In a histogram with unequal class widths, the height of a bar is frequency density, and the area of the bar is the frequency. So to find how many values are in a class, multiply the bar height by the class width.
This skill belongs to data displays and cumulative reasoning. It appears whenever a question gives a histogram, or asks you to draw one from a grouped frequency table.
Why does a histogram use frequency density?
Take a grouped table where one class is 10 units wide and the next is 20 units wide. The wider class collects more values just because it covers more of the scale. If you drew bars with height equal to frequency, the wide class would look busier than it really is.
Frequency density fixes this by sharing the frequency out per unit of width:
frequency density = frequency ÷ class width
It follows that frequency = frequency density × class width, which is the area of the bar.
How to read and draw one, step by step
- Find each class width by subtracting the lower boundary from the upper boundary.
- To draw, divide each frequency by its class width and use the result as the bar height.
- To read, multiply each bar height by its class width to get the frequency.
- For part of a bar, multiply the height by the width of only the part you need. This assumes the values are spread evenly across the class, so the answer is an estimate.
- Check the total against the number of values in the data, when the question gives it.
Worked example
A histogram shows the time, in minutes, that 85 students spent on homework one evening. The bar heights (frequency density) are:
| Time t (minutes) | Class width | Frequency density |
|---|---|---|
| 0 < t ≤ 10 | 10 | 1.2 |
| 10 < t ≤ 30 | 20 | 1.5 |
| 30 < t ≤ 40 | 10 | 2.5 |
| 40 < t ≤ 60 | 20 | 0.9 |
(a) Find the frequency of each class.
Step 1: 1.2 × 10 = 12.
Step 2: 1.5 × 20 = 30.
Step 3: 2.5 × 10 = 25.
Step 4: 0.9 × 20 = 18.
Step 5, check: 12 + 30 + 25 + 18 = 85, which matches the 85 students.
(b) Estimate how many students spent more than 20 minutes but at most 40 minutes.
The part of the 10 to 30 class from 20 to 30 is 10 wide, so its frequency is 1.5 × 10 = 15. The whole 30 to 40 class adds 25.
Estimate: 15 + 25 = 40 students. It is an estimate because we assumed the 30 students in the 10 to 30 class are spread evenly.
The mistake to watch for
The usual slip is to read the height of the bar as the frequency.
Mistaken answer: “The 10 to 30 class has a frequency of 1.5.”
The student read 1.5 straight off the vertical axis. But the axis shows frequency density, so this gives a value per minute, not a count of students.
The correction is to multiply by the class width: 1.5 × 20 = 30. A quick habit that catches this is to read the vertical axis label before you read any bar.
Check yourself
Try these without a calculator, then open each answer.
1. A bar covers 5 < x ≤ 15 and has height 3.2. What is the frequency of that class?
Show answer
The class width is 15 − 5 = 10. Frequency = 3.2 × 10 = 32.
2. The class 20 < x ≤ 50 has frequency 45. What bar height should be drawn?
Show answer
The class width is 50 − 20 = 30. Frequency density = 45 ÷ 30 = 1.5.
3. A class 0 < x ≤ 4 has frequency 10 and the next class 4 < x ≤ 10 has frequency 18. Find both bar heights and say which bar is taller.
Show answer
First bar: 10 ÷ 4 = 2.5. Second bar: 18 ÷ 6 = 3. So the second bar is taller (3 against 2.5). It holds 18 values over 6 units, which is more crowded than 10 values over 4 units.
Where this leads next
With the frequency of each class in hand, you can build a running total and estimate a median from cumulative frequency. Test the whole module with the data displays practice set. The non-calculator working trainer helps with the divisions and multiplications, and the percentage-base explorer helps when a question asks for a share of the total.
Some students follow every step in class but still read the wrong thing off a chart under time pressure. In online one-to-one Mathematics tuition, a teacher can see which step goes astray and fix it.