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Frequency Distribution
FREQUENCY DISTRIBUTION 4 3 4 5 7
6 5 4 5 6
5 4 5 3 5
4 5 6 3 5
4 3 4 5 4
Table 1:- shows the daily number of car accident in a certain city recorded over the period of 30 days.
Simple series or ungrouped data: It is a series of observations recorded without any systematic arrangement e.g. Table 1.The ungrouped data will be so numerous that their significance will not be readily comprehended. In such cases it will become necessary to summaries the data to an easily manageable form. We may summaries the raw data with the help of Frequency Distribution.
FREQUENCY DISTRIBUTION
Frequency Distribution is a statistical table which shows the
values of the variable arranged in order of magnitude and also the corresponding frequencies side by side. Q) What is Frequency? Number of times the particular value is repeated is known as frequency. In order to facilitate counting we may prepare a column of Tally. In another column, places all possible values of the variable from the lowest to highest and count the number of tally bar corresponding to each value of the variable and place it in the column of frequency, as below
Daily no of carTally Bar accidents 3 IIIII 4 IIIIIIIII 5 IIIIIIIIIII 6 IIII 7 I
Frequency (No of days) 5 9 11 4 1
Total
30
Table 2:Simple frequency distribution of car accidents.
Types of frequency distribution
There are two types of frequency distribution 1) Simple frequency Distribution-shows the value of the variable individually, e.g. Table 2 2) Grouped Frequency distribution-shows the value of the variable in groups. Generally when the number of variable is too large using grouped frequency distribution would be more advantageous than that of simple frequency distribution. For example: consider the following data showing the age (in years) of 200 unemployed persons who ed their names in an Employment Exchange on a certain day.
33
26
33
30
28
25
28
22
17
30
26
17
29
22
18
40
20
33
21
25
25
20
21
29
27
26
19
37
32
21
20
40
20
20
27
19
20
20
29
20
21
49
34
26
29
31
25
33
23
16
21
20
23
29
23
17
23
27
43
22
24
26
29
29
20
17
32
23
24
25
36
19
28
26
31
52
25
57
16
26
27
19
24
19
39
23
24
25
17
19
31
18
26
23
19
30
20
19
50
40
22
27
17
22
43
33
22
17
21
25
24
46
27
20
26
20
21
24
22
21
25
22
24
32
24
20
22
15
34
22
22
20
37
19
24
21
28
27
19
31
22
24
17
29
24
17
23
21
21
31
31
18
22
19
33
25
21
25
23
21
27
18
15
19
24
19
22
22
16
21
17
18
25
31
26
29
24
31
23
18
31
24
23
21
24
18
21
19
30
24
24
22
30
19
25
21
21
25
45
24
The first step towards summarization of data is to arrange the figure in the form of a simple frequency distribution as follows: Ages(year) Frequency Ages(year) 15 2 26 16 3 27 17 10 28 18 8 29 19 30 20 14 31 21 15 32 22 19 33 23 16 34 24 11 36 25 20 37 14
Frequency 10 7 3 9 5 8 3 6 2 1 2
Ages(year) 39 40 43 45 46 49 50 52 57
Frequency 1 3 2 1 1 1 1 1 1
Total
200
Although the table gives a somewhat clear picture under the simple frequency distribution but since the number of observations are too large it will be more representable if we group the observations(age) into number of intervals i.e. 15-19, 19-24 etc. and shows the frequency of occurrence (here the number of persons falling in the particular age group)separately for each interval. Such frequency table is known as Group Frequency Distribution. Age in year
Frequency(No of persons)
15-19 20-24 25-29 30-34 35-44 45-49
37 81 43 24 9 6
Total
200
Different component of Grouped frequency distribution
a) Class Interval
When the numbers of observations are too large i.e. varying in wide range, these are usually classified in several groups according to their size of values. Each of these groups defined as class interval or simply class. In our example we have classified the ages into 6 different interval i.e. 15-19,…….., 45-49 since the ages are varying in wide range from 17 to 57.
Although there is no hard and fast rule regarding the number of class intervals, It is generally agreed that the number classes neither be too large then the basic purpose of classification (i.e. summarization of data to an easily manageable form.) will not be served nor be too small then the true nature of the distribution will be obscured. As a working rule this number should lie between 5 and 15 depending upon the number of observations available.
