Measures of central tendency

1. Prepared by:
Ms. Nishat Anjum
Asst. Professor
Springdale Mahila Mahavidyalaya,
Bareilly
EDUCATIONAL
STATISTICS

2. MEASURES OF CENTRAL TENDENCY
 MEAN
 MEDIAN
 MODE

3. Types of series
1.Individual series
2.Frequency series:
a.Discrete series/ frequency array.
b.Frequency distribution/ Continuous
series:
1. Exclusive series 2. Inclusive
series.

4. Individual Series
 Contains only one variable.
 e.g. Marks, height, weight etc.
Roll no. Marks
1
2
3
4
5
09
07
10
05
06
6
7
8
9
10
06
10
09
08
05

5. Discrete series or frequency array
 Variable + frequency
Marks (x) No. of
students (f)
06
07
08
10
15
18
09
10
12
05
N=60

6. Frequency distribution or continuous
series
 Variable + frequency
 Variable in the form of class interval.
Class interval (Marks) No. of students
0 - 10
10 - 20
20 - 30
30 - 40
10
30
40
20
N= 100

7. Continuous series:
i. Exclusive series ii. Inclusive series
 Exclusive series: Upper limit is excluded.
Class interval (Marks) No. of students
(f)
0 - 10
10 - 20
20 - 30
30 - 40
10
30
40
20
N= 100
Lower
class
limit
Upper
class
limit

8. Class interval (Marks) No. of students (f))
11 - 20
21 – 30
31 - 40
41 - 50
10
30
40
20
N= 100
Inclusive series: Upper limit is included.
Conversion of inclusive series to exclusive series
(i) Find the difference between the upper limit of class interval and the lower limit of the next
class interval. For example the upper limit of the class interval 11-20 is 20.The lower limit of
the next class interval 21-30 is 21. The difference is 21 minus 20 =1.
(ii) Secondly, half of that difference is added to the upper limit of a class interval and half is
subtracted from the lower limit of the class interval.
Half of the difference found in first step will be 0.5.
Add 0.5 to the upper limit and subtract 0.5 from the lower limit for each class interval.
e.g. 11-0.5= 10.5; 20+0.5= 20.5 & so on.
Using these two steps, inclusive series csn be converted into an exclusive series

9. MEAN
“Mean is the centre of gravity of a
distribution.”
-F.C. Mills
 Generally known as ‘Average’.
 Denoted by

12. Methods of
Calculation

13. Individual series (Direct/Long Method)
Formula: Mean=Σx/N
Σx=sum of frequencies
N= No. of items.
Roll no. Marks (x)
1
2
3
4
5
09
07
10
05
04
Σx=35
Mean=Σx/N
=35/5
= 7

14. Individual series (Shortcut Method)
Formula: Mean=A+ Σd/N
A= Assumed Mean, Σ=Summation
d = deviations from mean, N= No. of items.
Roll no. Marks (x) d(x-a)
1
2
3
4
5
09
07
10
05
04
04
02
05
00
-1
Σd=10
Mean=A+ Σd/N
= 5+ 10/5
= 5 + 2
= 7

15. Mean=Σfx/N
=467/60
=7.7
Discrete Series (Direct Method)
Formula: Mean=Σfx/N
A= Assumed Mean, Σ=Summation
N= No. of items.
Marks
(x)
No. of students
(f)
fx
06
07
08
10
15
18
60
105
144
09
10
12
05
N= 60
108
50
Σfx=467

16. Discrete Series (Shortcut Method)
Formula: Mean=A+ Σfd/N
A= Assumed Mean, Σ=Summation
d = deviations from mean, N= No. of items.
Marks
(x)
No. of students
(f)
d(x-a) fd
06
07
08
10
15
18
-2
-1
0
-20
-15
0
09
10
12
05
N= 60
1
2
12
10
fd=-13
Mean=A+ Σfd/N
= 8 + (-13)/ 60
= 8 - 0.21
= 7.79

17. Continuous series:
i. Exclusive series ii. Inclusive series
 Exclusive series: Upper limit is excluded.
Class interval
(Marks)
No. of students
(f)
Mid value
(x)
x=(U+L)/2
fx
0 - 10
10 - 20
20 - 30
30 - 40
10
30
40
20
05
15
25
35
50
450
1000
700
N= 100 Σfx=2200
Mean=Σfx/N
=2200/100
=22
Long method/
Direct method

18. Inclusive series: Upper limit is included.
Class interval
(Marks)
No. of students
(f))
Mid value (x) fx
11 - 20
21 – 30
31 - 40
41 - 50
10
30
40
20
15.5
25.5
35.5
45.5
155
765
1420
910
N= 100 Σfx= 3250
Mean=Σfx/N
=3250/100
= 32.5
Long method/
Direct method

19. Continuous Series (Shortcut Method)
Formula: Mean=A+ Σfd/N
A= Assumed Mean, Σ=Summation
d = deviations from mean, N= No. of items.
Class interval
(Marks)
Marks
(x)
No. of
students
(f)
d(x-a)
a=35
fd
10-20
20-30
30-40
15
25
35
10
15
18
-20
-10
0
-200
-150
0
40-50
50-60
45
55
12
05
N= 60
10
20
120
100
fd=-130
Mean=A+ Σfd/N
= 35+ (-130)/ 60
= 35 – 2.1
= 32.9

20. Continuous Series (Step-deviation Method)
Formula: Mean=A+( Σfd’/N)×i
A= Assumed Mean, Σ=Summation
d = deviations from mean, N= No. of items.
Class
interval
(Marks)
Mid value
(x)
No. of
students
(f)
d(x-a)
a=35
d’=d/i
i=10
fd’
10-20
20-30
30-40
15
25
35
10
15
18
-20
-10
0
-2
-1
0
-20
-15
0
40-50
50-60
45
55
12
05
N= 60
10
20
1
2
12
10
fd’=-13
Mean=A+ (Σfd’/N) ×i
= 35+ (-13/ 60 )×10
= 35 + (-0.21) × 10
= 35- 2.1
= 32.9

21. Questions for practice.
 Calculate mean from the following ungrouped data.
निम्िलिखित अवर्गीकृ त आँकड़ो से मध्यमाि ज्ञात कीजिए।
Name of
students
छात्रों क
े नाम
Vijay Veer David Alok Sam Ram Amir Shyam Varun
Marks 21 45 32 38 23 28 30 50 43
Calculate mean from the following data by direct method and step deviation method.
C.I 70-
71
68-
69
66-
67
64-
65
62-
63
60-
61
58-
59
56-
57
54-
55
52-
53
50-
51
ƒ 2 2 3 4 6 7 5 1 2 3 1

22. MEDIAN:
•Median would divide the series in two parts.
•Median is the value of the middle item of a series
arranged in ascending or descending order of magnitude.
•Denoted by (M).

24. Individual Series:
Step 1: Arrange the observations in ascending or descending
order.
Step 2: Determine the total no. of observations, ‘n’.
Step 3:
a. If ‘n’ is odd, then
Median = {(n+1)/2}thterm
b. If ‘n’ is even, then
Median = [(n/2)th term + {(n/2)+1}th]/2

25. Individual series (odd)
Formula: Median={(n+1)/2}th term
n= No. of observations.
Roll no. Marks (x)
1
2
3
4
5
6
7
09
07
10
05
04
06
08
Arrange the data in ascending order.
04, 05, 06, 07, 08, 09, 10
Apply the formula,
Median={(n+1)/2} th term
={(7+1)/2} th term
={8/2} th term
={4} th term
=07
So, M=07

26. Individual series (even)
Formula:
Median= [(n/2)th term + {(n/2)+1}th term]/2
Roll no. Marks (x)
1
2
3
4
5
6
09
07
10
05
04
06
Arrange the data in ascending order.
04, 05, 06, 07, 09, 10
Apply the formula,
Median=[(n/2)th term + {(n/2)+1}th term]/2
=[(6/2)th term + {(6/2)+1}th term]/2
=[(3)rd term + {(3)+1}th term]/2
=[(3)rd term + {4}th term]/2
=[6+7]/2
=13/2
=6.5
So, M=6.5

27. Discrete Series
 Step-1:Arrange the data in ascending or descending
order.
 Step-2: Compute cumulative frequencies (CF).
 Step-3: Apply the formula:
Median =Size of [(n+1)/2]th item.
 Step-4: Look at the cumulative frequency column,
Just equal to [(n+1)/2]th item or greater than
[(n+1)/2] th item.
And determine the value of the variable
corresponding to this .This gives the value of median .
Here ‘n’ means CF .

