Measures of Central Value

1. MEASURES OF
CENTRAL
TENDENCY
Dr. R. M. K. V
Assistant Professor

2. MEASURES OF CENTRAL
TENDENCY
In the study of a population with respect to one in which we are
interested we may get a large number of observations. It is not
possible to grasp any idea about the characteristic when we look at all
the observations. So it is better to get one number for one group. That
number must be a good representative one for all the observations to
give a clear picture of that characteristic. Such representative number
can be a central value for all these observations. This central value is
called a measure of central tendency or an average or a measure of
locations. There are five averages. Among them mean, median and
mode are called simple averages and the other two averages
geometric mean and harmonic mean are called special averages.

3. MEASURES OF CENTRAL
TENDENCY
“A measure of central tendency is a typical value around which other
figures congregate.”
“An average stands for the whole group of which it forms a part yet
represents the whole.”

4. CENTRAL VALUE
Central
Value
Mean
Arithmetic
Mean
Simple
Weighted
Geometric
Mean
Harmonic
Mean
Median
Mode

5. FREQUENCY
Raw
data
Grouped Data Continuous series
Exclusive method
Inclusive method
Open-end classes
Ungrouped Data Discrete Series
Individual
Observation

6. RAW DATA/ INDIVIDUAL
OBSERVATION

7. DISCRETE SERIES
Raw Data
Ungrouped Data(Discrete
Series)

8. CONTINUOUS SERIES
Raw Data Continuous series

9. EXCLUSIVE AND INCLUSIVE CLASS
Exclusive class Inclusive Class

10. OPEN-END CLASS

11. MEAN
Arithmetic mean or simply the
mean of a variable is defined as
the sum of the observations
divided by the number of
observations. If the variable x
assumes n values x1, x2 … xn
then the mean, x, is given by

12. ARITHMETIC MEAN
Direct method
Individual observation
X =
𝚺𝑿
𝑵
Discrete series
X =
𝚺𝐟𝑿
𝑵
Continuous series
X =
𝚺𝐟𝒎
N
Short Cut method
Individual observation
X = 𝑨 +
𝚺𝒅
𝑵
(A = Assumed Mean
d= x-A)
Discrete series
X = 𝑨 +
𝚺𝐟𝒅
𝑵
Continuous series
X = 𝑨 +
𝚺𝐟𝒅
𝑵

13. INDIVIDUAL OBSERVATION –
DIRECT METHOD
1. Calculate the mean for 2, 4, 6, 8, 10
X =
𝚺𝑿
𝑵
=
2+4+6+8+10
5
=
30
5
= 6

14. INDIVIDUAL
OBSERVATION
(SHORT-CUT
METHOD)
A student’ s marks in 5 subjects
are 75, 68, 80, 92, 56. Find his
average mark.
Let us take marks as X; A is an
Assumed Mean; d = X- A
Average Mark is
74.2

15. DISCRETE SERIES
Illustration: 1
In a survey of 40 families in a village, the number of children per family
was recorded and the following data obtained. 1, 0, 3, 2, 1, 5, 6, 2,2, 1, 0,
3, 4, 2, 1, 6, 3, 2, 1, 5, 3, 3, 2, 4, 2, 2, 3, 0, 2, 1, 4, 5, 3, 3, 4, 4, 1, 2, 4, 5
Find average children per family
Solution:1
Represent the data in the form of a discrete frequency distribution. Frequency
distribution of the number of children

16. Number of Children Tally Marks Number of Family
(Frequency)
0 ||| 3
1 |||| || 7
2 |||| |||| 10
3 |||| ||| 8
4 |||| | 6
5 |||| 4
6 || 2
1, 0, 3, 2, 1, 5, 6, 2,2,
1, 0, 3, 4, 2, 1, 6, 3, 2,
1, 5, 3, 3, 2, 4, 2, 2, 3,
0, 2, 1, 4, 5, 3, 3, 4, 4,
1, 2, 4, 5

17. X f fX
0 3 0
1 7 7
2 10 20
3 8 24
4 6 24
5 4 20
6 2 12
𝑓 = 40 𝑓𝑋 = 107
Let us take Number of
children as X and
Number of family as f
(Frequency)
𝑓𝑋 = 107 𝑓 = 𝑁 = 40
X =
𝚺𝐟𝑿
𝑵
=
107
40
= 2.675
Average number of
children per family is
2.675

18. DISCRETE SERIES – ILLUSTRATION
2
Given the following frequency distribution, calculate the arithmetic
mean
Discrete series
X =
𝚺𝐟𝑿
𝑵
Direct Method
X = 𝑨 +
𝚺𝐟𝒅
𝑵
Short-cut Method
Marks 64 63 62 61 60 59
Number
of
Students
8 18 12 9 7 6

19. SOLUTION - DM
Let us take marks as X and Number
of students as f (Frequency)
𝑓𝑋 = 3713 𝑓 = 𝑁 = 60
X =
𝚺𝐟𝑿
𝑵
=
3713
60
= 61.88
Average mark of Students is 61.88
X f fX
64 8 512
63 18 1134
62 12 744
61 9 549
60 7 420
59 6 354
𝑓 = 60 𝑓𝑋 = 3713

20. SOLUTION (SM) X f D = X-
A(62)
fd
64 8 2 16
63 18 1 18
62 12 0 0
61 9 -1 -9
60 7 -2 -14
59 6 -3 -18
𝑓 = 60 𝑓𝑑 = −7
Let us take marks as X and
Number of students as f
(Frequency)
X = 𝑨 +
𝚺𝐟𝒅
𝑵
= 62+
−7
60
= 62+(-0.117)
=62 – 0.117
= 61.88
Average mark of
Students is 61.88

21. CONTINUOUS SERIES
Illustration 1
Let us consider the weights in kg of 50 college students.
42, 62, 46, 54, 41, 37, 54, 44, 32, 45, 47, 50, 58, 49, 51, 42, 46, 37, 42, 39,
54, 39, 51, 58, 47, 64, 43, 48, 49, 48, 49, 61, 41, 40, 58, 49, 59, 57, 57, 34, 56,
38, 45, 52, 46, 40, 63, 41, 51, 41 Find Average weights.
Here the size of the class interval as per Sturges rule is obtained as follows
Size of class interval = C =
𝑅𝑎𝑛𝑔𝑒
1+3.322 log 𝑁
Range = Highest Value – Lowest Value = 64 -32 = 32
N= 50

22. C =
32
1+3.322 log 50
=
32
1+3.322 𝑥 1.6990
=
32
1+5.64
=
32
6.64
= 4.8 = 5
Class interval is 5
Lower limit = nearest smaller number divisible by 5 of lowest
number (32) = 30
Higher limit = nearest larger number divisible by 5 of highest
number(64) = 65
First class Lower limit(Starting point) 30 ending point = Starting
point + C = 30+5 =35
Last class Higher limit(ending point) 65 staring point = ending
point – C = 65- 5 = 60
Class 30 – 35 , 35 – 40, 40 – 45, 45 – 50, 50 – 55, 55 – 60, 60 –
65

23. Class (Weight) Tally Marks Number of Students
30 – 35 || 2
35 – 40 |||| | 6
40 – 45 |||| |||| || 12
45 – 50 |||| |||| |||| 14
50 – 55 |||| | 6
55 – 60 |||| | 6
60 – 65 |||| 4
42, 62, 46, 54, 41, 37, 54, 44, 32, 45, 47,
50, 58, 49, 51, 42, 46, 37, 42, 39, 54, 39,
51, 58, 47, 64, 43, 48, 49, 48, 49, 61, 41,
40, 58, 49, 59, 57, 57, 34, 56, 38, 45, 52,
46, 40, 63, 41, 51, 41

24. X m (mid
point)
f fm
30 – 35 (
30+35
2
)=
32.5
2 65
35 – 40 (
35+40
2
)=
37.5
6 225
40 – 45 (
40+45
2
)=
42.5
12 510
45 – 50 (
45+50
2
)=
47.5
14 665
50 – 55 (
50+55
2
)=
52.5
6 315
55 – 60 (
55+60
2
)=
57.5
6 345
60 – 65 (
60+65
2
)=
62.5
4 250
Let us take class of
weight as x and number
of students as f
X =
𝚺𝐟𝒎
N
=
2375
50
= 47.5
Average Weight of the
50 students is 47.5 Kg

25. CONTINUOUS SERIES –
ILLUSTRATION 2
Given the following frequency distribution, calculate the arithmetic
mean
Continuous series
X =
𝚺𝐟𝑚
𝑵
Direct Method
X = 𝑨 +
𝚺𝐟𝒅
𝑵
Short-cut Method d= m – A
Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Number
of
Students
5 10 25 30 20 10

26. X m (mid
point)
f
fm
0 – 10 (
0+10
2
)= 5 5 25
10 – 20 (
10+20
2
)= 15 10 150
20 – 30 (
20+30
2
)= 25 25 625
30 – 40 (
30+40
2
)= 35 30 1050
40 – 50 (
40+50
2
)= 45 20 900
50 - 60 (
50+60
2
)= 55 10 550
𝑓 = 100 𝑓𝑚 =3300
Let us take class of
marks as x and number
of students as f
X =
𝚺𝐟𝒎
N
=
3300
100
= 33
Average Mark of the
100 students is 33

27. X m (mid
point)
d (m-
A) A=
25
f
fd
0–10 (
0+10
2
)= 5 -20 5 -100
10–20 (
10+20
2
)=
15
-10 10 -100
20–30 (
20+30
2
)=
25
0 25 0
30–40 (
30+40
2
)=
35
10 30 300
40–50 (
40+50
2
)=
45
20 20 400
50–60 (
50+60
2
)=
55
30 10 300
𝑓 = 100 𝑓d = 800
Let us take marks as X and
Number of students as f
(Frequency)
X = 𝑨 +
𝚺𝐟𝒅
𝑵
= 25 +
800
100
= 25 + 8
= 33
Average mark of Students
is 33