2. It is simple average of all items in a series. It is the simplest measure
of central tendencies.
No. of Matches = 5 + 6 + 7 + 8 + 9 = 35
No. of Boys = 5
Mean = Total Value of the Items = 35 = 7
No. of Items 5
FORMULA
= X1 + X2 + X3 + …….. + Xn / N = ∑X / NX
3. TYPES OFARITHMETIC MEAN
Arithmetic mean is of two types :
1. Simple arithmetic mean
2. Weighted arithmetic mean
METHODS OF CALCULATING SIMPLE ARITHMETIC
MEAN
We know, there are three types of statistical series :
1. Individual series
2. Discrete series
3. Frequency distribution
4. CALCULATION OF SIMPLE ARITHMETIC MEAN
In case of individual series, arithmetic mean may be calculated by 2
methods :
1. Direct method
2. Short-cut method
1. Direct method
= ∑X / N = Total value of the items / No. of items
ILLUSTRATION.
Pocket allowance of 10 students is rupees 15,20,30,22,25,18,40,50,
55,65. Find out the average pocket allowance.
X
6. 2 . SHORT-CUT METHOD :
This method is used when the size of item is very large.
d(deviation) = X-A
Formula:-
= A+ ∑d/N
CALCULATION OF SMALLARITHMETIC MEAN IN DISCRETE
SERIES OR FREQUENCY ARRAY
1. Direct method
2. Short-cut method
3. Step-deviation method
X
7. 1. Direct method
Formula:- = ∑fX / ∑f
ILLUSTRATION.
Following is the weekly wage earnings of 19 workers:
Solution:
= ∑fX / ∑f = 560/19 = 29.47
Mean wage earnings of 19 workers = Rs 29.47
X
Wages (Rs) 10 20 30 40 50
No. of workers 4 5 3 2 5
Wages (Rs)
(X)
No. of Workers or
Frequency (f)
Multiple of the
Value of X & Frequency (fx)
10 4 4 x 10 = 40
20 5 5 x 20 = 100
30 3 3 x 30 = 90
40 2 2 x 40 = 80
50 5 5 x 50 = 250
∑𝑓 = 19 ∑𝑓𝑋 = 560
X
8. 2 . Short-cut method
short-cut method of estimated mean of the discrete frequency series
user the following formula
Formula:-
=A+ ∑fd / ∑f
3 . Step-deviation method
(i) Step deviation d’ is obtained by dividing the deviation (of the
actual value from the assumed average) by the common factor.
d’ = X-A/C = d/C
Formula:-
= A+ ∑fd’ / ∑f x C
X
X
9. CALCULATION OF SIMPLE ARITHMETIC MEAN
IN CASE OF FREQUENCE DISTRIBUTION
1. Direct method
2. Short-cut method
3. Step-deviation method
1 . Direct method
Formula:-
= ∑fm / ∑f
2 . Short-cut method
Formula:-
= A+ ∑fd / ∑f
X
X
10. 3 . Step-deviation method
(i) Find out deviation of the mid value form some assumed average
That is,
d=m-A
Formula:-
= A+ ∑fd’ / ∑f x CX
11. CALCULATION OF ARITHMETIC MEAN IN CASE OF
CUMULATIVE FREQUENCY DISTRIBUTION
ILLUSTRATION.
Marks in Statics of student of Class XI are given below. Find out
arithmetic mean.
Solution:
A cumulative frequency distribution should first be converted into a
Simple frequency distribution, as under:
Marks No. of students
Less than 10 5
Less than 20 17
Less than 30 31
Less than 40 41
Less than 40 49
12. Conversion of a Cumulative Frequency Distribution
Into a simple frequency distribution
Now, mean value of the data is obtained using Direct method
Calculation of Mean
= ∑fm / ∑f = 1265 / 49 = 25.82
Arithmetic Mean = 25.82 marks
Marks No. of students
0-10 5
10-20 17 – 5 = 12
20-30 31 – 17 = 14
30-40 41 – 31 = 10
40-50 49 – 41 = 8
Marks(x) Mid-value
(m=l1+l2/2)
Number of students or
frequency(f)
Multiple of mid-value
&frequency (fm)
0-10 5 5 25
10-20 15 12 180
20-30 25 14 350
30-40 35 10 350
40-50 45 8 360
∑𝒇 = 49 ∑𝑓𝑚 = 1265
X
13. CALCULATION OF ARITHMETIC MEAN IN A MID-VALUE SERIES
ILLUSRTATION.
Following table given marks in statistics of the students of a class. Find out the
marks.
Solution:- In this services ,mid-values are already given. The calculation of
arithmetic mean involves the same procedure as in case of exclusive series.