One should have class intervals of either five or multiple of five like 10, 20, 25 etc.The reason is that the human mind is accustomed more to think in of certain multiples of 5,10 and the like. Classes must be mutually exclusive means that each of the given values is included in one of the classes . For Example:- 15-19 , 20-24 etc. (not like 15-20, 20-25 etc )
b) Class Frequency
The number of observations falling within a class is called class frequency. Sum of all class frequency is total frequency. In our example the class frequencies are 37,81 etc.
c) Class Limit
The two numbers used to specify the limits of a class interval are called Class Limits. The smaller of the pair is known as lower class limit and the larger as upper class limit. In our example, column 3 is the lower class limits and column 4 is the upper class limits as shown in the Table-3.
Class Interval (1) 15-19 20-24 25-29 30-34 35-44 45-49 Total
Class frequency (2) 37 81 43 24 9 6
Class Limit Lower Upper (3) (4) 15 20 25 30 35 45
200 Table-3
19 24 29 34 44 59
d) Class Boundaries
In our present example the variable is discrete since the age is expressed only in years. But if we consider age as continuous variable, then it may be 14.5 years of 19.5 years, then the concept of class limit will not suffice.(In which class the observation 14.5 or19.5 will be included?). So in case of continuous variable, certain adjustment in the class interval is needed to obtain the continuity.
The extreme values which would ever be included in the class interval in order to obtain the correct class interval in case of continuous variable is known as Class Boundaries. The lower extreme point is known as the lower class boundary and the upper extreme point is known as the upper class boundary with reference to the particular class. In our example, column (5) shows the lower class Boundaries and column (6) shows the upper class Boundaries (as shown in the Table-4)
Correction Factor (d) (Lower limit of the second class - Upper limit of the first class)
= ----------------------------------------------------------------2 Lower Class Boundary= Lower Class Limit - d Upper Class Boundary= Upper Class Limit + d
Class Interval (1) 15-19 20-24 25-29 30-34 35-44 45-49 Total
Class frequency (2) 37 81 43 24 9 6
Class Boundaries
Lower (5) 14.5 19.5 24.5 29.5 34.5 44.5
200 Table-4
Upper (6) 19.5 24.5 29.5 34.5 44.5 59.5
e) Class Mark The value exactly at the middle of a class interval is called Class Mark. Class mark is the representative value of the class interval, for the calculation of mean, SD etc.In our example column (7) of table -5 shows the class marks.
Class Mark= (Lower Class Limit+ Upper Class Limit) ÷ 2 Or (Lower Class Boundary+ Upper Class Boundary) ÷ 2
Class Interval (1)
Class frequency (2)
Class Mark (7)
15-19 20-24 25-29 30-34 35-44 45-59
37 81 43 24 9 6
15+19 = 34 ÷2= 17 20+24 = 44 ÷2= 22 25+29 =54 ÷2 = 27 30+34 = 64 ÷2 = 32 35+44 =79 ÷2 = 39.5 45+59 =104 ÷2= 52
Total
200
e) Width of class or Size of the class 1) Width of class (in case of continuous variable) = Upper Class Boundary - Lower Class Boundary. or Upper Class limit - Lower Class limit. For example:Class boundaries
Width
14.5 – 19.5 19.5 – 24.5
5 5
Class limits
Width
15 – 20 20 - 25
5 5
2) Width of class (in case of discrete variable) = Upper Class limit of the second class - Lower Class limit of the first class. For example:Class limits
Width
15-19 20-24 25-29 30-34 35-44 45-59
20 -15 = 5 25 – 20 = 5 30 -25 = 5 35 – 30 = 5 45 – 35 = 10 60 – 45 = 15
How to choose the number of classes and class limits Step I :- the following points The number of classes should be preferably between 5 and 15. The width of the classes should be preferable be 5 or its multiple e.g., 15-19,2024 etc. One of the class limit being preferable a multiple of 5.
Step II :Apply the Sturges formula for determining the approximate number of classes,i.e.
k = 1 + 3.322 log N Where k = The approximate number of classes N = Total number of observation
Step III :The third step is to find the maximum and minimum of the observations and find the “range” i.e. Range = Maximum - Minimum value If we like to have classes of equal width, then the approximate width of the class may be obtained on dividing the range by the approximate number of classes i.e.
Range Approximate width = -------------------------------
1 + 3.322 log N
For example:Problem 1:The profits (in lakh rupees) of 30 companies for the year 2008 is given below. 20,22,35,42,37,42,48,53,49,65,39,48,67, 18,16,23,37,35,49,63,65,55,45,58,57,69,25,29,58,65. Classify the above data taking suitable class-interval
Solution:Step I :- Find the approximate number of classes. k = 1 + 3.322log N Where N = 30 K = 1 + 3.322 log 30 = 1 + 3.322 x 1.4771 = 7 + 4.91 = 5.91 or 6
Step II :-Width of the classes Maximum - Minimum Width of the classes = ----------------------------------------------------1 + 3.322log N ( 69 – 16) = ------------------------------5.91 53 = --------------5.91 = 8.97 or 9 But since width of the classes preferable be 5 or multiple of 5 i.e. 10,15 We consider width as 10 instead of 9.
Frequency distribution of the profits
Profit (Rs. Lakh) 15 – 24 25 – 34 35 – 44 45 - 54 55 – 64 65 – 74 Total
No of companies 5 2 7 6 5 5 30
The lower limit of the first class may not coincide with the minimum observation,since the width of the classes and the one of the class limit being preferable multiple of 5. For example:- Here lowest value of the data is 16 and we have to take the classes of width 10 and one of the class limit as multiple of 5, then the lower limit of the first class may not coincide with the minimum observation i.e. it should be 15 – 24 not 16 – 25.
Problem 2:The following set of numbers presents the mutual fund prices at the end of the week for the selected 40 nationally sold funds 11,17,15,22,11,16,19,24,29,18 25,26,32,14,17,20,23,27,30,12 15,18,24,36,18,15,21,28,33,38 34,13,11,16,20,22,29,29,23,31 Classify the above data taking suitable class-interval
Solution Ans) k = 1 + 3.322 log 40 = 1+ 3.322 x 1.602 = 1 + 5.32 = 6.32 or 6 (38 – 11) Width = ------------------ = 4.27 or 5 6.32
Class Interval
Frequency
10 – 14
6
15 - 19
11
20 – 24
9
25 - 29
7
30 – 34
5
35 - 39
2
Total
40
FREQUENCY DISTRIBUTION 4 3 4 5 7
6 5 4 5 6
5 4 5 3 5
4 5 6 3 5
4 3 4 5 4
Table 1:- shows the daily number of car accident in a certain city recorded over the period of 30 days.
Simple series or ungrouped data: It is a series of observations recorded without any systematic arrangement e.g. Table 1.The ungrouped data will be so numerous that their significance will not be readily comprehended. In such cases it will become necessary to summaries the data to an easily manageable form. We may summaries the raw data with the help of Frequency Distribution.
FREQUENCY DISTRIBUTION
Frequency Distribution is a statistical table which shows the
values of the variable arranged in order of magnitude and also the corresponding frequencies side by side. Q) What is Frequency? Number of times the particular value is repeated is known as frequency. In order to facilitate counting we may prepare a column of Tally. In another column, places all possible values of the variable from the lowest to highest and count the number of tally bar corresponding to each value of the variable and place it in the column of frequency, as below
Daily no of carTally Bar accidents 3 IIIII 4 IIIIIIIII 5 IIIIIIIIIII 6 IIII 7 I
Frequency (No of days) 5 9 11 4 1
Total
30
Table 2:Simple frequency distribution of car accidents.
Types of frequency distribution
There are two types of frequency distribution 1) Simple frequency Distribution-shows the value of the variable individually, e.g. Table 2 2) Grouped Frequency distribution-shows the value of the variable in groups. Generally when the number of variable is too large using grouped frequency distribution would be more advantageous than that of simple frequency distribution. For example: consider the following data showing the age (in years) of 200 unemployed persons who ed their names in an Employment Exchange on a certain day.
