Use frequency density for unequal-width histogram classes
Calculate class width and frequency density · Use bar area to represent frequency · Distinguish density peaks from the largest class count
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Grouped continuous data
- Area of a rectangle
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Calculate class width and frequency density
- Use bar area to represent frequency
- Distinguish density peaks from the largest class count
Area, not height, gives frequency
Frequency density=frequency/class width. The illustrative classes 0≤t<10, 10≤t<20, 20≤t<40 have frequencies 4,8,8, widths 10,10,20 and densities 0.4,0.8,0.4 observations per minute. On a frequency-density histogram, rectangle area in axis units is width×density=frequency.
Unequal widths require different heights
The final two classes each contain eight observations, so their bars have equal area. The last bar is twice as wide and half as high. Adjacent continuous classes meet at their boundary. The peak density is in 10≤t<20; tied raw frequencies do not imply tied densities when widths differ. Do not draw height equal to frequency while keeping unequal widths.
Sources for this lesson
Pause and explain
Must two unequal-width classes with equal frequencies have equal histogram heights?
Worked example
Find the density for eight observations in 20≤t<40.
Show the worked solution
- Class width=20 minutes.
- Density=8/20.
Answer 0.4 observations per minute.
Try a topic test
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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.
For 0≤t<10, 10≤t<20, 20≤t<40 with frequencies 4,8,8, calculate all densities.
Give me a hint
Divide each frequency by its own width.
Compare my reasoning
- The widths are 10,10,20.
- Densities are 4/10,8/10,8/20.
0.4,0.8,0.4 observations per minute.
Look for these in your work
- Use the actual widths
- Label the density units
Explain why the last two illustrated bars have equal area but different heights.
Give me a hint
Each represents eight observations.
Compare my reasoning
- The last width is double the previous width.
- Its density is half, preserving width×height=8.
Both areas represent frequency 8; width 20 needs density 0.4 versus width 10 and density 0.8.
Look for these in your work
- Use area for frequency
- Compare widths and densities together
Which illustrated class has the highest density? Does it have strictly more observations than the final class?
Give me a hint
Compare heights, then counts.
Compare my reasoning
- The peak density 0.8 is in 10≤t<20.
- Both that class and the final class have frequency 8.
10≤t<20 has highest density; its frequency is tied with the final class.
Look for these in your work
- Separate density from count
- Avoid a height-only count comparison
Common mistakes
- Use class width, not upper boundary, as the divisor.
- Height alone is not frequency on a density histogram.
What can you explain now?
Must two unequal-width classes with equal frequencies have equal histogram heights?
Compare with the explanation
No; equal frequencies require equal areas.
Density changes inversely with width for a fixed frequency.
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Make it stick
Must two unequal-width classes with equal frequencies have equal histogram heights?
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.