Cambridge O Level Statistics: Scatter diagrams, correlation and lines of best fit (4040) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers sections 9.1 and 9.2 of the Cambridge O Level Statistics (4040) syllabus for examination in 2025, 2026 and 2027: correlation, semi-average lines of best fit, y = mx + c, and prediction. For an odd number of pairs, leave out the middle pair and say so.

Useful links: the study guide, the revision notes, the Cambridge O Level Statistics hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. State whether you would expect positive correlation, negative correlation or no correlation between each pair of variables, for a group of adults.

(a) The age of a used car and its selling price. [1] (b) The number of hours of sunshine in a day and the number of sunglasses sold. [1] (c) The height of the adult and the number of letters in the surname. [1] (d) The height above sea level of a town and the average air temperature there. [1]

2. Eight people record their weekly television time in hours (x) and a general knowledge score (y).

x	2	5	8	10	12	15	18	20
y	70	55	75	58	68	52	66	57

Describe the correlation between x and y. [3]

3. A line of best fit for the fuel left in a lorry tank (y litres) after travelling x hundred kilometres passes through the points (2, 52) and (6, 22).

(a) Work out the equation of the line, written as y = mx + c. [3] (b) Say what the gradient tells you about the fuel use. [1] (c) Use the line to estimate the fuel left after 500 km. [1]

4. The height of a seedling (y mm) is measured x days after sowing.

x	3	5	7	9	11	13	15	17
y	12	15	17	22	23	28	29	34

(a) Plot the data as a scatter diagram on graph paper. [2] (b) Use the method of semi-averages to find the equation of the line of best fit. [5] (c) Estimate the height after 10 days. [1] (d) A pupil uses the line to estimate the height after 40 days. Comment on this estimate. [2]

5. A shop records its monthly advertising spend (x, in thousands of rupees) and the number of new customers (y).

x	40	10	70	25	55	85	20	60	35
y	45	10	80	24	60	88	16	68	30

Find the equation of the line of best fit by the method of semi-averages. [6]

6. For a block of flats, daily outdoor temperature (x °C) and heating units used (y) were recorded for days with temperatures from 2 °C to 16 °C. The line of best fit is y = 48 − 2.5x.

(a) Estimate the units used when the temperature is 10 °C. [1] (b) Estimate the units used when the temperature is 25 °C, and comment on your answer. [2] (c) Estimate the temperature on a day when 30 units are used. [2]

7. The mean of the x values in a data set is 12 and the mean of the y values is 40. A line of best fit passes through the mean point and through the point (20, 56).

(a) Find the equation of the line. [3] (b) Estimate y when x = 15. [1]

8. The weekly number of book loans (y, in hundreds) at a library is recorded for weeks x = 1 to 9.

x	1	2	3	4	5	6	7	8	9
y	11	13	16	18	40	23	26	28	31

(a) Identify the outlier and suggest one possible reason for it. [2] (b) Ignore the outlier. Find the equation of the line of best fit by semi-averages for the other eight points. [5] (c) Use your line to estimate y for week 5. [1]

9. A café varies the price of a sandwich (x, in rupees) over ten weeks and records the number sold (y).

x	5	10	15	20	25	30	35	40	45	50
y	98	90	85	77	70	66	57	50	44	38

(a) Describe the correlation. [2] (b) Find the equation of the line of best fit by semi-averages. [5] (c) Estimate the number sold at a price of 28 rupees. [1] (d) Estimate the number sold at a price of 80 rupees, and comment on the result. [2] (e) Interpret the gradient in this context. [1]

10. Fire brigade records show strong positive correlation between the number of fire engines sent to a fire and the cost of the damage. Explain why this does not mean that sending more engines causes more damage. [3]

11. Six pairs of values are given: (1, 8), (2, 12), (3, k), (4, 22), (5, 28), (6, 31). The line of best fit found by the method of semi-averages is y = 5x + 2. Find the value of k. [5]

Answers

1. (a) Negative correlation [1]. (b) Positive correlation [1]. (c) No correlation [1]. (d) Negative correlation [1]. [4] Examiner insight: give one clear type for each pair, since offering two options may not be credited.

2. The points slope downwards slightly, so it is negative [1]. They are widely scattered, so it is weak [1]. People who watch more television tend to score slightly lower [1]. Weak negative correlation. [3] Examiner insight: give type, strength and a statement in context; two words alone may leave marks unclaimed.

3. (a) Gradient = (22 − 52)/(6 − 2) = −30/4 [1] = −7.5 [1]. Using (2, 52): 52 = −7.5 × 2 + c, so c = 67, giving y = −7.5x + 67 [1]. (b) The lorry uses 7.5 litres of fuel for every 100 km [1]. (c) 500 km is x = 5, so y = −37.5 + 67 = 29.5 litres [1]. [5] Examiner insight: x is in hundreds of kilometres, so substituting 500 instead of 5 gives a meaningless answer.

4. (a) Points plotted correctly [1], with axes labelled and scales suitable [1]. (b) Lower half x = 3, 5, 7, 9: mean x = 24/4 = 6 and mean y = 66/4 = 16.5 [1]. Upper half x = 11, 13, 15, 17: mean x = 56/4 = 14 and mean y = 114/4 = 28.5 [1]. m = (28.5 − 16.5)/(14 − 6) = 12/8 = 1.5 [1]. 16.5 = 1.5 × 6 + c, so c = 7.5 [1]. y = 1.5x + 7.5 [1]. (c) y = 15 + 7.5 = 22.5 mm [1]. (d) Day 40 is far outside the data, which stop at day 17, so this is extrapolation [1]. The line gives 67.5 mm, but growth is unlikely to stay linear, so the estimate is unreliable [1]. [10] Examiner insight: name the problem (extrapolation) and give a reason in context; “it is not accurate” says little.

5. Order by x: 10, 20, 25, 35, 40, 55, 60, 70, 85 [1]. Nine pairs, so leave out the middle pair (40, 45) [1]. Lower half: x = 10, 20, 25, 35 gives mean x = 22.5, and y = 10, 16, 24, 30 gives mean y = 80/4 = 20 [1]. Upper half: x = 55, 60, 70, 85 gives mean x = 67.5, and y = 60, 68, 80, 88 gives mean y = 296/4 = 74 [1]. m = (74 − 20)/(67.5 − 22.5) = 54/45 = 1.2 [1]. 20 = 1.2 × 22.5 + c, so c = −7, giving y = 1.2x − 7 [1]. [6] Examiner insight: keep each y value paired with its x value when you re-order; a slip spoils every later step.

6. (a) y = 48 − 25 = 23 units [1]. (b) y = 48 − 62.5 = −14.5 [1]. Negative use is impossible, and 25 °C is outside the data range (2 to 16) [1]. (c) 30 = 48 − 2.5x [1], so x = 18/2.5 = 7.2 °C [1]. [5] Examiner insight: to find x from y, rearrange before you calculate, and give the answer with its unit.

7. (a) The line passes through (12, 40) and (20, 56). m = (56 − 40)/(20 − 12) [1] = 16/8 = 2 [1]. 40 = 2 × 12 + c, so c = 16, giving y = 2x + 16 [1]. (b) y = 30 + 16 = 46 [1]. [4] Examiner insight: write the mean point as a pair of coordinates before you start the gradient.

8. (a) The outlier is (5, 40) [1]. A possible reason: a recording error, or a book fair that week [1]. (b) Lower half x = 1, 2, 3, 4: mean x = 2.5 and mean y = 58/4 = 14.5 [1]. Upper half x = 6, 7, 8, 9: mean x = 7.5 and mean y = 108/4 = 27 [1]. m = (27 − 14.5)/(7.5 − 2.5) = 12.5/5 = 2.5 [1]. 14.5 = 2.5 × 2.5 + c, so c = 8.25 [1]. y = 2.5x + 8.25 [1]. (c) y = 12.5 + 8.25 = 20.75 (about 2,075 loans) [1]. [8] Examiner insight: if you exclude a point, say so and say why; the remaining pairs must still split into equal halves.

9. (a) As the price rises, the number sold falls, so the correlation is negative [1]. The points lie very close to a straight line, so it is strong [1]. (b) Lower half x = 5 to 25: mean x = 15 and mean y = 420/5 = 84 [1]. Upper half x = 30 to 50: mean x = 40 and mean y = 255/5 = 51 [1]. m = (51 − 84)/(40 − 15) = −33/25 = −1.32 [1]. 84 = −1.32 × 15 + c, so c = 103.8 [1]. y = −1.32x + 103.8 [1]. (c) y = −36.96 + 103.8 = 66.84, so about 67 sandwiches [1]. (d) y = −105.6 + 103.8 = −1.8 [1]. A negative number sold is impossible, and 80 rupees is outside the range 5 to 50, so the line should not be used [1]. (e) Each extra rupee in price reduces sales by about 1.32 sandwiches [1]. [11] Examiner insight: interpret a negative gradient as a fall, in the units of the question.

10. The correlation shows an association only, not cause [1]. Large fires cause both more engines to be sent and more damage, so the size of the fire is a lurking variable [1]. The engines do not make the fire bigger [1]. [3] Examiner insight: name the third factor and say how it affects both variables.

11. Six pairs, so each half has 3 pairs: x = 1, 2, 3 and x = 4, 5, 6 [1]. The line passes through both semi-average points. Lower mean x = 2, so the lower mean y = 5 × 2 + 2 = 12 [1]. Lower mean y = (8 + 12 + k)/3 [1], so (20 + k)/3 = 12 and k = 36 − 20 = 16 [1]. Upper half check: mean y = (22 + 28 + 31)/3 = 27 = 5 × 5 + 2 [1]. [5] Examiner insight: checking the other half confirms your working and shows the method.

Where marks are usually lost

• Describing correlation in one word without strength or context.
• Splitting into halves before putting the pairs in order of x.
• Subtracting in different orders on the top and bottom of the gradient.
• Giving a prediction outside the data range with no comment on reliability.
• Accepting an impossible value, such as negative sales, without comment.
• Saying that correlation proves cause.

Next steps

• Re-read the revision notes or the study guide.
• Open the Cambridge O Level Statistics hub and tick off the checklist.
• Take the free 10-minute diagnostics to find weak spots.
• Book a free trial class for help with this topic.

Official syllabus

Cambridge O Level Statistics (4040) syllabus, for examination in 2025, 2026 and 2027, published by Cambridge University Press & Assessment (Cambridge International Education). Sections 9.1 and 9.2.