This document discusses various statistical measures of central tendency, including the mean, median, and mode. It provides definitions and formulas for calculating the arithmetic mean using direct, shortcut, and step deviation methods for individual, discrete, and continuous data series. It also discusses how to calculate the median and weighted mean. The document compares the merits and demerits of the arithmetic mean and provides examples to illustrate the different calculation techniques for central tendencies.

1. STATISTICAL
ANALYSIS USING
CENTRAL
TENDENCIES
CELIA SANTHOSH
SLS,CUSAT

2. RESEARCH PROCESS
PRESENTING THE RESULTS
ANALYSIS OF DATA
PROCESSING OF DATA
COLLECTION OF DATA
SELECTION OF METHOD OF DATA COLLECTION
FORMULATION OF HYPOTHESIS
IDENTIFICATION OF OBJECTIVE OF THE STUDY
CLARIFICATION OF CONCEPT
LITERATURE REVIEW
DEFINE RESEARCH PROBLEM
IDENTIFICATION OF RESEARCH PROBLEM
CENTRAL TENDENCIES

3. Central Tendencies
• It is very difficult for everybody to understand or remember large set
of data.
• Therefore one would like to know certain values which will represent
or summarise all the data.
• Thus, the basic purpose of central tendencies i.e to summarise the
data using a single significant figure.
• Which is why central tendencies are also known as AVERAGES.

4. MEASURES OF
CENTRAL
TENDENCIES
MEAN MEDIAN MODE

5. You want to find
the raise in no.
Accidents for the
years 2011-2015
and 2016-2020 in
Ekm
2011-2015 : GROUP A
2016-2020 : GROUP B

6. 2016 600
2017 700
2018 800
2019 900
2020 1000
2011 100
2012 200
2013 300
2014 400
2015 500
GROUP A GROUP B

7. MEAN
Ø Mean = Average Value of the dataset
Ø That is the sum of all the values in the dataset divided by the
number of values.
FORMULA:
Wherein, M = MEAN
SUM OF ALL TERMS
NO. OF TERMS
=

8. Or,
MEAN
SUM OF VALUES
NO. OF VALUES

9. 1500
5
=
ARITHMETIC MEAN OF GROUP A
= 300
ARITHMETIC MEAN OF GROUP B
4000
5
=
= 800
Individual series – driect method - AM

10. ARITHEMATIC MEAN
• It is a method of representing the whole data by one
single figure
• We can also say that the A.M is the mathematical
average A.M
SIMPLE WEIGHTED

11. ARITHEMETIC
MEAN
DIRECT METHOD
x/n
SHORT CUT
METHOD
STEP DEVIATION
METHOD
a+ d/n a+ d’/n*c
M
M
M

12. 1
• Take one of the values from the series and call it
assumed average. Denote it by “a”
2
• Subtract this assumed average from all the given
values. These difference are denoted by “d”. d = x - a
3
• Find the sum of all these difference and call it d
4
• Now apply the formula a+ d/n. To find the AM of
the series
SHORT CUT METHOD

13. NO. OF LOST MOBILE PHONES FROM JAN TO OCT
45,48,50,52,55,58,60,61,63,65
- TAKE 55 AS ASSUMED AVERAGE
- i.e a = 55
x d (x-55)
45 -10
48 -7
50 -5
52 -3
55 0
58 3
60 5
61 6
63 8
65 10
7
d = 7
a = 55
n = 10
X = a+ d/n
= 55+ 7/10
= 55+.7
= 55.7
Shortcut method- individual series - AM

14. FIND THE ASSUMED MEAN (a)
FIND d(x-a)
FIND c (common factor)
FIND d’ (c-d)
Apply the formula
STEP DEVIATION METHOD
THERE MUST BE A COMMON FACTOR

