Histogram (NonUniform Widths)
When constructing a histogram with nonuniform (unequal) class widths, we must ensure that the areas of the rectangles are proportional to the class frequencies.
Remember that the histogram differs from a bar chart in that it is the area of the bar that denotes the value, not the height. This means that we would need to consider the widths in order to determine the height of each rectangle.
Example
The following frequency distribution gives the masses of 48 objects measured to the nearest gram. Draw a histogram to illustrate the data.
Mass (g) 
10 – 19 
20 – 24 
25 – 34 
35 – 50 
51 – 55 
Frequency 
6 
4 
12 
18 
8 
Solution:
Evaluate each class widths.
Mass (g) 
10 – 19 
20 – 24 
25 – 34 
35 – 50 
51 – 55 
Frequency 
6 
4 
12 
18 
8 
Class width 
10 
5 
10 
15 
5 
Since the class widths are not equal, we choose a convenient width as a standard and adjust the heights of the rectangles accordingly.
We notice that the smallest width size is 5. We can choose 5 to be the standard width. The other widths are then multiples of the standard width.
The table below shows the calculations of the heights of the rectangles.
Mass (g) 
10 – 19 
20 – 24 
25 – 34 
35 – 50 
51 – 55 
Frequency 
6 
4 
12 
18 
8 
Class widths 
10 
5 
10 
15 
5 
2 × standard 
standard 
2 × standard 
3 × standard 
standard 
Rectangle’s height in histogram 
6 ÷ 2 = 3 
4 
12 ÷ 2 = 6 
18 ÷ 3 = 6 
8 
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