28. Marks (x) No. of students (f) Cumulative frequency
10 2 2
20 8 2+8=10
30 16 10+16=26
40 26 26+26=52
50 20 52+20=72
60 16 72+16=88
70 7 88+7=95
80 4 95+4=99
N=99
M= size of [(N+1)/2]th item.
= size of [(99+1)/2]th item.
= size of [100/2] th item.
=size of 50th item.
Median= 40 marks
Discrete series

29. Marks (x) No. of students (f) Cumulative frequency
50 20 20
70 15 20+15=35
80 12 35+12=47
100 15 47+15=62
110 18 62+18=80
Discrete series
N=80
M= size of [(N+1)/2]th item.
= size of [(80+1)/2]th item.
= size of [81/2] th item.
=size of 40.5th item.
Median= 80 marks
Marks(x) 100 50 70 110 80
No.of
students (f)
15 20 15 18 12

30. Questions for practice
Marks 3 5 6 7 8 12
No.of
students
5 4 2 6 6 7
Weight
(in kg)
50 70 45 55 75 80 65 60
No. of
students
5 4 2 16 3 1 20 19
Items 145 70 45 36 14 105
Freq. 0 3 1 5 2 12
Items 70 50 10 5 20 40 60 30
Freq. 7 0 5 2 1 12 5 3

31. Continuous Series
 Step-1: Prepare cumulative frequencies (CF).
 Step-2: Calculate Σƒ=n,
 Step-3: Find out Size of (n/2)th item.
 Step-4: Look at the cumulative frequency column, Just equal to (n/2)th item or
greater than (n/2) th item. Determine the corresponding class (Median class).
 Step-5: Apply formula:
Here,
L= lower limit of median class.
f= frequency of median class.
Cf= cf of the class preceding to median class.
i= size/ difference of median class.

32. Compute median from the following data:
Marks 0-10 10-20 20-30 30-40 40-50 50-60
No. of students 12 18 27 20 17 6
Marks
No. of
students
cf
0-10 12 12
10-20 18 30
20-30 27 57
30-40 20 77
40-50 17 94
50-60 6 100
N=100
Median class= size of (N/2)th item
= size of (100/2)th item.
=size of 50th item.
Median class= 20-30
= L+[(n/2-cf)/f] × i
= 20+[(50-30)/27] × 10
=20+ [20/27] × 10
=20+ [0.74]×10
=20+7.4
=27.4
L=20, lower limit of mdn class
Cf=30,cf of class preceding class
interval
i=10, size of class
n/2= 50
F=27

33. MODE:
 “Value of observations which occur the
greatest no. of times or with the greatest
frequency”.
 Denoted by ‘Z.’

34. Individual series:
 (i) Data Array: The data are arrayed- the value of item which
occurs most frequently is the mode of the distribution.
Marks 4 6 5 7 9 10 4 7 6 5 8 7 7 8 9
Marks 4 4 5 5 6 6 7 7 7 7 8 8 9 9 10
Modal value
Mode = 7 Marks

35. Individual series:
(ii) Discrete Series: The data are converted into discrete
series to identify quickly the most frequently repeated value.
Ques. Find out the mode of the given data:
Marks: 4,6,5,7,9,8,10,4,7,6,5,8,7,7,9.
Marks Tally Bars Frequency
4 II 2
5 II 2
6 II 2
7 IIII 4
8 II 2
9 II 2
10 I 1
Modal freq.
Mode = 7 Marks

36. Discrete series
 (i) By inspection: The mode can be determined just by
inspection in discrete series, the size around which the items
are most heavily concentrated will be decided as ‘Mode’.
Marks Number of students
125 3
175 2
225 21
275 6
325 4
375 2
By inspection we can determine that the model mark is 225
because this value occurred the maximum number of times that
is 21 times

