Here we will learn how to construct frequency distribution tables.
Following is a collection of raw data showing the ages (in years) of 30 students of class.
14, 15, 14, 16, 15, 13, 17, 16, 17, 16, 17, 12, 13, 12, 14, 15, 16, 15, 18, 12, 17, 17, 18, 13, 14, 13, 16, 15, 16, 17, 16, 19
1. Frequency Distribution Table for Unclassified Data: Observe the values of the variable, and look for the least value and the greatest value. They are 12 and 19 respectively. The frequency table is constructed as shown below.
Explanation: As 14 is the first value, put a tally mark ( ∕ ) against 14 in the column of tall marks.
The next one is 15. So, put a tally mark against 15. Since the next value is 14, put a tally mark against 14 again and so on till you reach the last value. The fifth stroke against any number is given by \ across the earlier four strokes, ∕∕∕∕. Now count the tally marks in each case and write the corresponding frequency.
2. Frequency Distribution Table for Classified Data: In the above example, the range of raw data is 19 – 12 or 7.
(i) Take the overlapping class intervals 11 – 13, 13 – 15, 15 – 17, 17 – 19, 19 – 21 which cover all the values of the variable. The frequency distribution table with overlapping class intervals will be as shown below.
Explanation:
As the range is short, take closer pairs of numbers for class intervals. There should be 4 or more class intervals. Here the class intervals are overlapping. In the class 11 – 13, all the data greater than or equal to 11 but less than 13 fall. Other class intervals will also group data similarly.
The first value of the variable is 14. So, it falls in the class interval 13 – 15. Put a tally mark against this class interval in the table. The next value is 15. So, it falls in the class interval 15 – 17 (not in 13 – 15). Put a tally mark against this interval in the table, and so on. Put the fifth tally mark \ for any class interval across the tally marks ∕∕∕∕ in the corresponding class interval.
(ii) Take the nonoverlapping class intervals 10 – 12, 13 – 15, 16 – 18, 19 – 21 which cover all values of the variable. Then the frequency distribution table with nonoverlapping class intervals will appear as shown below.
Explanation:
In the class interval 10 – 12 will fall all those values which are greater than or equal to 10 but less than or equal to 12. Rest of the data will fall into other class intervals in the same way.
The first value of the variable is 14. So, it falls in the class interval 13 – 15. Put a tally mark against this class interval in the table. The next value is 15. So, it falls in the class 13 – 15. Put another tally mark against this class interval, and so, on.
Solved Example on Constructing Frequency Distribution Tables:
The age of 300 students of a school are collected and tabulated as below.
Age in Years
Below 10
Below 12
Below 14
Below 16
Below 18
Number of Students
45
116
210
275
300
Construct a frequency table of the data. Also answer the following.
(i) How many students are at least 12 years old but less than 14 years old?
(ii) How many students are at least 14 years in age?
Solution:
The frequency table is as given below.
Class Interval (Age in Years) 0 - 10 10 - 12 12 - 14 14 - 16 16 - 18 |
Frequency (Number of Students) 45 71 94 65 25 |
(= 116 - 45) (= 210 - 116) (= 275 - 210) (= 300 - 275) |
(i) 94 (= the frequency of the class interval 12 - 14)
(ii) 90 (= The sum of the frequencies of the class intervals 14 - 16 and 16 - 18)
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