Read grouped and cumulative frequency distributions
Create grouped summaries, choose class boundaries, calculate frequency density and use cumulative frequency to estimate values.
Section Paper 1 · Written · calculator required
Exam scope and assessment notes
This lesson maps the Read grouped and cumulative frequency distributions section of Cambridge IGCSE Statistics 0479 (2027 and 2028–2030 syllabus versions). This is original StudyForA teaching and self-check practice, not Cambridge material, an official mark scheme, a timed mock or an official grade. Charts and written working are not automatically assessed; human subject-teacher review remains pending. Use the syllabus version for your exam year. Cambridge offers 0479 in the June series only and does not make it available in every administrative zone. Check the current Cambridge 0479 page and confirm your centre's entry eligibility with your exams officer.
Before you begin
- Use the earlier 0479 ideas where needed; write units, working and a contextual interpretation for each result.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many observations are in classes with frequencies 6, 9 and 5?
What you will learn
- Compare grouped and ungrouped distributions
- Identify limits, boundaries, midpoints and widths for discrete and continuous groups
- Construct and interpret histograms and frequency polygons
- Form cumulative frequencies and interpret cumulative-frequency polygons or curves
Grouping makes patterns clearer but removes detail
An ungrouped distribution retains every value or value frequency. Grouping can make a large dataset easier to summarise, but exact values within a class are lost. A class such as 30 ≤ x < 40 includes 30 and excludes 40.
For a discrete inclusive class such as 10–14, class boundaries are 9.5 and 14.5 when the measured values are whole units; midpoint 12 and width 5. For continuous intervals 10 ≤ x < 20, the boundaries are already 10 and 20; midpoint 15 and width 10.
Histograms use area for frequency
When class widths differ, frequency density = frequency ÷ class width. The bar height is frequency density, so its area represents frequency. Label the vertical axis and use continuous class boundaries. A frequency polygon joins points at class midpoints; the syllabus comparison task uses equal class widths.
Cumulative frequency answers ‘at most’ questions
Add frequencies progressively to form a cumulative-frequency table. Plot cumulative frequency against the upper class boundary to make a polygon or smooth curve. Read medians, quartiles and percentiles at the appropriate cumulative count, using interpolation only when the question and grouped information support an estimate.
Pause and explain
Why does a histogram use frequency density for unequal class widths?
Worked example
A class interval 20 ≤ x < 30 has frequency 18. Find its midpoint, width and frequency density.
Show the worked solution
- Midpoint = (20 + 30)/2 = 25.
- Width = 30 − 20 = 10.
- Frequency density = 18/10 = 1.8.
Answer Midpoint 25, class width 10, and frequency density 1.8. In a histogram the bar area is 1.8 × 10 = 18, the class frequency.
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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.
A grouped table has intervals 0 ≤ t < 10, 10 ≤ t < 20, 20 ≤ t < 30 with frequencies 4, 10 and 6. Find the cumulative frequencies and estimate the median class.
Give me a hint
Add counts from the first class onward; locate half of the total count.
Compare my reasoning
- Cumulative frequencies are 4, 14, 20.
- There are 20 values, so the median position is around the 10th value.
- The 10th value lies in 10 ≤ t < 20.
The cumulative frequencies are 4, 14, 20, and the median class is 10 ≤ t < 20. This identifies a class, not the exact median value.
Look for these in your work
- Cumulative totals correct
- Total count stated
- Estimate labelled as grouped
A histogram class has width 5 and frequency 15. Another has width 10 and frequency 24. Find both bar heights and explain which bar is taller.
Give me a hint
Use frequency density for each width.
Compare my reasoning
- First density = 15/5 = 3.
- Second density = 24/10 = 2.4.
- The first bar is taller, even though its class has fewer observations.
The heights are 3 and 2.4. The first bar is taller because its frequency is concentrated in a narrower interval; its area remains 15, while the second bar area is 24.
Look for these in your work
- Both densities correct
- Heights compared
- Area-frequency distinction explained
A measurement recorded to the nearest whole centimetre is grouped as 12–16. State the class boundaries, midpoint and width.
Give me a hint
Half-unit boundaries make adjacent recorded-value classes continuous.
Compare my reasoning
- Lower boundary = 11.5 and upper boundary = 16.5.
- Midpoint = (11.5 + 16.5)/2 = 14.
- Width = 16.5 − 11.5 = 5.
The class boundaries are 11.5 and 16.5, midpoint 14, width 5. The stated limits 12–16 are inclusive recorded values.
Look for these in your work
- Boundaries include half units
- Midpoint calculated
- Width calculated
Common mistakes
- Using frequency as histogram height when widths differ
- Treating a discrete inclusive class limit as its continuous boundary
- Plotting cumulative frequency at the class midpoint
- Interpreting a grouped estimate as an exact original observation
What can you explain now?
A class has width 4 and frequency 28. What is its frequency density?
Compare with the explanation
7
Frequency density is 28 ÷ 4 = 7; the class bar's area is 7 × 4 = 28.
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Make it stick
What does the area of a histogram bar represent, and how do you choose its height when widths differ?
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.