Choose and calculate a measure of centre
Find means, medians and modes for small or ungrouped data, estimate grouped values, and choose a measure for the context.
Section 4 · Choose and calculate a measure of centre
Exam scope and assessment notes
This lesson maps the Choose and calculate a measure of centre 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
If five values have mean 7, what is their total?
What you will learn
- Calculate means and find medians and modes for small sets and ungrouped frequency tables
- Solve problems where the mean is given
- Estimate grouped means and medians, including by cumulative frequency or interpolation
- Identify modal classes and justify a suitable measure of centre
Calculate centre from values or frequencies
For n values, mean = total ÷ n. In a frequency table, multiply each value x by its frequency f, then mean = Σfx ÷ Σf. To find a missing value from a stated mean, first calculate the required total n × mean, then subtract the known total.
Order data before finding the median. For an odd number of values, use the middle value; for an even number, average the two middle values. The mode is the most frequent value; a grouped distribution may have a modal class rather than a single modal value.
Estimate when values are grouped
For grouped mean, use each class midpoint as a representative value and calculate Σ(f × midpoint) ÷ Σf. It is an estimate because the exact observations are unknown.
For grouped median, locate the middle cumulative count. Read it from a cumulative-frequency curve or estimate within the median class by linear interpolation, assuming values are spread evenly within that class. State that assumption and retain sensible accuracy.
Justify the measure in context
The mean uses every value but is affected by extreme values. The median is resistant to extremes and suits a skewed distribution. The mode identifies a common value or category. Choose based on the question, data type and shape; explain what the selected measure says in context.
Pause and explain
A highly skewed salary distribution contains a few very large values. Which centre is usually more resistant to those extremes?
Worked example
Values 2, 4 and 8 have frequencies 3, 2 and 1. Find the mean and median.
Show the worked solution
- Total count = 3 + 2 + 1 = 6.
- Weighted total = 2×3 + 4×2 + 8×1 = 22, so mean = 22/6 ≈ 3.67.
- The ordered list is 2, 2, 2, 4, 4, 8; median = (third + fourth)/2 = 3.
Answer Mean = 11/3 ≈ 3.67 and median = 3. The repeated values are accounted for using their frequencies.
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 frequency table has values 1, 2, 3 with frequencies 2, 5, 3. Find the mean and median.
Give me a hint
Calculate Σf and Σfx; then locate the fifth and sixth observations.
Compare my reasoning
- Count = 2 + 5 + 3 = 10; weighted total = 1×2 + 2×5 + 3×3 = 21.
- Mean = 21/10 = 2.1.
- The fifth and sixth ordered observations are both 2, so median = 2.
Mean = 2.1 from the weighted total 21 divided by 10 observations; median = 2 because the fifth and sixth ordered values are both 2.
Look for these in your work
- Frequency total correct
- Weighted total correct
- Median positions correct
A grouped distribution has classes 0–10, 10–20, 20–30 with frequencies 4, 8, 8. Estimate the mean using class midpoints.
Give me a hint
Multiply each midpoint by its class frequency, then divide by the total count.
Compare my reasoning
- Midpoints are 5, 15 and 25.
- Σfx = 4×5 + 8×15 + 8×25 = 340; total frequency = 20.
- Estimated mean = 340/20 = 17.
Estimated mean = 17. It is an estimate because each class value is represented by its midpoint.
Look for these in your work
- Midpoints correct
- Weighted total correct
- Estimate limitation stated
A dataset has 12 ordered observations. Explain how to find its median, and why the mean might tell a different story if the final value is unusually large.
Give me a hint
Use positions for the median and compare how the mean uses all magnitudes.
Compare my reasoning
- For 12 values, average the 6th and 7th observations.
- The median depends on those middle positions.
- A very large last value raises the total and therefore can pull up the mean.
Median = (6th value + 7th value)/2. An unusually large final value may raise the mean because the mean includes every magnitude, while it usually leaves the median unchanged.
Look for these in your work
- Even-count positions correct
- Median method stated
- Outlier effect explained
Common mistakes
- Ignoring frequency when calculating a mean
- Finding the median before ordering values
- Calling a grouped estimate exact
- Choosing the mean despite a strong outlier without discussing its effect
What can you explain now?
A grouped class has frequency 12 and midpoint 25. What contribution does it make to Σfx?
Compare with the explanation
300
The midpoint represents each of the 12 values, so 12 × 25 = 300.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
When values are grouped, why is the estimated mean based on class midpoints rather than exact observations?
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.