Cambridge O Level Mathematics: Histograms and Frequency Density with Unequal Class Widths (4024) – Revision Notes

These are quick-recall notes for section 9.7, Histograms, of the Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027. The content is the same for every candidate. If a point here is unclear, go back to the study guide. To test yourself, use the practice questions. The course hub lists the other topics, the printable checklist tracks your progress, and the free 10-minute diagnostics show where to focus.

The two outcomes

Syllabus	Outcome
9.7 (1)	Draw and interpret histograms. The vertical axis is labelled “Frequency density”.
9.7 (2)	Calculate with frequency density = frequency ÷ class width.

Core ideas

• A histogram shows grouped continuous data, with bars that touch.
• The area of a bar shows the frequency. The height shows the frequency density.
• Heights are needed because class widths can differ. Plotting frequency itself would make wide classes look too large.
• Frequency density tells you how many values there are per one unit on the horizontal axis.

Formulas

You know	You want	Use
Frequency, class width	Bar height	frequency density = frequency ÷ class width
Bar height, class width	Frequency	frequency = frequency density × class width
Bar width, bar height	Frequency	frequency = area of bar
All frequencies	Total	add the frequencies (not the heights)

Class width = upper boundary − lower boundary. Work it out for every class.

Method boxes

Table to histogram

1. Find the width of each class (upper − lower).
2. Divide each frequency by its width. This is the frequency density.
3. Draw the horizontal axis as a continuous scale and mark the class boundaries.
4. Label the vertical axis Frequency density and choose an equal-step scale.
5. Draw touching bars, each at its own height.

Histogram to table

1. Read each bar’s height and work out its width from the boundaries.
2. Multiply: frequency = height × width.
3. Add the frequencies if the total is needed.
4. Check that each frequency is a whole number.

No vertical scale

1. Take a bar whose frequency you know and measure its width and height in cm (or count squares).
2. Area = width × height.
3. Frequency per cm² = known frequency ÷ area.
4. For any other bar, frequency = area × frequency per cm². To find a height, reverse it: height = (frequency ÷ frequency per cm²) ÷ width.

Missing values

1. Use every bar you can read to get a frequency.
2. Subtract the known frequencies from the stated total to find the remaining one.
3. Divide by that class’s width to get its height.
4. Check that the final frequencies add to the total.

Small worked reminders

Reminder 1: table to height. The class 2 ≤ x < 8 has frequency 9. Width = 6, so frequency density = 9 ÷ 6 = 1.5.

Reminder 2: height to frequency. A bar from 30 to 45 has height 0.8. Width = 15, so frequency = 15 × 0.8 = 12.

Reminder 3: comparing classes. Class A has width 5 and density 4. Class B has width 20 and density 1.5.

A: 5 × 4 = 20
B: 20 × 1.5 = 30

Class B holds more values even though its bar is shorter.

Reminder 4: part of a bar. A bar from 40 to 60 has density 1.5. The values between 50 and 60 are about 10 × 1.5 = 15, if they are spread evenly through the class. Always state that assumption.

Must-know distinctions

Idea	Do	Do not
Vertical axis	Frequency density	Frequency
Bar from table	Divide by width	Multiply by width
Bar to frequency	Multiply by width	Divide by width
Tallest bar	Most crowded class	Necessarily the biggest class
Largest area	Largest frequency	Compare heights alone
Bars	Touch each other	Leave gaps
Estimating part of a class	Take a fraction of the width	Take a fraction of the height

If every class has the same width, bar heights are proportional to frequencies. That is a special case. With unequal widths, always divide by the width first.

One full example to remember

Ages of 60 members: 0 ≤ a < 10 has 9 members, 10 ≤ a < 18 has 16, 18 ≤ a < 30 is unknown, and 30 ≤ a < 50 has a bar of height 0.7.

Last class:   frequency = 0.7 × 20 = 14
Third class:  frequency = 60 − 9 − 16 − 14 = 21
Heights:      9 ÷ 10 = 0.9,  16 ÷ 8 = 2,  21 ÷ 12 = 1.75,  0.7

The order matters. Use a bar to get a frequency, use the total to get the rest, and only then find the missing height.