15. STEP DEVIATION METHOD
10,20,30,40,50,60,70,80,90,100
Formula:
a+ d’/n x c
x d (x-a) d’ (c-d)
10 -40 -4
20 -30 -3
30 -20 -2
40 -10 -1
50 0 0
60 10 1
70 20 2
80 30 3
90 40 4
100 50 5
5 d’
a (assumed mean) = 50
d’ = 5
n = 10
c (common factor) = 10
= 50+5/10x10
= 55
10

16. TYPES OF SERIES
SERIES
INDIVIDUAL DISCRETE CONTINUOUS

18. INDIVIDUAL SERIES
• You simple write down the mark of each student like
1,2,5,7,8,3,2,5,7,8,8,5,6,10,9,8,10………still the 100th student
….so individually we write down the mark of each and every
student it is called individual series.
For eg - when the teacher writes down the mark of each student
against his/her roll no or name that is know as individual series.
ROLL
NO
MARK
1 3
2 5
3 6
4 6
5 7
6 8
100 10

19. DISCRETE SERIES
• Now one major problem with individual series is that you will not be
able to understand the character of the data by simple looking at it
• For eg – simple by looking that the pervious table the teacher will not
be able to understand the performance of the class.
• So what one must to is, convert the individual series to discrete series
Note : In discrete series the data must
repeat itself and the variety of data
must be limited
Mark (x) Students (f)
4 5
5 20
6 25
7 10
8 20
9 10
10 10
Total 100

20. CONTINOUS SERIES
• Now if the characteristics of the data is such that there is a lot variety
then go for continuous series
i.e if the marks of the students are 2.5,3.75,9.5 etc
Marks students
0-1 5
1-2 20
2-3 25
3-4 10
4-5 10
5-6 10
6-7 10

21. ARITHMETIC MEAN IN DISCRETE
SERIES
• DRIECT METHOD
FORMULA
X = fx/N Where, N = f
EXAMPLE
Value (MARK) 5 15 25 35 45 55 65 75
freq. (STUDENTS) 15 20 25 24 12 31 71 52
(Repeats)

22. VALUES
(x)
FREQUENCY
(f)
fx = f * x
5 15 75
15 20 300
25 25 625
35 24 840
45 12 540
55 31 1705
65 71 4615
75 52 3900
250 12,600
FORMULA
X = fx/N
= 12600/250
= 50.4
N fx
arithmetic mean - discrete series – direct method

23. SHORT
CUT
METHOD
Mean is obtained by adding *sigma*fd/n with a
Divide *sigma*fd by N
Find the sum of the frequencies ie *sigma*f take *sigma*f as N
find the sum of fd call it *sigma*fd
Multiply 'd' values with corresponding frequencies to get 'fd' values
From all the values subtract an assumed average 'a', so that we can
get d i.e d = x-a
Call the Values as ‘x’

24. X = a + f d
N

25. EXAMPLE
Value (MARK) 5 15 25 35 45 55 65 75
freq. (STUDENTS) 15 20 25 24 12 31 71 52
Taking 35 as ‘a’. Then d = x - 35
VALUES
(x)
FREQUENCY
(f)
d = x - 35 fd = f x d
5 15 (5-35) -30 -450
15 20 -20 -400
25 25 -10 -250
35 24 0 0
45 12 10 120
55 31 20 620
65 71 30 2130
75 52 40 2080
250 3850
Here,
a = 35 and fd = 3850
x = a + fd/N
= 35 + 3850/250
= 50.4
N arithmetic mean - discrete series – shortcut method

26. STEP DEVIATION METHOD
X = a + f d’ x c
N
If the values have a
common difference ‘c’
the take d’ = x – a/ c
and modify the formula
as
arithmetic mean - discrete series – step deviation method

27. EXAMPLE
Value (MARK) 5 15 25 35 45 55 65 75
freq. (STUDENTS) 15 20 25 24 12 31 71 52
Taking 35 as ‘a’ and ‘c’ as 10
VALUES
(x)
FREQUENCY
(f)
d (x-a) d’ = d/c fd’ (f x d’)
5 15 5-35 = -30 -3 (-30/10) -45
15 20 15-35= -20 -2 -40
25 25 25-35= -10 -1 -25
35 24 35-35= 0 0 0
45 12 45-35= 10 1 12
55 31 55-35= 20 2 62
65 71 65-35= 30 3 213
75 52 75-35= 40 4 208
250 385
X = a + f d’ x c
N
fd’
N arithmetic mean - discrete series – step deviation method
10

