Draw a frequency polygon for the following frequency distribution:
Class interval1−1011−2021−3031−4041−5051−60Frequency8361227
Sol:
The given frequency distribution table is as below:
Class intervals | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 |
Frequency | 8 | 3 | 6 | 12 | 2 | 7 |
This table has inclusive class intervals and so these are to be converted into exclusive intervals (i.e. true class limits).
These are (0.5-10.5), (10.5-20.5),(20.5-30.5),(30.5-40.5),(40.5-50.5), and (50.5-60.5).
In order to draw a frequency polygon, we need to determine the class marks. Class marks of a class interval = upper limit+lower limit2
Take imaginary class interval (-9.5 – 0.5) at the beginning and (60.5 – 70.5) at the end, each with frequency zero. So we have the following table
Class intervals | True class Intervals | Class marks | Frequency |
(-9)-0 | (-9.5)-0.5 | -4.5 | 0 |
1-10 | 0.5-10.5 | 5.5 | 8 |
11-20 | 10.5-20.5 | 15.5 | 3 |
21-30 | 20.5-30.5 | 25.2 | 6 |
31-40 | 30.5-40.5 | 35.5 | 12 |
41-50 | 40.5-50.5 | 45.5 | 2 |
51-60 | 50.5-60.5 | 55.5 | 7 |
61-70 | 60.5-70.5 | 65.5 | 0 |
Now, take class marks along x-axis and their corresponding along y-axis.
Mark the points and join them.
Thus, we obtain a complete frequency polygon as shown below: