June 2025 Paper 3 Q17
17 A maths teacher holds a revision session every Wednesday.
In a random sample of eight Wednesdays, the number of students who attended is listed below.
| 2 | 4 | 3 | 1 | 4 | 5 | 2 | 3 |
(a) Find the mean of these eight values. [1 mark]
(b) Find the variance of these eight values. [1 mark]
(c) The teacher believes that the number of students who attended a revision session each Wednesday can be modelled by the binomial distribution \(\mathrm{B}(30, 0.1)\).
Comment on whether the mean and variance found in parts (a) and (b) support the teacher’s belief.
Fully justify your answer.
[3 marks](d) The teacher wanted to decide if a Saturday morning revision session was worth doing.
The teacher asked the first 10 students who entered the classroom if they would attend a Saturday session.
(i) Name this method of sampling. [1 mark]
(ii) Describe one advantage of this method of sampling. [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Obtains 3 | B1 | 1.1b |
| (1) |
Typical solution
Mean = 3
| Scheme | Marks | AO |
|---|---|---|
| Obtains 1.5 Accept AWRT 1.7 | B1 | 1.1b |
| (1) |
Typical solution
Variance = 1.5
| Scheme | Marks | AO |
|---|---|---|
| Calculates the mean or variance of the given model \(\mathrm{B}(30, 0.1)\) or forms a correct equation in one variable to find \(n\) or \(p\) for binomial model using their mean from part (a) and their variance from part (b) | M1 | 3.1a |
| Obtains the correct mean 3 and variance 2.7 or obtains the correct \(n = 6\) and \(p = 0.5\) for binomial model Accept \(n = 7\) and \(p = 0.43\) for binomial model if using variance 1.7 | A1 | 1.1b |
| Infers that it is not suitable because of different variance or infers teacher’s belief not supported because of different variance or infers that it is not suitable because of different \(n\) and \(p\) from \(n = 30\) and \(p = 0.1\) Must make a single overarching conclusion Must have obtained correct mean from part (a) and correct variance from part (b) To be awarded R1, marks M1A1 must be scored | R1 | 2.2b |
| (3) |
Typical solution
\[\text{Mean} = 30 \times 0.1\]\[= 3\]\[\text{Variance} = 30 \times 0.1 \times 0.9\]\[= 2.7\]The mean is the same but the variance is different, so this is not a suitable model.
| Scheme | Marks | AO |
|---|---|---|
| (i) Recalls ‘opportunity’ or ‘convenience’ or ‘opportunistic’ sampling method | B1 | 1.2 |
| (1) | ||
(ii) States one of the following advantage of sampling method
Do not accept ‘convenient’ | E1 | 3.5a |
| (1) | ||
| (7 marks) |
Typical solution
(d)(i)
Opportunity sampling.
(d)(ii)
Easy to do.