quartile Deviation =
𝑸𝟑−𝑸𝟏
𝟐

83. Problem 1
› Find the QD & Coefficient of QD
A) 23, 25, 8, 10, 9, 29, 45, 85, 10, 16, 24

84. QD in discrete frequency distribution
Q1= 𝑺𝒊𝒛𝒆 𝒐𝒇
𝑵+𝟏
𝟒
th item (N= Σf)
Q3= 𝑺𝒊𝒛𝒆 𝒐𝒇
𝑵+𝟏
𝟒
×3th item
Quartile Deviation =
𝑸𝟑−𝑸𝟏
𝟐

85. P1: Find QD
Size Frequency
5 3
8 10
10 15
12 20
19 8
20 7
32 6

86. P2: Find QD
Age Number
57 15
59 20
60 32
61 35
62 33
63 22
64 20
66 10
68 8

87. QD in Continuous frequency Distribution
Q1= 𝑺𝒊𝒛𝒆 𝒐𝒇
𝑵
𝟒
th item (N= Σf)
Q3= 𝑺𝒊𝒛𝒆 𝒐𝒇
𝑵
𝟒
×3th item
Quartile Deviation =
𝑸𝟑−𝑸𝟏
𝟐

88. Interpolation formula for calculating QD
Q1=
Q3=

89. P1: Find QD
Class Frequency
0-10 15
10-20 30
20-30 53
30-40 75
40-50 100
50-60 110
60-70 115
70-80 125

90. P2 Find QD & Coefficient of QD
Class Frequency
10-19 5
20-29 8
30-39 17
40-49 29
50-59 30
60-69 20
70-79 10
80-89 1

91. P2: Find QD & Coefficient of QD
Wages Workers
Below 5 4
Below 10 10
Below 15 13
Below 20 21
Below 25 33
Below 30 40

92. 3. Mean Deviation
› Mean deviation is defined as the Arithmetic mean of the
deviations of all the values in a series from their average.
› For the purpose of the calculation of MD, all the
deviations are considered positive irrespective of their
sign.
│𝒅│= deviations from the average, without sign
Mean Deviation =
𝜮 │𝒅│
𝒏

93. › Coefficient of Mean Deviation =
𝑴𝒆𝒂𝒏 𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
𝑨𝒗𝒆𝒓𝒂𝒈𝒆

94. Mean Deviation in an Individual Series
› P1: Find the MD from Mean for the following values;
25, 63 85, 75, 62, 70, 83, 28, 30, 12

95. Mean Deviation in an Individual Series
› P2: Find the MD from the Median for the following values;
5, 28, 33, 44, 83, 87, 96, 99, 25, 35, 82

96. Mean Deviation in Discrete Frequency Distribution
Mean Deviation =
𝜮 𝒇│𝒅│
𝑵

97. P1: Find MD from Mean & Coefficient of MD
No. of Children (x) 0 1 2 3 4 5 6
No. of families (f) 171 82 50 25 13 7 2

98. P2: Find MD from mean & Coefficient of MD
X 10 11 12 13 14
Frequency 3 12 18 12 3

99. MD in Continuous Frequency Distribution
Mean Deviation =
𝜮 𝒇│𝒅│
𝑵

100. Find MD about Arithmetic Mean
Marks
(x)
0-10 10-20 20-30 30-40 40-50 50-60 60-70
Freq.
(f)
4 6 10 20 10 6 4

101. X F
0-10 18
10-20 16
20-30 15
30-40 12
40-50 10
50-60 5
60-70 2
70-80 2
Calculate the MD about Median

102. 4. Standard Deviation
Standard Deviation is the square root of the mean of the
squares of the deviations of all the values of a series from
their Arithmetic Mean
It is calculated as the square root of variance by
determining the variation between each data point
relative to the mean.
If the data points are further from the mean, there is a
higher deviation within the data set; thus, the more
spread out the data, the higher the standard deviation.

103. SD in Individual Series
SD =
Coefficient of Variance =
𝑺𝑫
𝑴𝒆𝒂𝒏
× 𝟏𝟎𝟎

104. Find the Variance and Standard Deviation
 5,8,7,11,9,10,8,2,4,6

105. SD in Discrete/Continuous Frequency Distribution
SD =

106. Marks Number of
Students
2 8
4 10
6 16
8 9
10 7
Find the Variance and Standard Deviation

107. Class Frequency
0-2 2
2-4 4
4-6 6
6-8 4
8-10 2
10-12 6
Find the Variance and Standard Deviation

108. From the above table, showing the runs scored by
two batsmen in their last 8 innings’, find out who is
more consistent and who is more efficient.
Batsman A 10 12 80 70 60 100 0 4
Batsman B 8 9 7 10 5 9 10 8

109. Measures of Skewness

110. Symmetry

111. Symmetric Distributions
› A frequency distribution is said to
be Symmetric, if the frequencies are
distributed symmetrically or evenly
on either side of the average.
› In other words, the number of items
above the mean and below the
mean would be the same.
› For such distributions, Q3and Q1
would be equidistant from median

112. Skewness
› Skewness means lack of Symmetry.
› The word skewness literally denotes assymetry.
› If a frequency distribution is Skewed, there will be more
items on one side of the average than the other side.
› Such distributions will have a long tail on one side and a
shorter one on the other side.
› Most of the Economic data have skewed distributions.

113. Positive Skewness
Skewness is said to be positive, when Mean is greater
than Median and Median is greater than Mode.
( Mean ˃ Median ˃ Mode )
Here, the curve is skewed to the right.
More than half the area falls to the right side of the
highest ordinate.

