| Class intervals | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 |
| Frequency | 3 | 5 | 8 | 7 | 6 | 2 |
Solution:
The given distribution is not continuous.
Adjustment factor = (11-10)/2 = ½ = 0.5
We subtract 0.5 from lower limit of the class interval and add 0.5 to upper limit.
So the new table in continuous form is given below.
We write the given data in cumulative frequency table.
| Class intervals | frequency | Cumulative frequency |
| 0.5-10.5 | 3 | 3 |
| 10.5-20.5 | 5 | 8 |
| 20.5-30.5 | 8 | 16 |
| 30.5-40.5 | 7 | 23 |
| 40.5-50.5 | 6 | 29 |
| 50.5-60.5 | 2 | 31 |
To represent the data in the table graphically, we mark the upper limits of the class intervals on
the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis).
Plot the points (10.5, 3), (20.5, 8), (30.5, 16), (40.5, 23) , (50.5, 29) and (60.5, 31) on the graph.
Join the points with the free hand. We get an ogive as shown:
