AQA GCSE·📚 Statistics·higher

AQA GCSE Statistics — Paper 1 (Higher)

⏱ 110 minutes📊 82 marks📄 Paper 1 (Higher)

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ℹ️ About this paper: This is an exam-board-aligned practice paper written in the style of AQA GCSE — not an official past paper. Use it for timed practice, then check against the mark scheme included below. For official past papers, see the exam board's website.

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AQA GCSE Statistics — Paper 1 (Higher)

Total marks: 82 · Duration: 1 hour 50 minutes · Tier: higher

Instructions to candidates

• Answer all questions. • Use black ink or black ball-point pen. You may use an HB pencil for graphs. • You must show all your working; marks may be given for a correct method even if the answer is wrong. • A scientific calculator is required. • Give your answers to an appropriate degree of accuracy where a rounded answer is needed. • The total number of marks for this paper is 82. • All data you need is given within the questions in tables; no separate insert is required, and any graph or chart referred to is one you are asked to describe or construct yourself.

Paper

1. A student records the number of text messages received by each of 20 people in one hour. The results are:

Messages	0	1	2	3	4	5
Frequency	3	5	4	4	2	2

(a) Write down the modal number of messages. (1 mark) (b) Find the median number of messages. (2 marks) (c) Calculate the mean number of messages. (3 marks)

2. The table shows the times, t minutes, taken by 60 students to travel to school.

Time (t min)	0 < t ≤ 10	10 < t ≤ 20	20 < t ≤ 30	30 < t ≤ 40	40 < t ≤ 50
Frequency	12	20	15	9	4

(a) Write down the modal class. (1 mark) (b) Calculate an estimate of the mean travel time. (4 marks) (c) Explain why your answer to part (b) is only an estimate. (1 mark)

3. The mean of five numbers is 14. When a sixth number is included, the mean becomes 15. Find the sixth number. (3 marks)

4. A company classifies its 240 employees by department. It wants to take a stratified sample of 40 employees.

Department	Sales	Production	Admin	Delivery
Employees	96	84	36	24

(a) Explain what is meant by a stratified sample. (1 mark) (b) Calculate the number of employees that should be sampled from the Production department. (2 marks) (c) Give one advantage of a stratified sample over a simple random sample here. (1 mark)

5. The cumulative frequency table shows the masses, m grams, of 80 apples.

Mass (m ≤)	100	120	140	160	180
Cumulative frequency	6	26	58	74	80

(a) Use the table to estimate the median mass. (2 marks) (b) Estimate the interquartile range. (3 marks) (c) An apple is classed as "large" if its mass is more than 165 g. Estimate the number of large apples. (2 marks)

6. The table shows the number of hours of sunshine, x, and the number of ice creams sold, y, on eight days.

x (hours)	2	3	5	6	7	8	9	10
y (sold)	14	20	30	34	42	48	52	60

(a) Describe the correlation between x and y. (1 mark) (b) The equation of the line of best fit is y = 5.6x + 3. Use it to estimate the number of ice creams sold when there are 4 hours of sunshine. (2 marks) (c) Explain why it would not be sensible to use this equation to estimate sales for 20 hours of sunshine. (1 mark) (d) Interpret the value 5.6 in the equation of the line of best fit. (1 mark)

7. A spinner has four sections coloured red, blue, green and yellow. The probability of each colour is shown, with the probability of yellow unknown.

Colour	Red	Blue	Green	Yellow
Probability	0.3	0.25	0.2	y

(a) Find the value of y. (2 marks) (b) The spinner is spun 200 times. Estimate the number of times it lands on blue. (2 marks) (c) The spinner is spun twice. Find the probability that it lands on red both times. (2 marks)

8. The table shows the amount spent, £s, by customers in a shop.

Spend (£s)	0 < s ≤ 20	20 < s ≤ 40	40 < s ≤ 80
Frequency	30	24	16

A histogram is to be drawn for this data.