28. Here,
a = 35
fd’ = 385
c = 10
f d’
N
= 35 + 385/250 x 10
= 50.4
X c
X = a +

29. ARTHEMATIC MEAN IN CONTINOUS SERIES
• In case of continuous series we can write the mid value of the classes as
’x’.
Therefore x = l1 + l2
2
Where l1= lower limit and l2 = upper limit of each class

30. • Find the A.M of the following data
Age : 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
No. of
Persons : 15 30 53 75 100 110 115 125
Dying

31. Using Direct Formula. fx/N
AGE F Mid Value (x) fx
0-10 15 (0+10)/2 = 5 75
10-20 30 (10+20)/2=15 450
20-30 53 (20+30)/2=25 1325
30-40 75 (30+40)/2=35 2625
40-50 100 (40+50)/2=45 4500
50-60 110 ( 50+60)/2=55 6050
60-70 115 (60+70)/2=65 7475
70-80 125 (70+80)/2=75 9375
623 31875
AM – CONTINOUS SERIES – DRIECT METHOD

32. Here
fx = 31875 and N = 623
AM = fx / N = 31875/623 = 51.16 years
AM – CONTINOUS SERIES – DRIECT METHOD

33. Applying Short Cut Formula A.M = a+ fd
N
AGE f Mid (x) [l1+l2/2] d = x - 35 fd = f*d
0-10 15 5 5 - 35 = -30 -450
10-20 30 15 15 – 35 = -20 -600
20-30 53 25 24-35 = -10 -530
30-40 75 35 0 0
40-50 100 45 45 – 35 = 10 1000
50-60 110 55 55 – 35 = 20 2200
60-70 115 65 65 – 35 = 30 3450
70-80 125 75 75 – 35 = 40 5000
623 10070
AM – CONTINOUS SERIES – SHORTCUT METHOD

34. • Here a = 35, fd = 10070, N = 623
A.M = a+ fd
N
= 35 + 10070/623
= 35 + 16.16
= 51.16 years
AM – CONTINOUS SERIES – SHORTCUT

35. Applying Step Deviation Formula
AGE Mid
(x)
f d (x-a) d’ = d/10 fd’
0-10 5 15 5-35 = -30 -30/10 = -3 -45
10-20 15 30 15-35= -20 -2 -60
20-30 25 53 25-35= -10 -1 -53
30-40 35 75 35-35= 0 0 0
40-50 45 100 45-35= 10 1 100
50-60 55 110 55-35= 20 2 220
60-70 65 115 65-35= 30 3 345
70-80 75 125 75-35= 40 4 500
623 1007
10

36. • Here, a = 35, fd’ = 1007 and N = 623
AM = a + fd’
N
= 35 + 1007/623*10
= 35 + 16.16
= 51.16 years
X C
AM – CONTINOUS SERIES – SHORTCUT

37. Calculation of Arithmetic mean at a Glance
• INDIVIDUAL SERIES
• DISCRETE SERIES
• CONTINOUS SERIES
DRIECT METHOD SHORT CUT METHOD STEP DEVIATION METHOD
X/n a + d / n a + d’/n * c
DRIECT METHOD SHORT CUT METHOD STEP DEVIATION METHOD
fx / n a + fd / N a + fd’/N* c
DRIECT METHOD SHORT CUT METHOD STEP DEVIATION METHOD
fx / n a + fd / N a + fd’/N* c

38. MERITS OF ARITHEMATIC MEAN
• AM is simple to understand
• AM can be easily calculated
• AM can be determined in most cases
• It is based on all observations of the series
• It is capable of more algebraic treatment
• AM is stable.

