Preparation of frequency
table:
The premise of data in the form of frequency distribution
describes the basic pattern which the data assumes in the mass. Frequency
distribution gives
a better picture of
the pattern of
data if the number of items is large. If the identity of the individuals about
whom a particular information is taken, is not relevant then
the first step of condensation is to divide the observed range of variable into a suitable number of class-intervals and to record the number of observations in each class. 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
|
Here the size of the class interval as per sturges rule is obtained as follows
Size of class interval = C =
Range
1+3.322
logN
= 64 - 32
1+3.322 log(50)
= 32 5
6.64
Thus the number of class interval is 7 and
size of each class is 5. The required size of each class is 5.
The required frequency distribution is prepared using tally marks as given below:
|
Class Interval
|
Tally marks
|
Frequency
|
|
30-35
|
|
2
|
|
35-40
|
|
6
|
|
40-45
|
 |
12
|
|
45-50
|
 |
14
|
|
50-55
|
|
6
|
|
55-60
|
|
6
|
|
60-65
|
|
4
|
|
Total
|
|
50
|
Example 2:
Given below are the number of tools produced by workers in a
factory.
|
43
|
18
|
25
|
18
|
39
|
44
|
19
|
20
|
20
|
26
|
|
40
|
45
|
38
|
25
|
13
|
14
|
27
|
41
|
42
|
17
|
|
34
|
31
|
32
|
27
|
33
|
37
|
25
|
26
|
32
|
25
|
|
33
|
34
|
35
|
46
|
29
|
34
|
31
|
34
|
35
|
24
|
|
28
|
30
|
41
|
32
|
29
|
28
|
30
|
31
|
30
|
34
|
|
31
|
35
|
36
|
29
|
26
|
32
|
36
|
35
|
36
|
37
|
|
32
|
23
|
22
|
29
|
33
|
37
|
33
|
27
|
24
|
36
|
|
23
|
42
|
29
|
37
|
29
|
23
|
44
|
41
|
45
|
39
|
|
21
|
21
|
42
|
22
|
28
|
22
|
15
|
16
|
17
|
28
|
|
22
|
29
|
35
|
31
|
27
|
40
|
23
|
32
|
40
|
37
|
Construct frequency distribution with inclusive type of class
interval. Also find.
1. How many workers produced more than 38 tools?
2. How many workers produced less than 23 tools?
Solution:
Using sturges formula for determining the number
of class
intervals, we
have
Number of class intervals = 1+ 3.322 log10N
= 1+ 3.322 log10100
=
7.6
Sizes of class interval =
Range
Number of class interval
= 46 - 13
7.6
5
Hence
taking
the
magnitude
of class intervals
as 5, we
have
7
classes 13-17, 18-22… 43-47
are the classes
by inclusive
type. Using tally marks, the required frequency distribution is obtain in the following table
Class
Interval
Tally Marks Number of tools
produced
(Frequency)
13-17
6
18-22
11
23-27
18
28-32
25
33-37
22
38-42
11
43-47
7
Total
100
Percentage frequency table:
The comparison becomes difficult and at times impossible
when the total number of items are large and highly
different one
distribution
to other.
Under
these
circumstances
percentage frequency distribution facilitates easy comparability. In percentage frequency table, we
have
to convert the actual frequencies
into percentages.
The percentages are calculated by using the formula given below:
Frequency percentage = Actual Frequency
Total Frequency
It is also called relative frequency table:
× 100
An example is
given below
to construct
a
percentage frequency table.
|
Marks
|
No. of
students
|
Frequency
percentage
|
|
0-10
|
3
|
6
|
|
10-20
|
8
|
16
|
|
20-30
|
12
|
24
|
|
30-40
|
17
|
34
|
|
40-50
|
6
|
12
|
|
50-60
|
4
|
8
|
|
Total
|
50
|
100
|
Cumulative frequency table:
Cumulative frequency
distribution has a running total of the
values. It is constructed by adding the frequency
of the first class interval to the frequency of the second class interval. Again add that total to the frequency in the third class interval continuing until the final total appearing opposite to the last class interval will be the total of all frequencies.
The cumulative frequency may be
downward or upward.
A downward cumulation results in a list presenting the number of frequencies “less than” any given amount as revealed by the lower limit of succeeding class interval and the upward cumulative results in a list presenting the number of frequencies “more than” and given amount is revealed by the upper limit of a preceding class interval.
Example 3:
|
Age
group
(in years)
|
Number
of women
|
Less
than
Cumulative
frequency
|
More than
cumulative
frequency
|
|
15-20
|
3
|
3
|
64
|
|
20-25
|
7
|
10
|
61
|
|
25-30
|
15
|
25
|
54
|
|
30-35
|
21
|
46
|
39
|
|
35-40
|
12
|
58
|
18
|
|
40-45
|
6
|
64
|
6
|
(a) Less than cumulative frequency distribution table
|
End values
upper
limit
|
less than
Cumulative
frequency
|
|
Less than 20
|
3
|
|
Less than 25
|
10
|
|
Less than 30
|
25
|
|
Less than 35
|
46
|
|
Less than 40
|
58
|
|
Less than 45
|
64
|
(b) More than cumulative frequency distribution table
|
End values
lower
limit
|
Cumulative frequency
more than
|
|
15 and above
|
64
|
|
20 and above
|
61
|
|
25 and above
|
54
|
|
30 and above
|
39
|
|
35 and above
|
18
|
|
40 and above
|
6
|
Conversion of cumulative frequency
to simple
Frequency:
If we have only cumulative frequency ‘ either less than or
more than’ , we can convert it into
simple frequencies.