(a) Explain why frequency density, rather than frequency, must be plotted on the vertical axis. (1 mark) (b) Calculate the frequency density for each of the three classes. (3 marks) (c) On a histogram, the bar for the class 0 < s ≤ 20 has a height of 1.5 units. State the height that the bar for 40 < s ≤ 80 should have on the same scale. (2 marks)

9. A quality inspector records the number of faulty items in samples of 50 taken from a production line each day for 10 days:

2, 3, 1, 4, 2, 6, 3, 2, 5, 2

(a) Find the range of the number of faulty items. (1 mark) (b) Find the median and the interquartile range. (4 marks) (c) On day 11 a sample contains 12 faulty items. Using an appropriate calculation, explain whether 12 should be regarded as an outlier. (3 marks)

10. The table shows the population of a town, in thousands, over five years, and the index numbers for the cost of living (2020 = 100).

Year	2020	2021	2022	2023	2024
Population (000s)	40	42	45	47	50
Cost-of-living index	100	104	111	118	126

(a) Calculate the percentage increase in population from 2020 to 2024. (2 marks) (b) Interpret the cost-of-living index value of 126 in 2024. (1 mark) (c) A worker earned £24 000 in 2020. To keep pace with the cost of living, what would their salary need to be in 2024? (2 marks)

11. A statistician investigates whether a six-sided die is fair by rolling it 120 times. The observed frequencies are:

Score	1	2	3	4	5	6
Frequency	15	22	18	25	20	20

(a) State how many times each score would be expected if the die were fair. (1 mark) (b) Give one reason why the observed frequencies differing from the expected value does not, by itself, prove the die is biased. (1 mark) (c) Calculate the mean score obtained across the 120 rolls. (3 marks)

12. In a town, 60% of households own a car. Of those that own a car, 40% also own a bicycle. Of those that do not own a car, 75% own a bicycle.

(a) Complete a tree diagram for this information by stating the four probabilities on the second branches (car → bicycle, car → no bicycle, no car → bicycle, no car → no bicycle). (2 marks) (b) Find the probability that a household chosen at random owns a bicycle. (3 marks) (c) Find the probability that a household owns a car, given that it owns a bicycle. (3 marks)

13. The daily maximum temperatures, in °C, recorded in a town over seven days were:

18, 21, 19, 24, 20, 22, 17

(a) Calculate the mean temperature. (2 marks) (b) Calculate the standard deviation of the temperatures, giving your answer to 2 decimal places. (4 marks) (c) A second town over the same week had a mean of 20 °C and a standard deviation of 4.8 °C. Compare the temperatures of the two towns. (2 marks)

14. The quarterly sales, in £000s, of a small business are shown.

	Q1	Q2	Q3	Q4
Year 1	12	18	24	14
Year 2	16	22	28	18

(a) Explain why a four-point moving average is appropriate for this data. (1 mark) (b) Calculate the first two four-point moving averages. (3 marks) (c) State what the trend in the moving averages suggests about the business. (1 mark)

📋 Mark Scheme & Sample Answers

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Mark Scheme

1. (a) Mode = 1 (most frequent, freq 5) (B1). [1] (b) 20 values → median is mean of the 10th and 11th values (M1). Cumulative: 3, 8, 12… the 10th and 11th both fall in "2", so median = 2 (A1). [2] (c) Σ(fx) = 0×3 + 1×5 + 2×4 + 3×4 + 4×2 + 5×2 = 0 + 5 + 8 + 12 + 8 + 10 = 43 (M1 M1). Mean = 43 ÷ 20 = 2.15 (A1). [3]

2. (a) Modal class = 10 < t ≤ 20 (B1). [1] (b) Midpoints 5, 15, 25, 35, 45 (M1). Σ(fx) = 12×5 + 20×15 + 15×25 + 9×35 + 4×45 = 60 + 300 + 375 + 315 + 180 = 1230 (M1 M1). Mean = 1230 ÷ 60 = 20.5 minutes (A1). [4] (c) The exact values within each class are unknown, so midpoints are assumed; the true total may differ (B1). [1]

3. Sum of five numbers = 5 × 14 = 70 (M1). Sum of six numbers = 6 × 15 = 90 (M1). Sixth number = 90 − 70 = 20 (A1). [3]

4. (a) A sample where the number taken from each group (stratum) is in proportion to the size of that group (B1). [1] (b) Production fraction = 84/240 (M1). Sample = 84/240 × 40 = 14 (A1). [2] (c) It ensures every department is represented in proportion, so the sample better reflects the whole workforce (B1). [1]

