OCR GCSE Mathematics: Analysing data: averages, spread, box plots, misrepresentation, scatter graphs and outliers (J560) – 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.

These questions follow section 12.03 of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026): summary statistics (12.03a), misrepresenting data (12.03b), bivariate data (12.03c) and outliers (12.03d). Questions 10 and 11 are (Higher tier only). All the others are Foundation and Higher content. Learn the methods first from the study guide and the revision notes. The OCR GCSE Mathematics hub, the checklist and the 10-minute diagnostics are linked for planning.

Questions

1. Ravi sent these numbers of texts over eight days: 14, 9, 22, 9, 17, 11, 9, 19. Work out the mean, the median, the mode and the range. [4]

2. The mean of five numbers is 12.

(a) Four of the numbers are 7, 15, 9 and 14. Find the fifth number. [2] (b) A sixth number is added and the mean of the six numbers is 13. Find the sixth number. [2]

3. Waiting times at a clinic, in minutes, for 50 patients:

Time w	0 < w ≤ 5	5 < w ≤ 10	10 < w ≤ 15	15 < w ≤ 20	20 < w ≤ 25
Frequency	9	18	14	7	2

(a) State the modal class. [1] (b) Work out an estimate for the mean waiting time. [3] (c) Explain why your answer to (b) is an estimate. [1] (d) Estimate the median waiting time. [3]

4. Cafe A has a mean of 64 customers a day and a range of 58. Cafe B has a mean of 59 customers a day and a range of 21. Compare the two cafes. [3]

5. Prices of six flats, in £1000s: 210, 225, 230, 240, 255, 820.

(a) Find the mean and the median. [2] (b) Which is the better summary of a typical price? Give a reason. [1]

6. A bar chart shows visitors to a museum in January, February and March as 1950, 2000 and 2050. The vertical axis starts at 1900 and has no label.

(a) Give two reasons why the chart is misleading. [2] (b) Work out the percentage increase in visitors from January to March, to 1 decimal place. [2]

7. The table shows minutes of screen time before bed (x) and hours of sleep (y) for eight people.

x	10	30	45	60	80	100	120	150
y	8.9	8.6	8.4	8.1	7.8	7.4	6.9	6.4

(a) Describe the correlation in context. [2] (b) A line of best fit passes through (30, 8.7) and (150, 6.3). Estimate the hours of sleep for someone with 90 minutes of screen time. [2] (c) Use the line to estimate the sleep for 300 minutes of screen time. Comment on the reliability of this estimate. [3]

8. A newspaper reports a positive correlation between daily umbrella sales and the number of road accidents. It says that carrying umbrellas causes accidents. Explain why this conclusion is not justified. [2]

9. Seven sprinters’ times, in seconds, are 12.1, 12.4, 11.9, 12.6, 12.2, 1.25 and 12.3.

(a) Write down the outlier. [1] (b) Suggest why it may have been recorded. [1] (c) Find the mean of the seven values as recorded, to 2 decimal places. [2] (d) Find the mean if the outlier should have been 12.5, to 2 decimal places. [1]

10. (Higher tier only) The time (t minutes) 60 students spent on homework is shown on a cumulative frequency graph. It passes through (0, 0), (20, 5), (40, 14), (60, 33), (80, 49), (100, 57) and (120, 60). Treat the graph as straight lines between points.

(a) Estimate the lower quartile, the median and the upper quartile, to the nearest minute. [3] (b) Work out the interquartile range. [1] (c) How many students spent more than 100 minutes? [1] (d) Another class has a median of 65 minutes and an interquartile range of 24 minutes. Compare the two classes. [3]

11. (Higher tier only) Times, in seconds, for nine runners in team A: 11.8, 12.1, 12.4, 12.6, 12.9, 13.1, 13.5, 14.2, 15.0. Team B has minimum 11.5, lower quartile 12.0, median 12.5, upper quartile 13.0 and maximum 14.0.

(a) Find the five-number summary for team A. [3] (b) Draw a box plot for each team on the same scale. [2] (c) Compare the teams using the median and the interquartile range. [2]

12. Distances travelled to work by 20 + x people:

Distance d km	0 < d ≤ 10	10 < d ≤ 20	20 < d ≤ 30	30 < d ≤ 40
Frequency	6	x	10	4

An estimate of the mean distance is 18 km. Find x. [5]

Answers

1. In order: 9, 9, 9, 11, 14, 17, 19, 22. Sum = 110 so mean = 110 ÷ 8 = 13.75 [1]. Median = (11 + 14) ÷ 2 = 12.5 [1]. Mode = 9 [1]. Range = 22 − 9 = 13 [1] [4] Examiner insight: Sort the data before taking the median. Writing the ordered list shows your method even if you slip on the arithmetic.

2. (a) Total = 5 × 12 = 60 [1]. Fifth = 60 − (7 + 15 + 9 + 14) = 60 − 45 = 15 [1] (b) Total of six = 6 × 13 = 78 [1]. Sixth = 78 − 60 = 18 [1] [4] Examiner insight: Work from the totals (mean × count). Subtracting means directly is a common slip that gives no useful working.