37. Discrete series
 (ii) By grouping: In grouping method data are arranged in
ascending order and the frequencies against each observation are
written against each item properly.
 A grouping normally consists of 6 columns.
Column-1: Greatest frequency is observed by putting a circle.
Column-2: Frequencies are grouped in two’s.
Column-3: Leaving the first frequency, other frequencies are grouped in two’s.
Column-4: Frequencies are grouped in threes.
Column-5: Leaving the first frequency, other frequencies are grouped in threes.
Column-6: Leaving the first two frequency, other frequencies are grouped in threes.
An analysis table is prepared after completing grouping table in order to find out the
item which is repeated maximum number of times.

38. Discrete series
 Ques. Find the mode of the following data:
Marks 125 175 225 275 325 375
Freq. 3 8 21 6 4 2
Marks Freq. (1) (2) (3) (4) (5) (6)
125 3
175 8
225 21
275 6
325 4
375 2
425 1
11
27
06
29
10
32
12
35
31

39. Analysis Table
Column 125 175 225 275 325 375
1 |
2 | |
3 | |
4 | | |
5 | | |
6 | | |
1 3 6 3 1
Since, the value 225 has been repeated maximum no. of times in the series,
so 225 is modal mark.

40. Marks 3 5 6 7 8 12
No.of
students
5 4 2 6 6 7
Weight
(in kg)
50 70 45 55 75 80 65 60
No. of
students
5 4 2 16 3 1 20 19
Items 145 70 45 36 14 105
Freq. 6 3 1 5 2 12
Items 70 50 10 5 20 40 60 30
Freq. 7 4 5 2 1 12 5 3
Questions for practice (Discrete series)
Solve the questions by grouping method.

41. Continuous Series(Grouping Method)
Marks Freq. (1) (2) (3) (4) (5) (6)
125-175 3
175-225 8
225-275 21
275-325 6
325-375 4
375-425 2

42. Colum
n
125-
175
175-
225
225-
275
275-
325
325-
375
375-
425
1 |
2 | |
3 | |
4 | | |
5 | | |
6 | | |
1 3 6 3 1
Analysis Table
Mode lies in
class 225-275.
lí= 225,lower limit of modal class.
F =21, freq.of modal class.
= = 8, freq. preceding modal class.
= =6, freq. succeeding modal class.
i=50, size of modal class.
Z= 225+[(21-8)/(2×21-8-6)]×50
=225+[(13)/(42-14)]×50
= 225+[13/28]×50
=225+0.46×50
=225+23
=248

43. Marks Freq.
(1)
125-175 3
175-225 8
225-275 21
275-325 6
325-375 4
375-425 2
Mode lies in
class 225-275.
Z= 225+[(21-8)/(2×21-8-6)]×50
=225+[(13)/(42-14)]×50
= 225+[13/28]×50
=225+0.46×50
=225+23
=248
Continuous Series(2nd Method)

44. Alternative Formula
Mode = 3 Median -2 Mean
=3×25- 2× 20
= 75-40
= 35

45. Marks 0-12 12-24 24-36 36-48 48-60 60-72
No. of students 10 5 25 10 11 19
Compute Mean, Median & Mode from the following data:
Marks 0-10 10-20 20-30 30-40 40-50 50-60
No. of students 10 15 25 20 16 9
Marks 70-75 75-80 80-85 85-90 90-95 95-100
No. of students 12 18 25 20 16 14
Marks 0-15 15-30 30-45 45-60 60-75 75-90
No. of students 20 15 13 17 25 10
Marks 0-10 10-20 20-30 30-40 40-50 50-60
No. of students 20 25 15 20 17 13
Marks 0-3 3-6 6-9 9-12 12-15 15-18
No. of students 2 6 3 4 8 7
Questions for practice
By: Nishat Anjum
Asst. Prof.
Springdale Mahila Mahavidyalaya.

46. Any Query?
Thank you
Presented by:
Ms.Nishat Anjum
Assistant Prof.
Springdale Mahila Mahavidyalaya, Bareilly.