For example if we have ‘ less than Cumulative frequency, we can convert this to
simple frequency by the method given below:
|
Class interval
|
‘ less than’
Cumulative frequency
|
Simple frequency
|
|
15-20
|
3
|
3
|
|
20-25
|
10
|
10 3 =
7
|
|
25-30
|
25
|
25 10 = 15
|
|
30-35
|
46
|
46 25 = 21
|
|
35-40
|
58
|
58 46 = 12
|
|
40-45
|
64
|
64 58 = 6
|
Method of converting ‘ more than’ cumulative frequency to simple
frequency is given below.
|
Class interval
|
‘ more than’
Cumulative frequency
|
Simple frequency
|
|
15-20
|
64
|
64 61 = 3
|
|
20-25
|
61
|
61 54 = 7
|
|
25-30
|
54
|
54 39 = 15
|
|
30-35
|
39
|
39 18 = 21
|
|
35-40
|
18
|
18 6
= 12
|
|
40-45
|
6
|
6 0
=
6
|
Cumulative percentage Frequency table:
Instead of cumulative frequency, if cumulative percentages
are given, the distribution is called cumulative percentage frequency distribution.
We can form this table either by converting the frequencies into percentages and then cumulate it or we can convert the given cumulative frequency into percentages.
Example 4:
|
Income (in Rs )
|
No. of
family
|
Cumulative
frequency
|
Cumulative
percentage
|
|
2000-4000
|
8
|
8
|
5.7
|
|
4000-6000
|
15
|
23
|
16.4
|
|
6000-8000
|
27
|
50
|
35.7
|
|
8000-10000
|
44
|
94
|
67.1
|
|
10000-12000
|
31
|
125
|
89.3
|
|
12000-14000
|
12
|
137
|
97.9
|
|
14000-20000
|
3
|
140
|
100.0
|
|
Total
|
140
|
|
|
Bivariate frequency distribution:
In the
previous
sections, we
described frequency
distribution involving
one variable only. Such frequency distributions are called univariate frequency distribution. In many
situations simultaneous study of two variables become necessary. For example, we want to classify
data
relating to the weights are height of a group of individuals, income and expenditure of a group
of
individuals, age
of
husbands and wives.
The data so classified on the basis of two variables give rise to the so called bivariate frequency distribution and it can be summarized in the form of a table is called bivariate (two-way) frequency table.
While
preparing
a
bivariate frequency distribution, the values of each variable are grouped into various classes (not necessarily the same for each variable) . If the data
corresponding to one variable, say
X is grouped into m classes and the data corresponding to the other variable, say Y is grouped into n classes then the bivariate table will consist of mxn cells.
By going through
the different pairs of
the values, (X,Y)
of the
variables and using tally marks we can find the frequency
of each
cell and thus, obtain the bivariate frequency table. The formate of a bivariate frequency table is given below:
Format of Bivariate
Frequency table|
x-series y-series
|
Class-Intervals
|
Marginal
Frequency of Y
|
|
|
Mid-values
|
|||
|
Class-intervals
|
MidValues
|
|
fy
|
|
Marginal
frequency of X
|
fx
|
Total
Ȉfx= Ȉfy=N
|
|
Here f(x,y) is the frequency
of
the pair (x,y). The frequency distribution of the values of the variables x together with
their
frequency total (fx) is called the marginal distribution of x and the
frequency distribution of the values of the variable Y together with
the total frequencies is known as the marginal frequency distribution of Y. The total of the values of manual
frequencies is called grand total (N)
Example 5:
The data given below relate to the height and weight of
20 persons. Construct a bivariate frequency
table with class interval of height as 62-64, 64-66…and weight as
115-125,125-135,
write down the marginal distribution of X and Y.
|
S.No.
|
Height
|
Weight
|
S.No.
|
Height
|
Weight
|
|
1
|
70
|
170
|
11
|
70
|
163
|
|
2
|
65
|
135
|
12
|
67
|
139
|
|
3
|
65
|
136
|
13
|
63
|
122
|
|
4
|
64
|
137
|
14
|
68
|
134
|
|
5
|
69
|
148
|
15
|
67
|
140
|
|
6
|
63
|
121
|
16
|
69
|
132
|
|
7
|
65
|
117
|
17
|
65
|
120
|
|
8
|
70
|
128
|
18
|
68
|
148
|
|
9
|
71
|
143
|
19
|
67
|
129
|
|
10
|
62
|
129
|
20
|
67
|
152
|
Solution:
Bivariate frequency table showing height and weight of persons.
|
Height(x)
Weight(y)
|
62-64
|
6
64-66
|
66-68
|
6
68-70
|
70-72
|
7
Total
|
|
115-125
|
II (2)
|
II (2)
|
|
|
|
4
|
|
125-135
|
I (1)
|
|
I (1)
|
II (2)
|
I (1)
|
5
|
|
135-145
|
|
III (3)
|
II (2)
|
|
I (1)
|
6
|
|
145-155
|
|
|
I (1)
|
II (2)
|
|
3
|
|
155-165
|
|
|
|
|
I (1)
|
1
|
|
165-175
|
|
|
|
|
I (1)
|
1
|
|
Total
|
3
|
5
|
4
|
4
|
4
|
20
|
The marginal distribution of height and weight are given in
|
|||||||||||||||||||||||||||||||||||||
the following table.
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