114. Negative Skewness
Skewness is said to be Negative, when Mean is less than
Median and Median is less than Mode.
( Mean ˂ Median ˂ Mode )
Here, the curve is skewed to the left.
More than half the area falls to the left side of the
highest ordinate.

116. What is Skewness?

117. X F FX
1 2 2
2 4 8
3 6 18
4 8 32
5 6 30
6 4 24
7 2 14
0
1
2
3
4
5
6
7
8
9
0 2 4 6 8
Mean
Median
Mode
Normal Distribution

118. X F FX
1 10 10
2 12 24
3 10 30
4 8 32
5 4 20
6 2 12
7 2 14
0
2
4
6
8
10
12
14
0 2 4 6 8 10
Mode
Mean
Positive Skewness

119. X F FX
1 1 2
2 2 4
3 4 12
4 6 24
5 10 50
6 12 72
7 10 70
0
2
4
6
8
10
12
14
0 2 4 6 8 10
Mode
Mean
Negative Skewness

120. Measures of Kurtosis

121. Kurtosis
› Kurtosis indicates whether a distribution in flat topped or
peaked.
› Thus, Kurtosis is a measure of peakedness.
When a frequency curve is more peaked than the normal
curve, it is called Lepto Kurtic
When it is more flat topped than the normal curve, it is
called Platy Kurtic.
When the curve is neither peaked nor flat topped, it is
called Meso Kurtic

123. Index Numbers

124. Meaning
› An Index number is a statistical device for
measuring changes in the magnitude of a group of
related variables during a specific period in
comparison to their level in some other period.

125. BSE SENSEX
› Published since 1 January 1986,
the S&P BSE SENSEX is regarded
as the pulse of the domestic
stock markets in India. The base
value of the SENSEX was taken as
100 on 1 April 1979 and its base
year as 1978–79.

127. Consumer Price Index
› A Consumer Price Index
measures changes in the price
level of market basket of
consumer goods and services
purchased by households. In
India the base year is 1982.

129. Methods for Index Number Construction
1. Simple Aggregative Method
2. Weighted aggregative Method
3. Simple average relative method
4. Weighted average of price relative

130. 1. Simple Aggregative Method
Simple Index Number =
𝜮𝑷𝟏
𝜮𝑷𝟎
× 𝟏𝟎𝟎

131. 2. Weighted Aggregative Method
 Laspeyer’s Index Number
 Paasche’s Index Number
 Fisher’s Index Number

132. Laspeyer’s Index number
Laspeyer’s Index number =
𝜮𝑷𝟏𝑸𝟎
𝜮𝑷𝟎𝑸𝟎
× 𝟏𝟎𝟎
 P0, P1 = Price in Base year and Current year
 Q0 = Quantity in the base year

133. Paasche’s Index number
Paasche’s Index number =
𝜮𝑷𝟏𝑸𝟏
𝜮𝑷𝟎𝑸𝟏
× 𝟏𝟎𝟎
 P0, P1 = Price in Base year and Current year
 Q1 = Quantity in the Current year

134. Find out Index number.
Commodity Unit of
consumption in
base year
Price in Base
year
Price in current
year
A 200 1.00 1.20
B 50 3.00 3.50
C 50 4.00 5.00
D 20 20.00 30.00
E 40 2.00 5.00
F 50 10.00 15.00
G 60 2.00 2.50
H 40 15.00 18.00

135. Find out Index number.
Commodity Quantity
Consumed
(2009)
Price
2005 2009
A 50 32 40
B 35 30 42
C 55 16 24
D 45 40 52
E 15 45 42

136. Fisher’s Index number
Fisher’s Index number =
𝜮𝑷𝟏𝑸𝟏
𝜮𝑷𝟎𝑸𝟏
×
𝜮𝑷𝟏𝑸𝟎
𝜮𝑷𝟎𝑸𝟎
× 𝟏𝟎𝟎
 P0, P1 = Price in Base year and Current year
 Q0,Q1 = Quantity in the Base year &Current year

137. Find the weighted aggregative index number
Commod
ity
Price Quantity
2009 2010 2009 2010
A 4 7 10 8
B 5 9 8 6
C 6 8 15 12
D 2 2 5 6

138. 3. Simple Average relative method
› Under this method, the price relative of each item is
individually calculated and the average of all such values
would be the Index number.
Here, Price Relative; I =
𝑷𝟏
𝑷𝟎
× 𝟏𝟎𝟎
Price Index =
Σ𝑰
𝒏

139. Calculate the Simple index number using
average relative method.
Item Base year price Current year price
1 5 7
2 10 12
3 15 25
4 20 18
5 8 9

140. 3. Weighted Average of Price relative method
› Here, we will be assigning some arbitrary numbers as
weight. The price relative would be found and then
multiplied by the concerned weight.
Here, Price Relative; I =
𝑷𝟏
𝑷𝟎
× 𝟏𝟎𝟎
Price Index =
Σ𝑰𝑽
Σ𝑽

141. Calculate the Price Index for the following
data.
Commodities V (Weight) Price (2008) Price (2009)
A 40 16 20
B 25 40 60
C 5 2 2
D 20 5 6
E 10 2 1

142. Consumer Price Index
› A Consumer Price Index
measures changes in the price
level of market basket of
consumer goods and services
purchased by households. In
India the base year is 1982.

143. Steps in the Construction of CPI number.
› Decisions about class of people & Scope
› Decisions about the items to be selected
› Family budget enquiry
› Obtaining price quotations
› Selection of base period and weightages
› Selection of suitable indexing number

144. Food
(35%)
Rent
(15%)
Clothing
(20%)
Fuel
(10%)
Misc
(20%)
Price
(2006)
150 30 75 25 40
Price
(2008)
145 30 65 23 35