CALCULATION OF ARITHMETIC MEAN
Mid-values 5 10 15 20 25 30 35 40
No. of students 5 7 9 10 8 6 3 2
Mid-values No. of students or frequency
(f)
Multiple of mid-values and frequency
(fm)
5 5 25
10 7 70
15 9 135
20 10 200
25 8 200
30 6 180
35 3 105
40 2 80
∑𝒇 = 50 ∑𝑓𝑚 = 995
14. = ∑fm / ∑f = 995 / 50 = 19.9
Mean marks = 19. 9
CALCULATION OF ARITHMETIC MEAN IN CASE OF INDUCTIVE
SERIES
ILLUSTRATION.
The following table show monthly pocket expense of the students of a class. Find out
the average pocket expenses.
Solution:-
Calculation of arithmetic mean of inclusive is the series is the same
exclusive series.
X
Pocket expenses(Rs) 20-29 30-39 40-49 50-59 60-69
No. of students 10 8 6 4 2
15. = A + ∑𝑓𝑑′ / ∑𝑓 x C = 44.5 + (-20) / 30 x 10
= 44.5 – 20 / 3 = 44.5 – 6.67 = 37.83
Average Pocket expenses = Rs 37.83
CALCULATION OF ‘CORRECTED’ARITHMETIC MEAN
=∑x(Wrong) + (Correct value) – (incorrect value) / N
Pocket expenses
(Rs)
Mid-values
(m=l1+l2/2)
No. of students or
frequency
(f)
Deviation
(d = m-A)
(A = 44.5)
Step-deviation
(d’= d/C)
(C = 10)
Multiple of
Step-deviation and
frequency
(fd)
20-29 24.5 10 - 20 - 2 - 20
30-39 34.5 8 - 10 - 1 - 8
40-49 44.5 6 0 0 0
50-59 54.5 4 + 10 + 1 + 4
60-69 64.5 2 + 20 + 2 + 4
∑𝑓 = 30 ∑𝑓𝑑′ = −20
X
X
16. WEIGHTED ARITHMETIC MEAN
CALCULATION OF WEIGHTED MEAN
Formula: W = ∑WX/ ∑W
COMBINED ARITHMETIC MEAN
Formula: 1,2 = 1N1 + 2N2 / N1 + N2
When there are more than 2 parts series, the following formula is used to
work out Combined Arithmetic Mean
Formula: 1,2,3……n = 1N1 + 2N2 +…+ nNn / N1 + N2 +…+N2
X
X X X
X X X X
17. MERITS ANS DEMERITS OF ARITHMETIC MEAN
Merits :- The following are some of the main merits of arithmetic mean:
1. SIMPLICITY
2. CERTAINTY
3. BASED ON ALL ITEMS
4. ALGEBRAIC TREATMENT
5. STABILITY
6. BASIS OF COMPARISON
7. ACCURACY TEST
Demerits:- Arithmetic mean suffers from following demerits:
1. EFFECT OF EXTREME VALUE
2. MEAN VALUE MAY NOT FIGURE IN THE SERIES AT ALL
3. UNSUITABILITY
4. MISLEADING CONCLUSIONS
18. MEASURES OF CENTRAL TENDENCY- MEDIAN AND MODE
1. MEDIAN: Median is a centrally located value of a series such that half of
the values (or items) of the series are above it and the other half below it.
Formula:-
M=Size of (N+1/2)th item
POSITIONAL
AVERAGES
MEDIAN PARTITION
VALUE
QUARTILE DECILE PERCEN
MODE
19. CALCULATION OF MEDIAN FOR DIFFERENT TYPE OF STATISTICAL
SERIES
a) INDIVIDUAL SERIES AND THE MEDIAN
b) DISCRETE SERIES OR FREQUENCY ARRAY AND THE MEDIAN
c) FREQUENCY DISTRIBUTION SERIESAND THE MEDIAN
Formula:- M=l1 +( N / 2-c.f.)/f x i
d) CUMULATIVE FREQUENCY SERIES AND THE MEDIAN
Illustration: Calculate median of the following series:
Solution:-
As a first step the cumulative frequency of ‘less than’ type is converted
into a simple frequency distribution as under :
Wage rate (Rs)
(lessthan)
10 20 30 40 50 60 70 80
No. of workers 15 35 60 84 96 127 198 250
20. CALCULATION OF MEDIAN
Solution:-
M = l1 + ((N/2 – cf ) / f) x i
M = 50 + ((250/2 – 96) / 31) x 10
M = 50 + ((125 – 96) / 31) x 10
M = 50 + (29 / 31) x 10
M = 50 + 9.32 = 59.32
Median wage rate = Rs 59.32
Wage rate (Rs) Cumulative Frequency Frequency (f)
0 - 10 15 15
10 - 20 35 35 – 15 = 20
20 - 30 60 60 – 35 = 25
30 - 40 84 84 – 60 = 24
40 - 50 96 (c.f.) 96 – 84 = 12
(l1) 50 - 60 127 127 – 96 = 31 (f)
60 - 70 198 198 – 127 = 71
70 - 80 250 250 – 198 = 52
∑𝑓 = 𝑁 = 250
21. e) INCULDING SERIES AND THE MEDIAN
ILLUSTRATION.