33
26
33
30
28
25
28
22
17
30
26
17
29
22
18
40
20
33
21
25
25
20
21
29
27
26
19
37
32
21
20
40
20
20
27
19
20
20
29
20
21
49
34
26
29
31
25
33
23
16
21
20
23
29
23
17
23
27
43
22
24
26
29
29
20
17
32
23
24
25
36
19
28
26
31
52
25
57
16
26
27
19
24
19
39
23
24
25
17
19
31
18
26
23
19
30
20
19
50
40
22
27
17
22
43
33
22
17
21
25
24
46
27
20
26
20
21
24
22
21
25
22
24
32
24
20
22
15
34
22
22
20
37
19
24
21
28
27
19
31
22
24
17
29
24
17
23
21
21
31
31
18
22
19
33
25
21
25
23
21
27
18
15
19
24
19
22
22
16
21
17
18
25
31
26
29
24
31
23
18
31
24
23
21
24
18
21
19
30
24
24
22
30
19
25
21
21
25
45
24
The first step towards summarization of data is to arrange the figure in the form of a simple frequency distribution as follows: Ages(year) Frequency Ages(year) 15 2 26 16 3 27 17 10 28 18 8 29 19 30 20 14 31 21 15 32 22 19 33 23 16 34 24 11 36 25 20 37 14
Frequency 10 7 3 9 5 8 3 6 2 1 2
Ages(year) 39 40 43 45 46 49 50 52 57
Frequency 1 3 2 1 1 1 1 1 1
Total
200
Although the table gives a somewhat clear picture under the simple frequency distribution but since the number of observations are too large it will be more representable if we group the observations(age) into number of intervals i.e. 15-19, 19-24 etc. and shows the frequency of occurrence (here the number of persons falling in the particular age group)separately for each interval. Such frequency table is known as Group Frequency Distribution. Age in year
Frequency(No of persons)
15-19 20-24 25-29 30-34 35-44 45-49
37 81 43 24 9 6
Total
200
Different component of Grouped frequency distribution
a) Class Interval
When the numbers of observations are too large i.e. varying in wide range, these are usually classified in several groups according to their size of values. Each of these groups defined as class interval or simply class. In our example we have classified the ages into 6 different interval i.e. 15-19,…….., 45-49 since the ages are varying in wide range from 17 to 57.
Although there is no hard and fast rule regarding the number of class intervals, It is generally agreed that the number classes neither be too large then the basic purpose of classification (i.e. summarization of data to an easily manageable form.) will not be served nor be too small then the true nature of the distribution will be obscured. As a working rule this number should lie between 5 and 15 depending upon the number of observations available.
One should have class intervals of either five or multiple of five like 10, 20, 25 etc.The reason is that the human mind is accustomed more to think in of certain multiples of 5,10 and the like. Classes must be mutually exclusive means that each of the given values is included in one of the classes . For Example:- 15-19 , 20-24 etc. (not like 15-20, 20-25 etc )
b) Class Frequency
The number of observations falling within a class is called class frequency. Sum of all class frequency is total frequency. In our example the class frequencies are 37,81 etc.
c) Class Limit
The two numbers used to specify the limits of a class interval are called Class Limits. The smaller of the pair is known as lower class limit and the larger as upper class limit. In our example, column 3 is the lower class limits and column 4 is the upper class limits as shown in the Table-3.
Class Interval (1) 15-19 20-24 25-29 30-34 35-44 45-49 Total
Class frequency (2) 37 81 43 24 9 6
Class Limit Lower Upper (3) (4) 15 20 25 30 35 45
200 Table-3
19 24 29 34 44 59
d) Class Boundaries
In our present example the variable is discrete since the age is expressed only in years. But if we consider age as continuous variable, then it may be 14.5 years of 19.5 years, then the concept of class limit will not suffice.(In which class the observation 14.5 or19.5 will be included?). So in case of continuous variable, certain adjustment in the class interval is needed to obtain the continuity.
The extreme values which would ever be included in the class interval in order to obtain the correct class interval in case of continuous variable is known as Class Boundaries. The lower extreme point is known as the lower class boundary and the upper extreme point is known as the upper class boundary with reference to the particular class. In our example, column (5) shows the lower class Boundaries and column (6) shows the upper class Boundaries (as shown in the Table-4)
Correction Factor (d) (Lower limit of the second class - Upper limit of the first class)
= ----------------------------------------------------------------2 Lower Class Boundary= Lower Class Limit - d Upper Class Boundary= Upper Class Limit + d
Class Interval (1) 15-19 20-24 25-29 30-34 35-44 45-49 Total
Class frequency (2) 37 81 43 24 9 6
Class Boundaries
Lower (5) 14.5 19.5 24.5 29.5 34.5 44.5
200 Table-4
Upper (6) 19.5 24.5 29.5 34.5 44.5 59.5
e) Class Mark The value exactly at the middle of a class interval is called Class Mark. Class mark is the representative value of the class interval, for the calculation of mean, SD etc.In our example column (7) of table -5 shows the class marks.