3. (a) 5 < w ≤ 10 [1] (b) Midpoints 2.5, 7.5, 12.5, 17.5, 22.5 [1]. Sum of fx = 22.5 + 135 + 175 + 122.5 + 45 = 500 [1]. Mean = 500 ÷ 50 = 10 minutes [1] (c) The exact waiting times are not known, only the class each lies in, so midpoints are used [1] (d) The median is the 25th value. Running totals are 9, 27, … so it is in 5 < w ≤ 10 [1]. It is the 16th of 18 values in that class, so 5 + (16/18) × 5 [1] = 9.4 minutes (1 d.p.) [1] [8] Examiner insight: Show the midpoints and the column of products. Divide by the total frequency, not the class count.

4. Cafe A has the higher mean, so it has more customers on a typical day [1]. Cafe B has the smaller range, so its daily numbers are more consistent [1]. Cafe A’s customer numbers vary much more from day to day (range 58 against 21) [1] [3] Examiner insight: A comparison needs context and one statement about average and one about spread. Quoting the numbers without saying what they mean is not a comparison.

5. (a) Sum = 1980, mean = 1980 ÷ 6 = £330 000 [1]. Median = (230 + 240) ÷ 2 = £235 000 [1] (b) The median, because the single price of 820 drags the mean up, so the mean is not typical of the other five [1] [3] Examiner insight: Choose an average and give a reason that refers to the data. A bare “median” with no reason leaves the explanation unproved.

6. (a) The vertical axis does not start at zero, so the increase looks much bigger than it is [1]. The axis has no label (or title), so the reader cannot tell what is measured [1] (b) Increase = 2050 − 1950 = 100 [1]. 100 ÷ 1950 × 100 = 5.1% [1] [4] Examiner insight: For each feature, say what effect it has. Naming a feature alone is thinner than naming it and its effect.

7. (a) Negative correlation [1]. The more screen time before bed, the fewer hours of sleep [1] (b) Gradient = (6.3 − 8.7) ÷ (150 − 30) = −0.02 [1]. At x = 90: 8.7 − 0.02 × 60 = 7.5 hours [1] (c) At x = 300: 8.7 − 0.02 × 270 = 3.3 hours [1]. 300 minutes is far outside the data (10 to 150) [1], so this is extrapolation: the pattern may not carry on, and the estimate is unreliable [1] [7] Examiner insight: Describe the relationship in terms of the two variables, not just “negative”. For reliability, say that the value is outside the range of the data.

8. The two variables may both depend on a third factor, such as rainy weather, which means more umbrellas and also more accidents [1]. The data show that the two quantities rise together, which does not prove that either one produces the other [1] [2] Examiner insight: Name a plausible other factor. “Correlation does not mean causation” on its own does not explain this particular case.

9. (a) 1.25 [1] (b) A decimal point was missed (12.5 recorded as 1.25), or a timer error [1] (c) Sum = 74.75 [1]. Mean = 74.75 ÷ 7 = 10.68 [1] (d) Sum = 74.75 − 1.25 + 12.5 = 86.0, mean = 86 ÷ 7 = 12.29 [1] [5] Examiner insight: Give a reason that fits the context. Notice how far the outlier moves the mean: 10.68 against 12.29.

10. (Higher tier only) (a) Heights 15, 30 and 45 [1]. Lower quartile: 40 + (1/19) × 20 = 41.05, so 41 [1]. Median: 40 + (16/19) × 20 = 56.8, so 57. Upper quartile = 75 [1] (b) 75 − 41 = 34 minutes [1] (c) 60 − 57 = 3 students [1] (d) This class has the lower median (57 against 65), so students typically spent less time [1]. Its IQR is larger (34 against 24) so the times are less consistent [1]. State both in context of homework time [1] [8] Examiner insight: Mark the reading lines on the graph and keep working visible. Readings from a drawn graph are estimates, so say that your answers are estimates and show where you read them.

11. (Higher tier only) (a) Minimum 11.8 and maximum 15.0 [1]. Median = 12.9 [1]. Q1 = (12.1 + 12.4) ÷ 2 = 12.25 and Q3 = (13.5 + 14.2) ÷ 2 = 13.85 [1] (b) A box from 12.25 to 13.85 with the median at 12.9 and whiskers to 11.8 and 15.0 [1]. A box from 12.0 to 13.0 with the median at 12.5 and whiskers to 11.5 and 14.0, on the same scale [1] (c) Team B has the lower median (12.5 against 12.9), so B is typically faster [1]. B’s IQR is 1.0 against 1.6, so B’s times are more consistent [1] [7] Examiner insight: Draw on a labelled scale with a ruler. In the comparison, give the medians and the IQRs and say what each means.

12. Midpoints 5, 15, 25, 35 [1]. Sum of fx = 30 + 15x + 250 + 140 = 420 + 15x [1]. Total frequency = 20 + x [1]. (420 + 15x) ÷ (20 + x) = 18, so 420 + 15x = 360 + 18x [1]. 60 = 3x, so x = 20 [1] [5] Examiner insight: Set up the equation carefully and keep the total frequency as 20 + x. Check by substituting: 720 ÷ 40 = 18.

Where marks are usually lost

• Taking the median from data that are not in order.
• Dividing a grouped total by the number of classes instead of the number of values.
• Quoting only numbers when asked to compare, with no meaning in context.
• Calling a chart misleading without saying what feature causes the effect.
• Describing correlation as “positive” or “negative” with no reference to the variables.
• Using extrapolation without noting that it is outside the data.
• Reading quartiles at the wrong heights of a cumulative frequency graph.
• Forgetting the unit or the context in a final answer.

Next steps

• Review the revision notes and the study guide.
• Open the OCR GCSE Mathematics hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, section 12.03 Analysing data (12.03a to 12.03d).