Use histogram density and area for unequal classes
Construct correctly scaled rectangles, recover missing frequencies and avoid interpreting the tallest class as the most frequent.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.1. Label variables, units, class boundaries and diagram keys. In these raw-data quartile exercises, use the median of each ordered half, excluding the overall median when the number of observations is odd; follow any convention supplied in an examination question. Cumulative-frequency quartiles use ranks n/4 and 3n/4, with a median rank n/2. Grouped-data interpolation and midpoint moments are estimates, not recovered raw measurements. Histogram area represents frequency, with height proportional to frequency density. Use descriptive variance with divisor n, not the sample-estimator divisor n − 1. Coded and combined totals must refer to the same units and compatible coding before addition. Retain unrounded working and distinguish variance units from standard-deviation units. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate interval widths and rectangle areas.
- Read continuous class boundaries.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A class has frequency 12 and width 10. What is its density?
What you will learn
- Calculate histogram density from class width and frequency.
- Recover frequencies or a scale factor from rectangle areas.
Frequency belongs to area
For times in classes [0,10), [10,20), [20,40), [40,50) minutes with frequencies 4, 12, 8, 6, widths are 10, 10, 20, 10. Frequency densities are 0.4, 1.2, 0.4, 0.6 observations per minute. Draw adjoining rectangles on the time scale; the third bar is twice as wide as the first and has the same height, so twice its frequency.
When the vertical axis is actual frequency density, frequency = width × density. Some questions use a scaled vertical axis whose heights are only proportional to density; determine the common factor from a known bar before finding other frequencies. Numerical area in diagram units is then proportional to frequency, not automatically equal to it.
Read boundaries and distinguish the modal class
Use stated interval boundaries consistently so a measurement is not counted twice. In the classes above, a time of 20 enters [20,40), not [10,20). The tallest bar shows the greatest density. With unequal widths, a wider lower bar can have more observations than a taller narrow bar; the modal histogram class by density is not a claim about an exact raw mode.
For [0,5) with frequency 15 and [5,20) with frequency 30, densities are 3 and 2. The first bar is taller even though the second class has more observations. The grouped histogram cannot identify where those 30 values lie inside their class. Estimated within-class counts require an additional uniform-spread assumption, not just a rectangle on a page.
Pause and explain
A bar has width 20 and actual frequency-density height 0.4. Its frequency is…
Worked example
Construct a density histogram for [0,10), [10,20), [20,40), [40,50) minutes with frequencies 4, 12, 8, 6.
Show the worked solution
- Widths are 10, 10, 20 and 10 minutes.
- Divide frequency by width to obtain heights 0.4, 1.2, 0.4 and 0.6 observations per minute.
- Use contiguous boundaries 0, 10, 20, 40, 50; areas sum to 4 + 12 + 8 + 6 = 30 observations.
Answer Density heights 0.4, 1.2, 0.4, 0.6; total frequency 30.
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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.
Find density heights for class widths 10, 10, 20, 10 minutes with frequencies 4, 12, 8, 6 and recover the total from areas.
Give me a hint
Divide each frequency by its own width.
Compare my reasoning
- Heights are 0.4, 1.2, 0.4 and 0.6 observations per minute.
- Areas are 10(0.4), 10(1.2), 20(0.4), 10(0.6), recovering 4, 12, 8, 6.
- The total frequency is 30 observations.
Heights 0.4, 1.2, 0.4, 0.6; total 30 observations.
Look for these in your work
- Use the width 20 for the third class.
- Check counts through areas.
Class [0,5) has frequency 15 and class [5,20) has frequency 30. Which is taller on an actual density histogram, and which contains more observations?
Give me a hint
Compare densities separately from frequencies.
Compare my reasoning
- The first width is 5 and density is 15/5 = 3.
- The second width is 15 and density is 30/15 = 2.
- The first bar is taller, while the second class contains twice as many observations.
First bar taller (density 3 versus 2); second class more frequent (30 versus 15).
Look for these in your work
- Separate density from frequency.
- Use both widths in the comparison.
Histogram heights are scaled. A width-5 class has plotted height 3 and frequency 30. A width-10 class has plotted height 4. Find its frequency.
Give me a hint
Find observations per plotted area unit using the known rectangle.
Compare my reasoning
- The known plotted area is 5 × 3 = 15.
- The frequency factor is 30/15 = 2 observations per plotted area unit.
- The second plotted area is 10 × 4 = 40, so its frequency is 2 × 40 = 80.
Scale factor 2 observations per plotted area unit; missing frequency 80.
Look for these in your work
- Calibrate the scale from the supplied frequency.
- Apply the same factor to area, not just height.
Common mistakes
- Unequal widths need density heights.
- Determine any vertical-axis scale factor before converting area to frequency.
- The tallest bar identifies largest density, not necessarily largest frequency.
What can you explain now?
A class is 20 minutes wide with actual density 0.4 observations per minute. How many observations does it contain?
Compare with the explanation
8 observations.
Frequency equals width × density = 20 × 0.4.
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Make it stick
Width × actual density recovers frequency; a scaled-height diagram first needs its common area-to-frequency factor.
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.