Histograms with unequal class widths
Use frequency density so rectangle area, rather than height alone, represents frequency.
Extended only. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Read continuous class boundaries.
- Calculate rectangle area and divide quantities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What represents frequency in an unequal-width histogram?
What you will learn
- Calculate frequency density and construct a histogram.
- Recover counts from bar areas and compare unequal-width classes.
Bar area represents frequency
For continuous data, histogram bars touch at class boundaries. Frequency density = frequency / class width. Classes 0–10, 10–15 and 15–30 with frequencies 20,15,30 have widths 10,5,15 and densities 2,3,2. Plot these heights on a vertical axis explicitly labelled frequency density.
The narrow middle bar is tallest, but contains only fifteen observations. The last bar has height two and width fifteen, so its area represents thirty. Comparing heights alone is invalid when widths differ; multiply width by density to recover frequency.
Treat boundaries and units consistently
Use the actual continuous interval widths, not the number of printed integer labels. For 10 ≤ x < 15 the width is five. Keep the same horizontal unit throughout the graph; density has frequency per unit of that measured variable.
To estimate a frequency for part of a class, take the corresponding fraction of its rectangle area, assuming values are spread uniformly within the interval. This estimate is not an exact count known from the grouped table. Histogram totals are sums of represented bar areas, and the overall distribution can reveal clusters and skew.
Pause and explain
A class has width 5 and frequency 15. What is its density?
Worked example
Classes 0–10,10–15,15–30 have frequencies 20,15,30. Find the densities.
Show the worked solution
- Class widths are ten, five and fifteen.
- Divide each frequency by its width: 20/10,15/5,30/15.
- Densities are two, three and two, representing a total of sixty-five observations.
Answer Densities 2,3,2; total frequency 65.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A class width is 8 and frequency is 24. Find its density.
Give me a hint
Divide frequency by width.
Compare my reasoning
- Density = 24/8.
- The density is three.
- Check 8×3 = 24 by recovering the bar area.
3.
Look for these in your work
- Divided by the class width.
- Verified the represented area.
A bar covers 20 ≤ x < 35 and has density 4. Find its frequency.
Give me a hint
Find the boundary difference first.
Compare my reasoning
- Width is 35−20 = 15.
- Frequency is width times density.
- The count is 15×4 = 60.
60.
Look for these in your work
- Used the continuous width.
- Recovered frequency from area.
A class 15 ≤ x < 30 contains 30 observations. Estimate the number with 20 ≤ x < 25, assuming uniform distribution within the class.
Give me a hint
Use the subinterval's fraction of the class width.
Compare my reasoning
- The full class has width fifteen and density two.
- The subinterval has width five.
- Estimated count is 5×2 = 10, subject to the uniform-within-class assumption.
Estimated 10 observations.
Look for these in your work
- Used a proportion of bar area.
- Stated the estimation assumption.
Common mistakes
- Using frequency as height for unequal widths.
- Omitting the density-axis label.
- Treating a partial-class estimate as exact.
What can you explain now?
Can the tallest histogram bar have fewer observations than a lower wider bar?
Compare with the explanation
Yes.
Height is density; a wider bar can have a larger area and therefore greater frequency.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, construct two unequal-width bars and recover each count from its area.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.