AQA A-Level Mathematics Paper 3, June 2025: Question 17

7 marks · Medium difficulty · Multi-step Problem

Calculate the mean and variance of attendance data, evaluate a proposed binomial distribution model, and identify and evaluate an opportunity sampling method.

Practise this question

Question

Question 17 shows attendance data from a random sample of eight Wednesdays: 2, 4, 3, 1, 4, 5, 2, 3. Part (a) asks to find the mean of these eight values for 1 mark. Part (b) asks to find the variance for 1 mark. Part (c) states the teacher believes attendance can be modelled by B(30, 0.1), and asks whether the sample mean and variance support this belief with full justification for 3 marks. Part (d) describes the teacher asking the first 10 students entering the room about a Saturday session, asking in (i) to name this sampling method for 1 mark, and in (ii) to describe one advantage of it for 1 mark.

Mark scheme

Show the mark scheme Mark scheme for Question 17: (a) Mean = 3, awarded B1. (b) Variance = 1.5 (accept AWRT 1.7 for sample variance), awarded B1. (c) M1 for calculating theoretical mean (30 × 0.1 = 3) or variance (30 × 0.1 × 0.9 = 2.7) of B(30, 0.1); A1 for both correct; R1 for reasoning that while the mean matches, the variance is different, so the model is not supported/suitable. (d)(i) B1 for 'opportunity', 'convenience', or 'opportunistic' sampling. (d)(ii) E1 for stating an advantage such as easy/simple, quick/fast, cheap/not expensive (excluding just 'convenient'). Total 7 marks.

How to answer it

Evaluating Binomial Models & Non-Random Sampling

📋 What this question tests

This question tests foundational statistics and mathematical modelling from the AQA A-Level specification:

  • Calculating sample summary statistics: mean (x̄) and variance (σ² or s²).
  • Knowing and applying the theoretical properties of the Binomial distribution: E(X) = np and Var(X) = np(1 − p).
  • Critically evaluating whether empirical data supports an assumed probability distribution model.
  • Identifying non-random sampling techniques (specifically opportunity / convenience sampling) and stating practical advantages.
Part 17 (a)

Sample Mean Calculation

Finding the mean of eight attendance values [1 Mark]

📐 Step-by-Step Calculation

Values: 2, 4, 3, 1, 4, 5, 2, 3

  1. Sum the values:
    Σx = 2 + 4 + 3 + 1 + 4 + 5 + 2 + 3 = 24
  2. Divide by sample size (n = 8):
    x̄ = 24 / 8 = 3

✅ Final Answer & Mark Scheme

Mean = 3

B1 (AO 1.1b): Correct value of 3 obtained.
Part 17 (b)

Sample Variance Calculation

Finding the variance of the attendance values [1 Mark]

📐 Step-by-Step Calculation

Using the population variance formula σ² = (Σx² / n) − x̄²:

  1. Calculate Σx²:
    4 + 16 + 9 + 1 + 16 + 25 + 4 + 9 = 84
  2. Compute variance:
    σ² = (84 / 8) − 3² = 10.5 − 9 = 1.5
  3. Alternative (unbiased sample variance s²):
    s² = (n / (n − 1)) × σ² = (8 / 7) × 1.5 ≈ 1.71

✅ Final Answer & Mark Scheme

Variance = 1.5 (or awrt 1.7)

B1 (AO 1.1b): Obtains 1.5 (AQA also accepts awrt 1.7 if calculated using divisor n − 1).

❌ Common Errors

  • Standard deviation trap: Writing √1.5 ≈ 1.22 instead of the variance. Always check whether the question asks for standard deviation (σ) or variance (σ²).
  • Inputting values incorrectly into the calculator's statistics list mode.

🧠 Exam Technique

Use the single-variable statistics function on your calculator (e.g. Casio fx-991EX/CW: Menu > 6: Statistics > 1: 1-Variable ). Read off σ²x (1.5) or s²x (1.714) directly to save time and prevent arithmetic slips.

Part 17 (c)

Hypothesis & Model Comparison

Evaluating the Binomial Model B(30, 0.1) [3 Marks]

💡 Key Knowledge: Binomial Properties

For a discrete random variable X ~ B(n, p):

  • Mean: E(X) = n × p
  • Variance: Var(X) = n × p × (1 − p)

Here, n = 30 and p = 0.1, so (1 − p) = 0.9.

📐 Calculations for B(30, 0.1)

  1. Model Mean:
    E(X) = 30 × 0.1 = 3
  2. Model Variance:
    Var(X) = 30 × 0.1 × 0.9 = 2.7

✅ Model Evaluation & Conclusion

Comparison:

  • The mean of the model is 3, which is identical to the sample mean from (a).
  • The variance of the model is 2.7, which is substantially higher than the sample variance of 1.5 (or 1.7).

Conclusion: The teacher's belief is not supported (the model is not suitable) because the variance of the data does not match the variance of the proposed binomial distribution.

🧠 Mark Breakdown & Examiner Commentary

M1 (AO 3.1a): Calculates the theoretical mean (3) or theoretical variance (2.7) for B(30, 0.1).

A1 (AO 1.1b): Correctly obtains both theoretical mean = 3 and theoretical variance = 2.7.

R1 (AO 2.2b): Infers that the belief is not supported specifically citing the difference in variance. (Must make a clear overarching conclusion to earn this mark).

❌ Common Misconceptions to Avoid

  • Incomplete comparison: Stating only "the mean matches so it is a good model" while completely ignoring the variance. Both statistics must be evaluated.
  • Vague conclusion: Writing "they are somewhat close" without explicitly answering whether the teacher's belief is supported or not.
Part 17 (d)

Sampling Methods & Properties

Identifying the sampling technique and stating an advantage [2 Marks]

(i) Name the Method of Sampling [1 Mark]

Opportunity sampling
(also accepted: convenience sampling or opportunistic sampling)

B1 (AO 1.2): Recalls the term 'opportunity', 'convenience', or 'opportunistic'.

Don't confuse with: Systematic sampling (e.g. every 5th person) or Quota sampling (stratifying by categories without a random sampling frame).

(ii) One Advantage of this Sampling Method [1 Mark]

Any one of the following:

  • Easy / simple to carry out
  • Quick / fast to collect data
  • Cheap / inexpensive / low cost
E1 (AO 3.5a): States an acceptable practical advantage (easy, quick, cheap).

Examiner Trap: Do NOT write "convenient"! Repeating the name of the method as its own advantage scores 0 marks.

Topics

Statistics · K: Statistical sampling · L: Data presentation and interpretation · N: Statistical distributions

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.