Draw a histogram to represent the following data:
Class interval10−1414−2020−3232−5252−80frequency5692521
Sol:
Class interval | 10-14 | 14-20 | 20-32 | 32-52 | 52-80 |
Frequency | 5 | 6 | 9 | 25 | 21 |
In the above table, class intervals are of unequal size, so we calculate the adjusted frequency by using the following formula:
Adjusted Frequency = Minimumclasssize×itsfrequencyClasssizeofthisclass
Thus, the adjusted frequency table is:
Class Intervals | Frequency | Adjusted Frequency |
10-14 | 5 | 45×5=5 |
14-20 | 6 | 46×6=4 |
20-32 | 9 | 412×9=3 |
32-52 | 25 | 420×25=5 |
52-80 | 21 | 428×21=3 |
Now, take class intervals along x-axis and adjusted frequency along y-axis and constant rectangles having their bases as class-size and height as the corresponding adjusted frequencies.
Thus, we obtain the histogram as shown below: