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###### Measure of Central Tendency (Average) PDF / PPT

Measure of Central Tendency (Average)

• Measures of central tendency are also usually called as the averages.
• They give us an idea about the concentration of the values in the

central part of the distribution.
• Measure of central tendency provides a very convenient way of

describing a set of scores with a single number that describes the
PERFORMANCE of the group.

• By calculating the measure of central tendency, we can find a single
value to represent whole data. It also help s to compare the value of
two or more groups.

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Types of Measure of Central Tendency

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Objective of Measure of Central Tendency

1. To obtain a single value that describe the characteristics of the
whole group.

2. To help for comparison.
3. To help to make quantitative relationship between different group

average.
4. To help in decision making.

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Mean

(1) Arithmetic Mean:
The arithmetic mean of a given set of observations is the ratio of

Sum of observations to the total number of observations. Symbolically,

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i) For ungrouped data

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ii) For ungrouped Frequency Distribution ( or Discrete series)

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iii) Continuous series ( grouped frequency Distribution)

• In grouped frequency distribution (continuous series) we are given class
intervals associated with the corresponding frequencies.

• In this case A.M. can be calculated by
(a) Direct Method
(b) Step Deviation Method

(i) Change of origin
(ii) Change of origin and scale

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(a) Direct method

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(b) Step-Deviation Method:
• Step – deviation method is also called short-cut method.
• This method reduces the variate value and thus facilitate the computation of

mean.

(i) Change of origin:
For ungrouped data

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• For Ungrouped Frequency distribution

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Example: From the following distribution of data, calculate the mean by change of origin
OR short cut method.

Class (x) 10 20 30 40 50 60

Frequencies (f) 3 2 5 10 11 8

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• For Grouped frequency Distribution

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(ii) Change of origin and scale both:
The labor of computation work is much more reduced by changing the origin and scale
both.
This is done by shifting the origin to an arbitrary point A and then dividing by the width of
the class interval.
Thus if

x: x1, x2, ….. xn

and we introduce a new variate
y : y1, y2, ….. yn

Where yi = 𝑥 𝐴

𝑖
𝑑

Where xi = Mid-point of the class interval
A = Point where origin is shifted
d = Width of the class intervals

So x̅ = A + ӯd
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− = 0
= 8 8 = 0 = 6−8 = −2 = -1

2 2 2 2

8
0
2

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Weighted Arithmetic mean

• Although arithmetic mean is a very good measure of central tendency and
gives equal importance to each item, but in real life situation this is not
always true.

• Some of them may be more important than others e.g. the importance of
consumption of wheat in our meals is not the same as that of sugar.
Similarly, the importance of edible oil is not the same as that of kerosene.

• So, if we want to find out the average price of such items, a simple
arithmetic mean will not be the true representative of the data.

• In such a case, the relative importance of items should be taken into
account for calculating the required average and such type of work is done
by calculating the weighted arithmetic mean.

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• Simple arithmetic mean is the same i.e.
72 for all the three universities.

• On comparing the weighted arithmetic
mean we find that for Mumbai the
mean value is the highest and hence we
conclude that in Mumbai the

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best.

Geometric Mean

• The Geometric mean of n variates is defined as then the root of the
product of n random variables x1, x2, … xn of the data.

• Symbolically, G.M. = 𝑛 x1 ,x2,…… . xn
1

• By taking logarithm log (G.M.) = (log x + og
𝑛 1 l x2+.. +log xn)
1

= Ʃ log x
𝑛 i

1
Taking antilog G.M = antilog ( Ʃ log x )

𝑛 i

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Example : Find the geometric mean of the following
(1) 15, 5415
G.M. = 15 × 5415

= 81255
= 285

𝟓
(2) 5,6,25,

𝟔
4 5

G.M. = 5 × 6 × 25 ×
6

4
= 5 × 25 × 5

4
= 54

= 5

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Harmonic Mean

• Let x : x1, x2,…… xn be n random variables and H be their harmonic
mean.

• Then H is defined by

1 1 1 1
= 1 +⋯
𝑛 x + +

𝐻 1 x2 xn
1

= Ʃ reciprocal of x
𝑛 i

H= 𝑛

Ʃ(𝑟𝑒𝑐𝑖𝑝𝑟𝑜𝑐𝑎𝑙 𝑜𝑓 xi)

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𝟓 𝟏𝟎
Example : Find the harmonic mean of , 5, 10 and

𝟐 𝟑

𝑛
H.M. = 1

Ʃ
𝑥
𝑖

4
= 2 1 1 3

+ + +
5 5 10 10

4
=
1

= 4

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Median

• The median is defined to be the middle most value in a distribution.
• It can also be defined as that value of the variate which occupies

middle most position in the series after rearranging the series in
ascending/descending order of magnitude.

• The median value splits the observation into two halves so that 50%
of observations have values less than the median value and the
remaining 50% of observations have values more than the median
value.

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Calculation of median

i) Case : Single series
Calculation of median involves the following steps:
(a) Rearrange the data in ascending/descending order of magnitude.
(b) If the number of observation in the series is odd.

The median value = size of 𝑛+1 𝑡ℎ item.
2

(c) If the number of observations is even then,

The median value = average of 𝑛 th item and 𝑛+1 𝑡ℎ item in the
2 2

arranged series.

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Exercise: Find the mean value of the data given below:
Income (Rs.) 1100, 1150, 1080, 1120, 1200, 1160, 1400

Solution : Rearranging the data in ascending order of magnitude the series is
1080, 1100, 1120, 1150, 1160, 1200, 1400
Here, n = 7 i.e. odd

Median value = Size of 7+1 𝑡ℎ item in the arranged series
2

= Size of 4th item in arranged series
= 1150

∴ Median value = Rs. 1150

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Exercise: Find out the median value of the following data
10, 9, 11, 12, 6, 8, 13 and 18.

Solution : The arranged series is
6, 8, 9, 10, 11, 12, 13 and 18

Here n = 8 i.e. even number

∴ Median value = average of 8 th item and 8+ 1 th item
2 2

= average of 4th item and 5th item
= 10+11 = 21

2 2

= 10.5

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ii) Case : Discrete series
In case of discrete series (i.e a frequency distribution with x values associated
with the corresponding frequencies). We calculate median as :

(a) Arrange the data in ascending/descending order of magnitude.
(b) Prepare the column of cumulative frequency.

(c) Calculated (𝑛) or 𝑛+ 1
2 2

(d) Run your eyes through cumulative frequency column. Find out the value
which is just more than 𝑛 or 𝑛+ 1

2 2

(e) Determine the corresponding x value. This variate value is the median
value.

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Example : From the data of heights of 100 persons given below calculate the median value if
the variate.

Heights (in inches) 58 60 61 62 63 64 65 66 68 70
x

No. of persons f 4 6 5 10 20 22 24 6 2 1

Heights Persons Cumulative freq.
𝑛 = 100• Solution: Here = 50 (x) (f) (c.f.)
2 2

58 4 4
• The cumulative frequency just above 50 is 67.

60 6 10
• The value of x corresponding to cumulative frequency

61 5 15
67 is 64.

62 10 25
• ∴ The required median height is 64.

63 20 45

64 22 67

65 24 91

66 6 97

68 2 99
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70 1 100

iii) Case: Continuous Frequency Distribution.
• In case of continuous frequency distribution the values of the variables are in the form of class-interval

associated with the corresponding frequencies.
• It is assumed that the class intervals are continuous and if it is not so we can make it continuous by using

the correction factor.
• The steps involved in calculating median are :
(i) Prepare the column of cumulative frequency distribution
(ii) Find out the value of 𝑛.

2

(iii) Find out cumulative frequency just greater than 𝑛
2

(iv) The class corresponding to this cumulative frequency is which the median value lies.
∴Locate the median class
(v) The median value is determined by using the formula.

Where L = Lower limit of the median class
n= Total of all frequencies

𝑛
−𝐶𝑝 Ср = Cumulative frequency preceding the median class

Median = L + 2 × ¡
𝑓𝑚 fm = Frequency corresponding the median class

i = Width of the class interval

Even if the width of the class intervals are unequal, the frequencies need not to be adjusted and the above
mentioned formula can be used directly

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Exercise: Find out the median value of the number of eggs for the data given below.

No. of eggs 10-12 12-14 14-16 16-18 18-20 20-22 22-24 24-26 26-28

No. of hens 3 11 22 25 32 37 13 10 7

No. of eggs No. of hens Cumulative We Know,
frequency

𝑛
10-12 3 3 −𝐶𝑝

Median = L + 2 × ¡
𝑓𝑚

12-14 11 14

14-16 22 36 Here
𝑛 = 160 = 80

16-18 25 61 2 2

18-20 32 93 The cumulative frequency just greater than 80 is 93.
∴ Median class is 18-20

20-22 37 130

22-24 13 143 Here L= 18, 𝑛 = 80, Cp= 61 , fm= 32 and ¡ = 2
2

24-26 10 153
Median = 18 + 80−61 × 2

32
26-28 7 160

= 18.2 eggs
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Exercise 19 : Calculate median from the data given below.

Marks below 10 20 30 40 50 60 70 80

No. of students 15 35 60 84 96 127 198 250

Solution : We known Marks No. of students Cumulative
(x) (f) frequency c.f.