39. DEMERITS OF ARITHEMATIC MEAN
• AM is affected by extreme values
• Usually mean does not coincide with any of the observed value
• It cannot be calculated for qualitative data which cannot be measured
numerically
• It may offer misleading and absurd results in some cases. For eg –
Average number of children per family in a village may work out as
3.2 which is impossible.

40. WEIGHTED MEAN
• Simple AM attached equal importance to all values
• Sometimes the item in a series may not have equal importance. So
simple AM is not suitable for those series
• In such cases weighed AM will be appropriate
• The weighed AM is used whenever the importance of the item in a
series differs.
• While calculating weighted AM each item is given a weight judged by
its relative importance.

41. PROBLEM
SUBJECTS MARK %
(X)
WEIGHTS
(W)
WX
CPL 60 1 60
ITL 75 1 75
ILO 63 2 126
RM 60 3 180
LAW AND JUSTICE 55 3 165
10 606
WEIGHTED AM =
wx
= 606/10 = 60.6
w

42. MEDIAN

43. MEDIAN = MIDDLE

44. • Median is the value of that item which occupies the central position when
items are arranged in the ascending or descending order of their magnitude.
• Therefore median is the value of that item which has equal number of items
above and below.
• The number of items greater than median and the number of item less than
median are equal.
• For eg – if there are five values
5,10,15,20,30
Then the median is 15.
That is there are two items whose values are more than median and two
items whose value is values are less than median

45. Definition
• According to L.R.Connor, (Author - Statistics in Theory and Practice.)
“The median is that value of the variable which divides the
values of the variable into two equal parts, one part containing all
values greater than the median value and the other part containing all
the values smaller than the median values.”
Thus MEDIAN is a POSTIONAL AVERAGE

46. MEDIAN IN INDIVIDUAL SERIES
• FORMULA
Median = Size of th item when the series are
arranged in the ascending or descending order of their magnitude.
STEPS
1. Arrange the values in the ascending order of their magnitude
2. Find out the value of the middle item. That is the median.
(n + 1)
2

47. PROBLEM
• Find the median of the following values:
4, 45, 60, 20, 83, 19, 26, 11, 27, 12, 52
No. of observation = 11
- To find median first arrange the values in ascending order
4, 11, 12, 19, 20, 26, 27, 45, 52, 60, 83
Median = Size of (n + 1) th item = 11+1/2 = 6th item = 26
2
INDIVIDUAL SERIES = ODD IN NUMBER

48. • When ’n’ is even there are two middle items. Take the average of
them.
Example 35, 23, 45, 50, 80, 61, 92, 40, 52, 61
- write the values in ascending order
23, 35, 40, 45, 50, 52, 61, 61, 80, 92
Median = Size of (n + 1) th item = 10+1 / 2 = 5.5th item
2
• Here there are two middle items 5th and 6th . Therefore take the
average of those two.
median = 50 + 52 / 2 = 51
INDIVIDUAL SERIES = EVEN IN NUMBER

49. MEDIAN IN DISCRETE SERIES
• FORMULA
MEDIAN = SIZE OF (N+1)th item
2
Where, N = f
Note : Prior to applying the formula find the cumulative frequency

50. PROBLEM
• VALUE : 5 8 10 15 20 25 (wages)
• FREQUENCY : 3 12 8 7 5 4 (No. of Workers) – Keeps Repeating/Changing
Solution - 1
VALUE FREQUENCY(f) CUMULATIVE
FREQUENCY
5 3 3
8 12 3 + 12 = 15
10 8 15 + 8 = 23
15 7 23 + 7 = 30
20 5 30 + 5 = 35
25 4 35 + 4 = 39
N = 39
MEDIAN - DISCRETE SERIES

51. • Median = Size of (N + 1)th item
2
= Size of 40 / 2 th item
= Size of 20th item
Therefore median corresponds to the 20th item of the series.
The first cumulative frequency which includes 20 is 23
The value of item for which cumulative frequency is 20 is 10
Therefore median = 10