Calculate median of the following data:
Solution:- This is an inclusive series given in the descending order. It should be
converted into an exclusive series and place in the ascending order, as in the
following tables:
Marks 46-50 41-45 36-40 31-35 26-30 21-25 16-20 11-15
No. of
students
5 11 22 35 26 13 10 7
Conversion into Exclusive
Series
Frequency
(f)
Cumulative frequency
10.5 – 15.5 7 7
15.5 – 20.5 10 17
20.5 – 25.5 13 30
25.5 – 30.5 26 56 (c.f.)
(l1) 30.5 – 35.5 35 (f) 91
35.5 – 40.5 22 113
40.5 – 45.5 11 124
45.5 – 50.5 5 129
N = 129
22. Median, M = Size of (N/2)th item; N=∑𝑓 = 129
= Size of (129/2)th item = Size of 64.5th item
Using the formula,
M = l1 + (N/2 - c.f.)/f x i
=30.5+(129/2-56)/35 x 5
=30.5+(64.5-56/35) x 5
=30.5+8.5/35 x 5
=30.5+1.2
=31.7
Median = 31.7 marks.
23. 6. MEDIAN OF THE SERIES WITH UNEQUAL CLASS INTERVALS
ILLUSTRATION:- Calculating median of the following distribution of data:
Solution :
Estimation of Median
CLASS INTERVAL 0-5 5-10 10-20 20-30 30-50 50-70 70-100
NUMBER OF
STUDENTS
12 15 25 40 42 14 8
Class Interval Frequency (f) Cumulative
Frequency
0 – 5 12 12
5 – 10 15 27
10 – 20 25 52 (c.f.)
(l1) 20 – 30 40 (f) 92
30 – 50 42 134
50 – 70 14 148
70 - 100 8 156
N = 156
24. M = Size of (N/2)th item; N = 156
= Size of (156/2)th item = Size of 78th item
this lies in 92th cumulative frequency and the corresponding median class is
20-30.
:- l1 = 20, c.f. = 52 , f = 40 and i = 10
substituting the values in the formula, we have
M = l1 + (N/2-c.f./f) x i
= 20 + (156/2-52/40) x 10
= 20 +( 78-52/40) x 10
= 20 +(26/40) x 10
= 20 +6.5
= 26.5
Median =26.5
25. MERITS OF MEDIAN
a. SIMPLICITY
b. FREE FROM THE EFFECT OF EXTREME VALUES
c. CERTAINTY
d. REAL VALUE
e. GRAPHIC PRESENTATION
f. POSSIBLE EVEN WHEN DATA IS INCOMPLETE
DEMERITS OF MEDIAN
a. LACK OF REPRESENTATIVE CHARACTER
b. UNREALISTIC
c. LACK OFALGEBRAIC TREATMENT
26. 2. PARTITION VALUE : QUARTILE
The value that divides the series into more than two parts is called partition value.
If a statistical series is divided into four equal parts, the end value of each part is
called a quartile .
a) Individual and discrete series
Formulae: Q1 = Size of (N+1/4)th item of the series
Q3 =Size of 3 (N+1/4)th item of the series
b) Frequency distribution series
Class interval of Q1 = Size of (N/4)th item
Class interval of Q3 = Size of 3 (N/4)th item
Formulae: Q1 = l1 + [N/4-c.f] / f x i
Q3 = l1 + [3(N/4)-c.f] / f x i
27. 3. MODE
Mode is another important measure of central tendency of statistical series. It
is the value which occurs most frequently in the series; that is ,model value has the
highest frequency in the series .
CALCULATION OF MODE
i) CALCULATION OF MODE IN INDIVIDUAL SERIES
a) By inspection
b) By converting individual series into discrete frequency series.
28. ii) CALCULATION OF MODE IN DISCRETE SERIES OR FREQUENCY
ARRAY
a) Inspection method
b) Grouping method
iii) CALCULATION OF MODE IN FREQUENCY DISTRIBUTION SERIES
a) Inspection method
b) Grouping method
Formula:-
Z = l1 + (f1 – f0 / 2f1 – f0 – f2) x i
29. MERITS OF MODE
1. Simple and popular
2. Less effect of marginal values
3. No need of knowing all the items or frequencies
DEMERITS OF MODE
1. Uncertain and vague
2. Difficult
3. Ignores extreme marginal frequencies