5. (a) Median is the 40th value (80 ÷ 2) (M1). From the table, cumulative frequency reaches 40 within the 120–140 class; interpolating, ≈ 40th value is about 128 g (allow 126–130) (A1). [2] (b) Lower quartile = 20th value ≈ 116 g; upper quartile = 60th value ≈ 142 g (M1 M1). IQR ≈ 142 − 116 = 26 g (allow 24–28) (A1). [3] (c) Number with mass ≤ 165 ≈ 76 (reading at 165, between cf 74 at 160 and 80 at 180, ≈ 74 + (5/20)×6 ≈ 75.5) (M1). Large apples ≈ 80 − 76 = about 4–5 (A1). [2]

6. (a) Positive correlation (as sunshine increases, sales increase) (B1). [1] (b) y = 5.6 × 4 + 3 = 22.4 + 3 = 25 (to nearest ice cream, 25) (M1 A1). [2] (c) 20 hours is outside the range of the data (extrapolation), so the linear relationship may not hold (B1). [1] (d) For each extra hour of sunshine, about 5.6 more ice creams are sold (B1). [1]

7. (a) Probabilities sum to 1: y = 1 − (0.3 + 0.25 + 0.2) = 1 − 0.75 = 0.25 (M1 A1). [2] (b) 0.25 × 200 = 50 (M1 A1). [2] (c) P(red, red) = 0.3 × 0.3 (M1) = 0.09 (A1). [2]

8. (a) The class widths are unequal, so area (not height) must represent frequency; frequency density = frequency ÷ class width keeps areas proportional (B1). [1] (b) Widths 20, 20, 40. Densities: 30/20 = 1.5; 24/20 = 1.2; 16/40 = 0.4 (M1 M1 A1). [3] (c) The class 0 < s ≤ 20 has density 1.5 shown as height 1.5, so 1 unit of height = 1 unit of density (M1). The class 40 < s ≤ 80 has density 0.4, so its height = 0.4 units (A1). [2]

9. (a) Range = 6 − 1 = 5 (B1). [1] (b) Ordered: 1, 2, 2, 2, 2, 3, 3, 4, 5, 6 (M1). Median = mean of 5th and 6th = (2 + 3)/2 = 2.5 (A1). LQ = 3rd value = 2; UQ = 8th value = 4 (M1). IQR = 4 − 2 = 2 (A1). [4] (c) Outlier boundary = UQ + 1.5 × IQR = 4 + 1.5 × 2 = 7 (M1 M1). Since 12 > 7, 12 is an outlier (A1). [3]

10. (a) Increase = 50 − 40 = 10; percentage = 10/40 × 100 (M1) = 25% (A1). [2] (b) The cost of living in 2024 is 26% higher than in 2020 (the base year) (B1). [1] (c) £24 000 × 126/100 (M1) = £30 240 (A1). [2]

11. (a) Expected = 120 ÷ 6 = 20 each (B1). [1] (b) Small differences are expected due to natural (random) variation even for a fair die; the sample is limited (B1). [1] (c) Σ(fx) = 1×15 + 2×22 + 3×18 + 4×25 + 5×20 + 6×20 = 15 + 44 + 54 + 100 + 100 + 120 = 433 (M1 M1). Mean = 433 ÷ 120 = 3.61 (to 3 s.f.) (A1). [3]

12. (a) Car → bicycle 0.4 and car → no bicycle 0.6; no car → bicycle 0.75 and no car → no bicycle 0.25 (B1 for the car branches, B1 for the no-car branches). [2] (b) P(bicycle) = P(car ∧ bike) + P(no car ∧ bike) = 0.6 × 0.4 + 0.4 × 0.75 (M1 M1) = 0.24 + 0.30 = 0.54 (A1). [3] (c) P(car | bicycle) = P(car ∧ bicycle) / P(bicycle) = 0.24 / 0.54 (M1 M1) = 0.444 (to 3 s.f.), or 4/9 (A1). [3]