52. MEDIAN IN CONTINOUS SERIES
• Steps followed in finding Median in continuous series
1. Form the cumulative frequency column
2. Find N/2. In order to find Median class
3. After getting the median class, find median by using the following
interpolation formula
Median = l1+ (N / 2 - cf) x c
f

53. Median = l1+ (N / 2 - cf) x c
f
Where, l1 = lower limit of the Median class
cf= cumulative frequency of the class
just preceding the median class
f = Frequency of the median class
c = interval of the median class
MEDIAN – CONTINUOUS SERIES

54. PROBLEM
• CLASS : 0-10 10-20 20-30 30-40 40-50 50-60 60-70
• FREQUENCY : 8 12 20 23 18 7 2
AN Median Class = Size of (N / 2)th item
= 90/2 = 45th item
45 is included in the
cumulative frequency 63
CLASS FREQUENCY CUMULATIVE
FREQUENCY
(1)
0-10 8 8
10-20 12 20 (8+12)
20-30 20 40 (20+20)
30-40 23 63 (40+23)
40-50 18 81 (63+18)
50-60 7 88 (81+7)
60-70 2 90 (88+2)
N= 90
The class having cumulative frequency 63 is 30 – 40
Therefore 30 – 40 is the median class
MEDIAN – CONTINUOUS SERIES

55. • Now apply the formula ,
(N/2 - cf)
f
X c
l1+
Median =
Here, l1 = 30
N / 2 = 45
Cf = 40
f = 23
c = 10
Median = 30+ (45-40) x 10
23
= 30 + 2.7
= 32.17
MEDIAN – CONTINUOUS SERIES

56. Calculation of Median at a Glance
• INDIVIDUAL SERIES
• DISCRETE SERIES
• CONTINOUS SERIES
SIZE OF (n + 1)/2th ITEM
SIZE OF (N + 1)/2th ITEM
[N/2] l1 + (N/2 – cf) / f x c
1st find - CUMULATIVE
FREQUENCY
ARRANGE
MEDIAN CLASS
value which corresponds to
cumulative frequency

57. MERITS OF MEDIAN
• It is a very simple measure
• Some times it can be located even my a mere glance (only in
individual series)
• It is not effect by extreme items (as only the middle value is
considered)

58. DEMERITS OF MEDIAN
• Median is not based on all observation
• It is not capable of algebraic treatment
• It requires arraying (ie writing in the ascending or descending order)
• In case of continuous series interpolation formula is to be used. The
value thus obtained may only give an approximate value.

59. M O D E

60. VALUE WHICH OCCURS
MOST NUMBER OF TIMES
MODE =

61. MODE
• Mode is the value of the item in a series which occur most frequently
DEFINITION
According to Kenny, “ the value of the variable which occurs most
frequently in a distribution is called the mode.”

62. MODE IN INDIVIDUAL SERIES
STEPS
• ARRANGE THE VALUES IN ASCENDING ORDER
• THEN BY INSPECTION IDENTIFY THE VALUE WHICH OCCURS MORE NUMBER OF
TIMES
Problem
23,45,28,42,62,53,35,28,42,35,23,42,35
Solution
Write the values in ascending order
23,23,28,28,35,35,35,35,42,42,42,53,62
35 appears more no. of times. Therefore mode = 35

63. MODE IN DISCRETE SERIES
• In discrete series the value having highest frequency is taken as Mode
• A glance at the series can reveal which is the highest frequency
• So we get the mode by mere inspection.
• So this method is called inspection method
VALUE
MARK
f
No.of st
5 3
8 12
10 25
12 40
29 31
35 20
40 18

64. MODE IN CONTINUOUS SERIES
• Mode lies in the class having highest frequency.
• This method of identifying model class is called inspection method.
• From model class mode is calculated by using the interpolation
formula
MODE = l1+
(f1 – f0) c
2f1 – f0 – f2
l1 Lower limit of model class
f0 Frequency of the class before
the model class
f2 Frequency of the class after
the model class
f1 Frequency of the model class
c Class Interval
f0
f1
f2

