Classes of different widths
A company asks 160 of its workers how long their journey to work takes, in minutes. Short journeys are common and long ones are spread thinly, so the times are grouped into classes of different widths. With t for the time in minutes, 40 workers are in the class , 60 are in , and 60 are in . The last class is 20 minutes wide, twice as wide as the others.
Draw a histogram with the frequencies as the heights, the way it is done when every class has the same width. Each bar stands over its own interval on the time axis, so the last bar is twice as wide as the others.
With frequency as the height, the 20–40 bar is as tall as the 10–20 bar and twice as wide, so it covers twice the area for the same 60 workers.
Why the height is not a fair comparison
The eye judges a bar by the space it covers, its area. The 20–40 bar covers twice as much of the chart as the 10–20 bar, so it looks like the largest group, but both hold 60 workers.
A wide class collects more values just by being wide. The 60 workers in the 20–40 class are spread over 20 minutes, which is 60 ÷ 20 = 3 workers for each minute of the class. The 60 workers in the 10–20 class are packed into 10 minutes, which is 60 ÷ 10 = 6 for each minute. Journeys of 10 to 20 minutes are twice as crowded, and the chart above hides that.
Frequency density
The fair height for a bar is the number of values for each unit of the scale, here for each minute. It is called the frequency density:
frequency density = frequency ÷ class width
For the journey times, the densities are 40 ÷ 10 = 4, 60 ÷ 10 = 6 and 60 ÷ 20 = 3 workers per minute. Draw each bar with its density as its height, and label the vertical axis “frequency density”. The 20–40 bar is now half as tall as the 10–20 bar, and the chart shows where the journeys are most crowded.
The heights are frequency densities: 4, 6 and 3 workers per minute. The 20–40 bar is twice as wide as the 10–20 bar and half as tall.
The area of a bar is its frequency
Multiply a bar’s width by its height: width × frequency density = width × (frequency ÷ width) = frequency. So the area of each bar is the number of values in its class. The first bar has area 10 × 4 = 40, the second 10 × 6 = 60, and the third 20 × 3 = 60. The areas add up to 40 + 60 + 60 = 160, every worker in the survey.
This is why a histogram shows frequency by area. When the classes are all the same width, the areas are in the same ratio as the heights, and the heights can be read as frequencies. When the widths differ, only the area gives the frequency.
height = frequency whatever the width: the class of width 20 is drawn as tall as it would be at width 10, so a wide class becomes a false peak
Switch to density and widen the last class to 40
The last class always holds 10 values. With frequency as the height, widening the class makes its bar cover more area, although no value was added. Switch to frequency density and widen it to 40: the height falls to 10 ÷ 40 = 0.25, and the area stays 40 × 0.25 = 10.
Reading frequencies from a histogram
To read a frequency from a histogram drawn with frequency density, work out the area of its bar: frequency = frequency density × class width. A bar from 40 to 60 minutes with a height of 1.5 is 20 minutes wide, so it stands for 1.5 × 20 = 30 values. Find each class width from the two ends of the class on the scale, not by counting bars.
The same idea gives an estimate for part of a class. How many of the 160 workers take less than 15 minutes? The first class holds all 40 of its workers. From 10 to 15 minutes is half of the 10–20 class, a strip 5 minutes wide and 6 high, with area 5 × 6 = 30. So about 40 + 30 = 70 workers take less than 15 minutes.
That answer is an estimate, because it assumes that the 60 workers in the 10–20 class are spread evenly across it. The histogram does not show where each value lies inside its class.
Worked example: Journey Times in a Histogram with Unequal Class Widths
Question A company drew a histogram of the journey times to work, in minutes, of its employees. The vertical axis shows frequency density, which is the frequency divided by the class width. The bar from 0 to 10 minutes has height 0.6, the bar from 10 to 20 has height 1.4, the bar from 20 to 40 has height 1.8, the bar from 40 to 60 has height 1.2 and the bar from 60 to 100 has height 0.5. (a) How many employees are shown in the histogram? (b) Estimate the number of employees whose journey takes more than 50 minutes.
1.Write the width of each class from its two ends. The widths are 10, 10, 20, 20 and 40 minutes.
The classes are 10, 10, 20, 20 and 40 minutes wide. 2.Multiply each frequency density by its class width: 0.6 × 10 = 6, 1.4 × 10 = 14, 1.8 × 20 = 36, 1.2 × 20 = 24 and 0.5 × 40 = 20.
The frequency is the area of a bar, its frequency density times its width: 6, 14, 36, 24 and 20. 3.(a) 6 + 14 + 36 + 24 + 20 = 100 employees.
(a) 6 + 14 + 36 + 24 + 20 = 100 employees. 4.A journey of more than 50 minutes is in the part of the 40 to 60 bar that lies between 50 and 60, or anywhere in the 60 to 100 bar. The part bar has area 1.2 × 10 = 12. The whole last bar is 20.
More than 50 minutes is the part of the 40 to 60 bar past 50, which is 1.2 × 10 = 12, and the whole of the last bar, 20. 5.(b) About 12 + 20 = 32 employees. It is an estimate, because it takes the 24 employees of the 40 to 60 class to be spread evenly across the class.
(b) About 12 + 20 = 32 employees.
Answer: (a) 100 employees; (b) about 32 employees
Common mistakes
- Reading the heights as frequencies and saying that the 20 to 40 class has 1.8 employees, or that the 60 to 100 class is the smallest because its bar is the lowest. The last bar is four times as wide as the first, and it holds 20 employees, more than the first two classes together.
- Taking half of the last bar as well as half of the 40 to 60 bar. Only the class that contains 50 minutes is cut. Every journey in the 60 to 100 class is longer than 50 minutes, so the whole of that bar counts.