A2 October 2021 Q2
2. On a weekday, a garage receives telephone calls randomly, at a mean rate of 1.25 per 10 minutes.
The manager of the garage randomly selects 150 non-overlapping 30-minute periods on weekdays.
She records the number of calls received in each of these 30-minute periods.
The manager of the garage decides to test whether the number of calls received on a Saturday is different from the number of calls received on a weekday. She selects a Saturday at random and records the number of telephone calls received by the garage in the first 4 hours.
The manager found that there had been 40 telephone calls received by the garage in the first 4 hours.
| Scheme | Marks | AO |
|---|---|---|
| \(C \sim \mathrm{Poisson}(3.75)\) | M1 | 3.3 |
| \(\mathrm{P}(C \geqslant 2) = 0.88829\ldots\)*. awrt 0.8883* | A1*cso | 1.1b |
| (2) |
Notes
M1: For calculating the mean and setting up the correct model. Poisson may be implied by 0.8883 or better or 1 – awrt 0.1117 but must see 3.75 or \(1.25 \times 3\)
A1*cso: \(\mathrm{P}(C \geqslant 2) =\) awrt 0.8883 or 1 – awrt 0.1117 = 0.888 Must see \(\mathrm{P}(C \geqslant 2)\) oe
| Scheme | Marks | AO |
|---|---|---|
| \(D \sim \mathrm{B}(6, \text{“}0.888\text{”})\) | M1 | 3.3 |
| \(\mathrm{P}(D \leqslant 3) = 0.02163\ldots\) awrt 0.0216 / 0.0215 | A1 | 1.1b |
| (2) |
Notes
M1: Setting up a new model using their answer to (a) Implied by correct answer
A1: awrt 0.0216 or awrt 0.0215
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(C = 8) = 0.02281\ldots\) | B1 | 1.1b |
| \(E \sim \mathrm{B}(150, \text{“}0.02281..\text{”}) \ \Rightarrow\ \text{mean} = 150 \times \text{“}0.02281\ldots\text{”}\ [= 3.4215\ldots]\) | M1 | 3.3 |
| \(E \sim \mathrm{Po}(\text{“}3.4215\ldots\text{”}) \ \Rightarrow\ \mathrm{P}(E \geqslant 3) = [1 - \mathrm{P}(E \leqslant 2)]\) | M1 | 3.4 |
| \(= 0.664\)* | A1*cso | 2.1 |
| (4) |
Notes
B1: awrt 0.0228
M1: Setting up a new model \(\mathrm{B}(150, \text{“}0.0228\text{”})\) and using \(np\) (working seen if incorrect)
M1: Using the model \(\mathrm{Po}(\text{their } np)\) Must be clearly stated and \(\mathrm{P}(E \geqslant 3)\) oe seen
A1*cso: Only award if the previous 3 marks have been awarded and 0.664 is stated.
NB Use of \(\mathrm{B}(150\ 0.02281)\) gives 0.668
| Scheme | Marks | AO |
|---|---|---|
| The number of periods is large and the probability of receiving 8 calls in 30-minutes is small. | B1 | 2.4 |
| (1) |
Notes
B1: Idea that \(n = 150\) (number of periods selected) is large and \(p\) is 0.022… (exactly 8 calls in the time period) is small.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0: \lambda = 30 \qquad \mathrm{H}_1: \lambda \neq 30\) | B1 | 2.5 |
| (1) |
Notes
B1: Both hypotheses correct using \(\lambda\) or \(\mu\) allow 1.25 or 3.75
| Scheme | Marks | AO |
|---|---|---|
| \(X \sim \mathrm{Po}(30)\) | B1 | 3.3 |
| \(\mathrm{P}(X \geqslant 40) = 1 - \mathrm{P}(X \leqslant 39)\) | M1 | 1.1b |
| \(= 0.04625\ldots\) | A1 | 1.1b |
| \(0.046\ldots \gt 0.025\) or no evidence to reject \(\mathrm{H}_0\) There is insufficient evidence at the 5% level of significance that the number of calls received is different on a Saturday | A1 | 2.2b |
| (4) | ||
| (14 marks) |
Notes
B1: Realising \(\mathrm{Po}(30)\) needs to be used. NB Implied by correct answer or \(\mathrm{P}(X = 40) = 0.0139\ldots\)
M1: Writing or using \(1 - \mathrm{P}(X \leqslant 39)\) or if CR method for \(\mathrm{P}(X \geqslant 42) = 0.0221\ldots\)
A1: 0.04… or awrt 0.05 or CR \(X \geqslant 42\) oe must be CR and not probability
A1: A fully correct solution and correct inference in context. Calls required
If put this prob but then give Cr X >= 40 M1A1A0