13. (a) Mean = (18 + 21 + 19 + 24 + 20 + 22 + 17) ÷ 7 = 141 ÷ 7 = 20.14 °C (allow 20.1) (M1 A1). [2] (b) Deviations from the mean (using 20.14…), or use Σx² method: Σx² = 18² + 21² + 19² + 24² + 20² + 22² + 17² = 324 + 441 + 361 + 576 + 400 + 484 + 289 = 2875 (M1). Variance = Σx²/n − (mean)² = 2875/7 − (141/7)² = 410.71 − 405.73 = 4.98 (M1 M1). SD = √4.98 = 2.23 °C (to 2 d.p.) (A1). [4] (c) The two towns have a similar mean (about 20 °C), but the second town has a much larger standard deviation (4.8 vs 2.23), so its temperatures were more spread out / less consistent (B1 for centre, B1 for spread). [2]

14. (a) The data has a repeating four-quarter (seasonal) cycle, so a four-point moving average smooths out the seasonal variation to show the trend (B1). [1] (b) First MA = (12 + 18 + 24 + 14) ÷ 4 = 68 ÷ 4 = 17 (M1 A1). Second MA = (18 + 24 + 14 + 16) ÷ 4 = 72 ÷ 4 = 18 (A1). So 17 and 18 (£000s) (M1 for method, A marks for values). [3] (c) The moving averages are increasing, suggesting the underlying sales trend is rising / the business is growing (B1). [1]

Sample Answers with Examiner Commentary

Question 8 — Sample Answers

Grade A response. "(a) The classes have different widths, so if I plotted frequency the wide class would look far bigger than it should. On a histogram the area of each bar represents the frequency, so I plot frequency density = frequency ÷ class width to keep the areas correct. (b) Widths are 20, 20 and 40. Densities: 30 ÷ 20 = 1.5, 24 ÷ 20 = 1.2, 16 ÷ 40 = 0.4. (c) The first bar has density 1.5 and height 1.5, so one height unit = one density unit. The last class has density 0.4, so its bar height is 0.4 units."

Mark: 6/6. Examiner commentary: A complete and precise answer. In (a) the candidate gives the essential reason — unequal class widths, with area representing frequency — rather than just quoting "frequency density" as a phrase. The densities in (b) are all correct, and critically the candidate divides the last frequency by the correct width of 40, which is where most marks are lost on this topic. Part (c) shows genuine understanding of the scale rather than guessing: they establish that height 1.5 corresponds to density 1.5 and scale the final bar accordingly.

Grade C response. "(a) Because the bars are different widths. (b) 30 ÷ 20 = 1.5, 24 ÷ 20 = 1.2, 16 ÷ 20 = 0.8. (c) 0.8 units."

Mark: 3/6. Examiner commentary: Part (a) earns the mark, just — the idea of unequal widths is there, though a stronger answer would mention that area represents frequency. The error that costs three marks is in (b): the candidate divides the final frequency by 20 instead of the correct class width of 40, giving 0.8 rather than 0.4. This single mistake then carries through to (c), where the follow-on value 0.8 is stated. Always read the class width straight from the interval: 40 < s ≤ 80 spans 40, not 20.

Question 9(c) — Sample Answers

Grade A response. "From part (b), LQ = 2, UQ = 4, so IQR = 2. The usual rule is that a value is an outlier if it is more than UQ + 1.5 × IQR above the upper quartile. That boundary is 4 + 1.5 × 2 = 7. Since 12 is greater than 7, 12 should be regarded as an outlier."

Mark: 3/3. Examiner commentary: Exactly what is required. The candidate quotes the 1.5 × IQR rule, computes the upper boundary of 7 from their own quartiles, and makes a clear comparison and conclusion. Stating the rule before using it secures the method even if the arithmetic had slipped, and the final sentence answers the question directly rather than leaving the examiner to infer the conclusion.

Grade C response. "12 is much bigger than all the other numbers, which only go up to 6, so yes it is an outlier."

Mark: 1/3. Examiner commentary: The conclusion happens to be correct and the candidate earns one mark for recognising that 12 is unusually large, but the question asks for "an appropriate calculation", and none is shown. Without the IQR boundary (UQ + 1.5 × IQR = 7) there is no objective test — a purely visual judgement is not enough at this level. The two method marks require the boundary calculation and an explicit comparison of 12 against it.

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