Cambridge O Level Mathematics: Scatter Diagrams, Correlation and Cumulative Frequency Diagrams (4024)

This study guide covers sections 9.5 (Scatter diagrams) and 9.6 (Cumulative frequency diagrams) of the Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027. The syllabus is not tiered, so every candidate meets all of this content. The drawing skills here are done by hand with a ruler, which the syllabus lists as required equipment for both papers. Work through it with the revision notes and the practice questions, and see the O Level Mathematics hub and the printable checklist for the rest of the course.

What this unit covers

Section	What you must be able to do
9.5 (1)	Draw and interpret scatter diagrams. Mark each point clearly, for example as a small cross (×).
9.5 (2)	Understand positive, negative and zero correlation.
9.5 (3)	Draw by eye, interpret and use a straight line of best fit.
9.6 (1)	Draw and interpret cumulative frequency tables and diagrams. Mark points as crosses and join them with a smooth curve.
9.6 (2)	Estimate and interpret the median, percentiles, quartiles and interquartile range from cumulative frequency diagrams.

Scatter diagrams (9.5 outcome 1)

A scatter diagram shows pairs of values for the same items, such as the age and value of each car in a sample. One variable goes on the horizontal axis and the other on the vertical axis. Each pair is one point. Use a scale that spreads the points over most of the grid, label both axes with units, and mark each point as a small cross.

Correlation (9.5 outcome 2)

Correlation describes the pattern of the points.

Type	Pattern	Example
Positive	Points rise from left to right: as one variable increases, the other tends to increase.	Height and arm length
Negative	Points fall from left to right: as one variable increases, the other tends to decrease.	Age of a bicycle and its resale price
Zero	No pattern, only a scatter of points.	Shoe size and test mark

When you describe correlation, name the type and say what it means in the context. “Negative correlation: older cars tend to be worth less” is a full answer. Strong or weak describes how tightly the points follow a line. Correlation shows that two variables are linked in the data. It does not prove that one causes the other.

Line of best fit (9.5 outcome 3)

The syllabus notes say a line of best fit:

• is a single ruled line drawn by inspection
• extends across the full data set
• does not need to pass through any particular point, but has a roughly even distribution of points either side of it along its whole length.

A helpful method is to find the mean of each variable, plot that mean point, and pivot your ruler about it until the points balance on both sides. Use a clear ruler so you can see the points underneath.

Worked example 1

The table shows the age (years) and value ($ thousand) of nine cars of one model.

Age	1	2	3	4	5	6	7	8	9
Value	18.5	16.9	15.1	13.2	12.6	10.4	9.9	7.8	7.2

(a) Describe the correlation.

The values fall as the age rises, so there is strong negative correlation: older cars tend to be worth less.

(b) Draw a line of best fit.

Mean age = 45 ÷ 9 = 5. Mean value = 111.6 ÷ 9 = 12.4. Plot the mean point (5, 12.4). Rule a line through it that runs from about (1, 18.1) to (9, 6.7). Five points lie above this line and four below, spread along its length, so the balance is good.

(c) Estimate the value of a car aged 6.5 years.

Go up from 6.5 on the age axis to the line, then across. Gradient = (6.7 − 18.1) ÷ (9 − 1) = −1.425.

value at 6.5 = 18.1 + (−1.425)(6.5 − 1)
             = 18.1 − 7.8375
             ≈ 10.3

The value is about $10 300. Your reading from your own line may differ a little, and a small tolerance is expected.

(d) Interpret the gradient.

The gradient is about −1.4. On average, the value falls by about $1 400 for each extra year of age.

(e) Comment on using the line for a car aged 12 years.

The line gives 18.1 − 1.425 × 11 ≈ 2.4, so $2 400. Age 12 is outside the data, which stops at 9 years. The pattern may not continue in a straight line, so this estimate is unreliable. Estimates inside the range of the data are more reliable.

Cumulative frequency tables (9.6 outcome 1)

Cumulative frequency is a running total. Each entry says how many values are less than or equal to the upper boundary of a class. The last entry equals the total frequency, which is a quick check on your adding.

Worked example 2

The lengths of 200 phone calls, t minutes, are recorded.

Length	0 < t ≤ 2	2 < t ≤ 4	4 < t ≤ 6	6 < t ≤ 8	8 < t ≤ 10	10 < t ≤ 12
Frequency	14	36	62	48	28	12

t ≤ 2:    14
t ≤ 4:    14 + 36 = 50
t ≤ 6:    50 + 62 = 112
t ≤ 8:   112 + 48 = 160
t ≤ 10:  160 + 28 = 188
t ≤ 12:  188 + 12 = 200   (matches the total)

Drawing the cumulative frequency diagram (9.6 outcome 1)

1. Variable along the bottom, cumulative frequency up the side. Label both.
2. Plot each cumulative frequency against the upper class boundary.
3. Add a start point where the first class begins, at height 0. Here that is (0, 0).
4. Mark every point with a small cross.
5. Join the points with a smooth curve, not ruled straight segments.

For example 2 the points are:

(0, 0) (2, 14) (4, 50) (6, 112) (8, 160) (10, 188) (12, 200)

The curve is S-shaped and never goes down.

Median, quartiles, percentiles and IQR (9.6 outcome 2)

With n values in total, read across from these heights and then down to the horizontal axis.

Measure	Height on the vertical axis
Lower quartile (LQ)	n/4
Median	n/2
Upper quartile (UQ)	3n/4
p-th percentile	(p/100) × n

The interquartile range is IQR = UQ − LQ. It shows how far apart the quartiles are, so extreme values do not affect it.

Worked example 3

Use the call lengths, with n = 200.

Median:  200 ÷ 2 = 100  →  about 5.6 minutes
LQ:      200 ÷ 4 = 50   →  about 4.0 minutes
UQ:      3 × 200 ÷ 4 = 150  →  about 7.6 minutes
IQR = 7.6 − 4.0 = about 3.6 minutes

Check each reading against the table. A height of 100 lies between cumulative frequencies 50 and 112, so the median must lie between 4 and 6 minutes. It does.

Worked example 4

(a) Estimate how many calls lasted more than 9 minutes.

Up from 9 on the horizontal axis to the curve, then across: about 174 calls lasted 9 minutes or less. So 200 − 174 = 26 calls lasted more than 9 minutes.

(b) The longest 10% of calls are charged at a higher rate. Estimate the length above which this applies.

The longest 10% start above the 90th percentile. Height = 0.9 × 200 = 180, which gives about 9.4 minutes.

A reading from the curve counts the values up to and including that point. For “more than”, take the reading away from the total.

Interpreting and comparing

When you compare two distributions, make one statement about the average and one about the spread, and write both in context.

Worked example 5

A year later, 200 calls had a median of 4.9 minutes and an IQR of 2.6 minutes. Compare them with the earlier calls.

• The median fell from about 5.6 to 4.9 minutes, so calls are typically shorter.
• The IQR fell from about 3.6 to 2.6 minutes, so call lengths are more consistent.

A comparison that gives only numbers, without saying what they mean, earns little credit.

Common errors

• Plotting cumulative frequency at the class midpoint instead of the upper boundary.
• Leaving out the starting point (lower boundary, 0).
• Joining the crosses with a ruler, or drawing a line of best fit as a curve or as a join-the-dots path.
• Reading the median at half of one class frequency instead of at n/2, half of the total.
• Stating “more than” counts straight from the curve without subtracting from n.
• Describing correlation without a context sentence.
• Extending a line of best fit far beyond the data and treating the result as reliable.

Next steps

Try the practice questions, then check your overall readiness with the free 10-minute diagnostics. The O Level Mathematics past papers page shows where to find official papers.

Official syllabus

Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, published by Cambridge University Press & Assessment (Cambridge Assessment International Education), sections 9.5 and 9.6.