Class Mark= (Lower Class Limit+ Upper Class Limit) ÷ 2 Or (Lower Class Boundary+ Upper Class Boundary) ÷ 2
Class Interval (1)
Class frequency (2)
Class Mark (7)
15-19 20-24 25-29 30-34 35-44 45-59
37 81 43 24 9 6
15+19 = 34 ÷2= 17 20+24 = 44 ÷2= 22 25+29 =54 ÷2 = 27 30+34 = 64 ÷2 = 32 35+44 =79 ÷2 = 39.5 45+59 =104 ÷2= 52
Total
200
e) Width of class or Size of the class 1) Width of class (in case of continuous variable) = Upper Class Boundary - Lower Class Boundary. or Upper Class limit - Lower Class limit. For example:Class boundaries
Width
14.5 – 19.5 19.5 – 24.5
5 5
Class limits
Width
15 – 20 20 - 25
5 5
2) Width of class (in case of discrete variable) = Upper Class limit of the second class - Lower Class limit of the first class. For example:Class limits
Width
15-19 20-24 25-29 30-34 35-44 45-59
20 -15 = 5 25 – 20 = 5 30 -25 = 5 35 – 30 = 5 45 – 35 = 10 60 – 45 = 15
How to choose the number of classes and class limits Step I :- the following points The number of classes should be preferably between 5 and 15. The width of the classes should be preferable be 5 or its multiple e.g., 15-19,2024 etc. One of the class limit being preferable a multiple of 5.
Step II :Apply the Sturges formula for determining the approximate number of classes,i.e.
k = 1 + 3.322 log N Where k = The approximate number of classes N = Total number of observation
Step III :The third step is to find the maximum and minimum of the observations and find the “range” i.e. Range = Maximum - Minimum value If we like to have classes of equal width, then the approximate width of the class may be obtained on dividing the range by the approximate number of classes i.e.
Range Approximate width = -------------------------------
1 + 3.322 log N
For example:Problem 1:The profits (in lakh rupees) of 30 companies for the year 2008 is given below. 20,22,35,42,37,42,48,53,49,65,39,48,67, 18,16,23,37,35,49,63,65,55,45,58,57,69,25,29,58,65. Classify the above data taking suitable class-interval
Solution:Step I :- Find the approximate number of classes. k = 1 + 3.322log N Where N = 30 K = 1 + 3.322 log 30 = 1 + 3.322 x 1.4771 = 7 + 4.91 = 5.91 or 6
Step II :-Width of the classes Maximum - Minimum Width of the classes = ----------------------------------------------------1 + 3.322log N ( 69 – 16) = ------------------------------5.91 53 = --------------5.91 = 8.97 or 9 But since width of the classes preferable be 5 or multiple of 5 i.e. 10,15 We consider width as 10 instead of 9.
Frequency distribution of the profits
Profit (Rs. Lakh) 15 – 24 25 – 34 35 – 44 45 - 54 55 – 64 65 – 74 Total
No of companies 5 2 7 6 5 5 30
The lower limit of the first class may not coincide with the minimum observation,since the width of the classes and the one of the class limit being preferable multiple of 5. For example:- Here lowest value of the data is 16 and we have to take the classes of width 10 and one of the class limit as multiple of 5, then the lower limit of the first class may not coincide with the minimum observation i.e. it should be 15 – 24 not 16 – 25.
Problem 2:The following set of numbers presents the mutual fund prices at the end of the week for the selected 40 nationally sold funds 11,17,15,22,11,16,19,24,29,18 25,26,32,14,17,20,23,27,30,12 15,18,24,36,18,15,21,28,33,38 34,13,11,16,20,22,29,29,23,31 Classify the above data taking suitable class-interval
Solution Ans) k = 1 + 3.322 log 40 = 1+ 3.322 x 1.602 = 1 + 5.32 = 6.32 or 6 (38 – 11) Width = ------------------ = 4.27 or 5 6.32
Class Interval
Frequency
10 – 14
6
15 - 19
11
20 – 24
9
25 - 29
7
30 – 34
5
35 - 39
2
Total
40
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