Reading a histogram safely

• Read the horizontal axis first and write down every class boundary.
• Write the width under each bar before you touch the heights.
• Check how many small squares make one unit on each axis. A scale of 0.2 per square is a common trap.
• If you are asked for a total, work out every frequency separately and then add.
• If you are asked which class has the most values, compare areas, not heights.
• Before you finish, ask whether the answer is sensible: a frequency must be a whole number, and a total should match the number of items the question says were recorded.
• Show each multiplication or division on its own line, with the width written out, so the method is clear to whoever reads your work.

Quick self-test

Do these without looking back. Answers are below.

1. Find the frequency density of the class 4 ≤ x < 10, which has frequency 15.
2. A bar has width 8 and height 1.5. How many values does it represent?
3. The class 20 ≤ x < 50 has frequency 21. What height should its bar have?
4. A histogram has two bars: 10 ≤ x < 30 with height 0.4, and 30 ≤ x < 40 with height 1.1. Find both frequencies.
5. Why is the vertical axis of a histogram with unequal class widths not labelled “Frequency”?
6. On a histogram, a bar 2 cm wide and 3 cm tall represents 30 values. How many values does a bar 4 cm wide and 1.5 cm tall represent?
7. A class 6 ≤ x < 15 has frequency 18. Find its frequency density, and say what the bar’s area equals.
8. The bar for 40 ≤ x < 60 has height 1.5. Estimate how many values lie between 50 and 60, and state the assumption you made.
9. A histogram has bars 0 ≤ x < 10 (height 0.9), 10 ≤ x < 20 (frequency 14) and 20 ≤ x < 40 (height 0.85). Find the total frequency.
10. A bar 3 cm wide represents 24 values, and a bar 2 cm wide and 2 cm tall represents 16 values. How tall is the first bar?

Answers

1. Width 6, so 15 ÷ 6 = 2.5.
2. 8 × 1.5 = 12 values.
3. Width 30, so 21 ÷ 30 = 0.7.
4. The first class has width 20, so 20 × 0.4 = 8. The second has width 10, so 10 × 1.1 = 11.
5. The bar heights show frequency density (frequency ÷ class width), and the areas show the frequencies. Plotting frequency would distort wide classes.
6. The first bar has area 6 cm², so 30 ÷ 6 = 5 values per cm². The second has area 4 × 1.5 = 6 cm², so it represents 6 × 5 = 30 values.
7. Width 9, so 18 ÷ 9 = 2. The area of the bar (9 × 2 = 18) equals the frequency.
8. The part from 50 to 60 is 10 wide, so 10 × 1.5 = 15 values. The answer assumes an even spread of values throughout the class.
9. First class: 10 × 0.9 = 9. Third class: 20 × 0.85 = 17. Total = 9 + 14 + 17 = 40.
10. The 2 cm by 2 cm bar has area 4 cm² and represents 16, so 4 values per cm². The first bar has area 24 ÷ 4 = 6 cm², so its height is 6 ÷ 3 = 2 cm.

Where marks are usually lost

• Plotting the frequency as the bar height when class widths are unequal.
• Using the wrong class width, or copying a width from a neighbouring class.
• Dividing by the width in place of multiplying (or the reverse) when moving between a bar and a frequency.
• Labelling the vertical axis “Frequency” instead of “Frequency density”.
• Leaving gaps between bars, or drawing bars that do not start and end at the class boundaries.
• Adding bar heights to find a total, rather than adding the frequencies.
• Choosing an uneven vertical scale, so heights are misread.
• Giving an estimate for part of a class with no mention of the even-spread assumption.
• Not showing the division or multiplication, so a single slip leaves nothing to credit.
• Missing the final check that the frequencies add to the stated total.

Official syllabus

Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, version 2, published by Cambridge University Press & Assessment (Cambridge Assessment International Education). The relevant section is 9.7 Histograms.