June 2025 Paper 2 Q11
11 A train company claims that a particular daily service arrives on time on 95% of days. Riley suspects that the true percentage is less than 95%. Riley tests the company’s claim by choosing a random sample of 50 days during the period November 2023 to March 2024. Riley finds that the service arrived on time on 44 of these days.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0: p = 0.95\) where \(p\) is the proportion of services that arrive on time \(\mathrm{H}_1: p \lt 0.95\) | B1 B1 | 1.1 2.5 |
| \(X \sim \mathrm{B}(50, 0.95)\) and \(X = 44\) | M1 | 3.3 |
| \(\mathrm{P}(X \leqslant 44) = 0.0378\) | A1 | 3.4 |
| \(0.0378 \lt 0.05\) | A1FT | 1.1 |
| Reject \(\mathrm{H}_0\) | M1 | 1.1 |
| There is sufficient evidence to suggest that the proportion of services arriving on time is less than 0.95 | A1 | 2.2b |
| [7] |
Notes
B1B1: Allow probability or percentage. Allow use of any letter as long as the parameter is defined correctly e.g. allow ‘\(p = 95\%\) where \(p\) is the % of services that arrive on time’ or ‘\(p = \mathrm{P}\)(service arrives on time)’
Deduct one B mark for each error as follows:
- 2-tail B1B0
- undefined \(p\) B1B0
- missing 0.95 or 95% B0B0.
- eg \(\mathrm{H}_0 = 0.95\) etc B0B0
M1: Correct distribution and value of \(X\), both may be stated or implied by an appropriate probability value e.g. by [\(\mathrm{P}(X \leqslant 44) =\) or \(\mathrm{P}(X \lt 45) =\)] 0.0378 or [\(\mathrm{P}(X \leqslant 43) =\) or \(\mathrm{P}(X \lt 44) =\)] 0.0118 or 0.962 or 0.988. Allow M1 for correct distribution and use of \(X = 44\) even if within an incorrect statement (e.g. \(\mathrm{P}(X = 44) = 0.026\))
A1: This whole statement must be correct (i.e. do not accept \(\mathrm{P}(X \lt 44)\))
Accept 0.038 (2sf) or 0.0378 to 3sf or 0.0377 ... rot
Accept \(\mathrm{P}(X \gt 44) = 0.96(22)\) or \(\mathrm{P}(X \lt 45) = 0.0378\) etc.
A1FT: FT their probability - correct comparison for their value e.g. \(0.962 \gt 0.95\) FT hypotheses (e.g. 2-tail 0.025)
M1: Dependent on \(\mathrm{P}(X \leqslant 44)\) or \(\mathrm{P}(X \lt 44)\) soi (i.e. 0.0378 or 0.0118 or 0.962 or 0.988). Condone ‘Accept H1’
A1: Conclusion must be in context, not definite and consistent with their comparison.
Dependent on all previous M and A marks, except the first A1 (i.e. this mark can be given if 0.0378 value is given incorrectly)
- Allow percentage or probability instead of proportion
- Allow 95% instead of 0.95
- Allow ‘sufficient evidence to suggest that Riley’s claim is correct’
- Condone ‘significant’ for ‘sufficient’ and ‘it is likely that…’
- Do not accept ‘Show’ or ‘Prove’ A0
| Scheme | Marks | AO |
|---|---|---|
| E.g.: The proportion of services arriving on time may be different [at a different time of year, or may depend on the weather] so \(p\) may not be constant | B1 | 3.5b |
| [1] |
Notes
B1: Must refer to one of the assumptions for using a binomial distribution i.e.
- The probability/proportion may not be constant
- The events may not be independent
- There may not be a fixed number of trials
- The trials may not have only 2 outcomes
And explain correctly in context by referring to e.g.
- Trains
- Service
- On time
But candidates need not suggest a ‘mechanism’ for how this could come about (i.e. ‘leaves on the line’ or ‘staff shortage’ etc.).
So accept e.g.:
- “the proportion/probability of services arriving on time may not be constant”
- “the event that a service arrives on time may not be independent”
- “the service may be delayed for related reasons on consecutive [or connected] days so the events are not independent”
- “the train service may not run on all 50 days sampled”
- “the train service may be cancelled [meaning it is neither on time nor delayed] and so there is a third outcome”
Do not accept statements about successive stations or successive trains (the question is about the same train service on successive days).
Do not accept just ‘the trains are not independent’ – must refer to ‘train times’ or ‘arrival on time’ oe.
Do not accept ‘equal chance’ without a reference to probability or proportion.