Describe relationships with bivariate data
Plot paired values, describe correlation, find a line of best fit and make cautious estimates within a supported range.
Section Paper 1 · Written · calculator required
Exam scope and assessment notes
This lesson maps the Describe relationships with bivariate data 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
Find the gradient between (1,4) and (3,10).
What you will learn
- Distinguish independent and dependent variables
- Describe positive, negative, weak, strong and absent correlation
- Plot bivariate data
- Draw a line of best fit by eye and by semi-averages
- Find and use y = mx + c
- Explain assumptions and the limits of extrapolation
Plot paired observations before describing them
The independent variable is used to explain or predict; the dependent variable is the response. Plot each pair (x,y) once, label both axes and choose a scale that makes the pattern visible. Correlation describes the direction and strength of a linear association: positive, negative, weak, strong or no correlation.
A best-fit line summarises the pattern
By eye, draw a straight line through the centre of the points, with roughly balanced points above and below. For semi-averages, sort pairs by x, split an even number into two equal groups, find each group's mean x and mean y, and draw the line through those two mean points.
For line y = mx + c, use two points to calculate gradient m = change in y ÷ change in x, then substitute a point to find c. Use the line to estimate the dependent value for a given x or rearrange it for x when appropriate.
Association is not a guarantee or a cause
A line assumes the observed linear pattern is suitable and that the context remains comparable. Estimates inside the observed range are interpolation; estimates outside it are extrapolation and are less reliable because the relationship may change. Correlation alone does not prove that one variable causes the other.
Pause and explain
Why is extrapolation usually less reliable than interpolation?
Worked example
A line of best fit passes through (2, 7) and (6, 15). Find its equation and estimate y when x=5.
Show the worked solution
- m = (15−7)/(6−2) = 2.
- Use y = 2x + c and (2,7): c = 3.
- At x=5, y=2×5+3=13.
Answer The line is y = 2x + 3 and the estimate at x = 5 is 13.
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.
For paired data (1,4), (2,6), (3,7), (4,11), find the two semi-average points and the equation of the line through them.
Give me a hint
Split the four pairs into two groups by x, average x and y in each, then find the line.
Compare my reasoning
- First group means: (1.5, 5). Second group means: (3.5, 9).
- Gradient = (9−5)/(3.5−1.5) = 2.
- Using (1.5,5): c=2, so y=2x+2.
The semi-average points are (1.5, 5) and (3.5, 9); the line is y = 2x + 2.
Look for these in your work
- Pairs ordered and split
- Mean points correct
- Equation correct
A fitted line is y = 1.5x + 4. Estimate y at x=8. The observed x-values ranged from 2 to 10. Explain why this estimate is more defensible than one at x=30.
Give me a hint
Calculate the estimate, then compare each x with the observed range.
Compare my reasoning
- At x=8, y=1.5×8+4=16.
- 8 is within the observed range 2 to 10, so this is interpolation.
- 30 is far outside the observed range, so the linear pattern may not continue there.
At x=8, the estimate is 16. This is interpolation within the observed range; x=30 would be extrapolation and depends on an unsupported continuation of the line.
Look for these in your work
- Substitution correct
- Interpolation identified
- Extrapolation risk explained
A study finds a positive correlation between time spent revising and exam scores. Give one reasonable interpretation and one reason not to claim that revision time alone caused the scores.
Give me a hint
Describe association first; then consider a third variable or how data were collected.
Compare my reasoning
- Students reporting more revision tend to have higher scores in this sample.
- Prior knowledge, support or motivation may affect both revision time and scores.
- The correlation does not isolate a causal effect.
The variables are positively associated in the data, but prior attainment or other factors may influence both. Correlation alone does not establish that revision time caused the higher scores.
Look for these in your work
- Direction interpreted
- Plausible confounder stated
- Causation not claimed
Common mistakes
- Swapping independent and dependent variables
- Claiming a line proves causation
- Splitting an odd number of pairs evenly without following a stated rule
- Extrapolating far beyond the observed data as if reliable
What can you explain now?
A scatter diagram rises from left to right with points close to a straight line. Describe the correlation.
Compare with the explanation
Strong positive correlation
The upward direction is positive; close clustering around a line indicates strength.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
What are the steps in the semi-averages method, and when is a prediction extrapolation?
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.