𝑛
−𝐶𝑝

Median = L + 2 × ¡ 0-10 15 15
𝑓𝑚

10-20 35-15= 20 35

Here n= 250 20-30 60-35= 25 60

30-40 84-60= 24 84

So 𝑛 = 250 = 125 40-50 96-84= 12 96
2 2

50-60 127-96 =31 127

The cumulative frequency just more than 60-70 198-127= 71 198

125 is 127. 70-80 250-198 = 52 250

So Median class is 50-60. Here L= 50 , n/2 = 125, Cp= 96 , fm= 31 and ¡= 10

Median = 5w0w w+. D
125−96
ulo3Mix.com× 10 = 50 + 9.3548 = 59.3548 marks

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Mode

• Mode represents the value occurring most often in the data.
• It may also be defined as the size of the item having maximum

frequency.
• Mode is the value which has greatest frequency density in its

immediate neighbourhood or, it may also be regarded as the most
typical value of a distribution.

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Calculation of mode:

Case (i): Individual observation or in case of Discrete distribution

• In case of individual observation (discrete distribution) modal value
can be found out by inspection.

• In case of ungrouped data/individual observations, we count the
repetition of every item.

• The value which occurs maximum number of times is the mode of the
distribution.

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Exercise: Find the mode of the following values of heights of a group of students
(in inches).
51, 53, 52, 51, 54, 53, 50, 54, 55, 53, 54, 55, 56 and 54.

Solution : Arranging the given values in the form of frequency table:

Values (x) 50 51 52 53 54 55 56

Frequency (f) 1 2 1 3 4 2 1

• Since the maximum frequency is 4 the corresponding value of the variate is 54.

• So Mode value = 54 inches

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Case (ii) Discrete Series

• In certain cases where two frequencies in the data are highest or higher
than the rest or the two values are very near to each other then little care
will be needed to decide the mode.

• Actually, that frequency shall be taken as modal frequency which has bulk
of them in its neighborhood.

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• In this data the highest frequency is
32 and the next higher frequency is
31.

• Since these two frequencies are more
or less equal as compared to the
other frequencies, it becomes
necessary to decide the mode out of
these values of high frequencies.
Following table gives the procedure
of grouping the data taking 2 at a
time, 3 at a time and so on.

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• To decide which variate value is involved in
maximum number of times, the following
table is made:

• The variable 56 occurs 5 times.
• So required mode value = 56

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Case (ii) : Continuous series
To determine the mode of a continuous distribution, we first determine the
class of maximum frequency (called the modal class) and then apply the
formula

f − f
Mode = L + m −1 × i

2fm −f −f
−1 1

Where L = Lower limit of the modal class
fm = Frequency of the modal class
f
− 1= Frequency preceding the modal class

f1 = Frequency following the modal class
i = Width of the class interval

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Exercise : Compute the mode from the table given below for the heights of the students with
the corresponding frequencies.

Heights Frequency Solution : Here the maximum frequency Is 43.
(inches)

57-58 17 The corresponding modal class is 62-63.

58-59 27
f − f
m 1

59-60 14 We know, Mode = L + − × i
2fm −f −f

1 1
60-61 19 −

61-62 22

62-63 43 Here L=62 , fm= 43, f-1 = 22 , f1= 37 and i = 1
63-64 37

64-65 14 Mode = 62 + 43−22 × 1
65-66 8 2 ×43 −22 −37

66-67 12
= 62 + 21

67-68 5 27

68-69 3

= 62 + 0.777 = 62.78 inches
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Examples
1. Compute the arithmetic mean, median and mode of the heights of 15patients. The

heights are61, 62, 63, 61. 63, 64, 64, 64, 60, 65, 63, 64, 65, 66, 64.
Solution : Let x= height of a patient’
here Ʃx = 949 and n= 15

Ʃ𝑥 949
So A.M = x̅ = = = 63.27

𝑛 15

We, now, arrange the given observations in ascending order of magnitudes.
60, 61, 61, 62, 63, 63, 63, 64, 64, 64, 64, 64, 65, 65, 66

𝑛+1 15+1
Median = th value = ( )th value = 8th value = 64

2 2
Mode is the most frequently occurring value..
Mode = 64
Thus, A.M. = 63.27. Median = 64 and Mode = 64.

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2. Find mean , median and mode for t he following data:
x 1 2 3 4 5 6 7

f 7 12 21 27 19 11 3

Solution : Here n= Ʃf = 100 and Ʃfx = 384.
Ʃfx 384

So Mean = = = 3.84
Ʃf 100

Now, we construct a table for cumulative frequencies.
x 1 2 3 4 5 6 7

f 7 19 40 67 86 97 100
𝑛

Median = th value = 50th value
2

It is clear that 50th value lies against x =4.
So median = 4
Maximum frequency 27 occurs against x = 4
So mode = 4

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