65. PROBLEM
• CALCULATE MODE OF THE FOLLOWING DATA
CLASS f
10-15 4
15-20 8
20-25 18
25-30 30
30-35 20
35-40 10
40-45 5
45-50 2
MODE = l1+ (f1 – f0) c
2f1 – f0 – f2
l1 = lower limit of model class = 25
f1 = Frequency of the model class =30
f0 = Frequency of the class before the model class = 18
f2 = Frequency of the class after the model class = 20
c = class interval = 5
MODE = 25 + (30-18) x 5/(60-18-20)
= 25 + 60/22 = 27.73
l1
f0
f1
f2
MODE – CONTINUOUS SERIES

66. Calculation of Mode at a Glance
• INDIVIDUAL SERIES
• DISCRETE SERIES
• CONTINOUS SERIES
INSPECTION METHOD
INSPECTION METHOD
l1 + (f1 – f0) c/ 2f1 – f0 – f2
ARRANGE
VALUE HAVING THE HIGHEST
FREQUENCY
FIRST FIND MODEL CLASS
BY INSPECTION METHOD

67. MERITS OF MODE
• Mode is very simple among the measure of central tendency. Just a
glance at the series is enough to locate the model value (not in
countinous series)
• Mode is less affected by extreme values in the series
• Usually mode coincide with one of the values in the series (unless it is
continuous series)

68. DEMERITS OF MODE
• Mode is not capable of further algebraic treatment
• Mode is not based on all the values of the series
• Since mode is based on frequency of occurrence it might not always
adequately represent the data. For eg – if 0 appears more number of
times then mode is 0 and that is not a representative value.

69. COMPARISON
MEAN MEDIAN MODE
IS A MATHEMATICAL AVG POSITIONAL AVG POSITIONAL AVG
BASED ON ALL
OBSERVATIONS
IS THE MIDDLE VALUE,
THUS NOT BASED ON ALL
OBSERVATIONS
IS THE FREQUENTLY FOUND
ITEM – NOT BASED ON ALL
OBSERVATIONS
CAPABLE OF FURTHER
MATHEMATICAL
TREATMENT
NOT CAPABLE OF MORE
MATHEMATICAL
TREATMENT
NOT CAPABLE OF MORE
MATHEMATICAL
TREATMENT
VALUES NOT ALWAYS
FOUND IN THE SERIES
VALUES MOSTLY FOUND IN
THE SERIES (EXCEPT IN
CONTINOUSE SERIES AND
EVEN NUMBERED
INDIVIDUAL SERIES)
VALUES MOSTLY FOUND IN
THE SERIES
(EXECPT IN CONTINOUS
SERIES)
CANNOT BE OBTAINED BY
MERE INSPECTION
CAN BE OBTAINED BY MERE
INSPECTION ( ONLY IN
INDIVIDUAL SEREIS)
CAN OBTAINED BY MERE
INSPECTION (INDIVIDUAL
AND DISCRETE)
IS AFFECTED BY EXTREME
VALUES
NOT MUCH/ALWAYS
AFFECTED BY EXTREME
VALUES
NOT MUCH/ALWAYS
AFFECTED BY EXTREME
VALUES

70. FEATURES OF CENTRAL TENDENCIES
• It is the single figure which represents the whole series and sums up
all the characteristics of the data
• It is neither the lowest or the highest it lies somewhere in the middle
of the distribution
• It is a part of the whole group

71. IMPORTANCE/USE OF CENTRAL
TENDENCIES
• Gives a general idea about the whole group
• Can be used for summarizing the data
• Helps in comparison eg – no. of accidents
• Helps in decision making eg – govt. policies
• Constitutes the basis of statistical analysis (It is the science of
collecting, exploring and presenting large amounts of data to discover
underlying patterns and trends) and central tendencies helps in giving
a strong basis for understanding the data in hand.