| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Hypothesis test on Poisson rate |
| Difficulty | Standard +0.3 This is a standard S2 Poisson question covering routine applications: basic probability calculations, scaling the parameter for different time periods, combining Poisson with binomial, and a straightforward hypothesis test. Part (d) requires conceptual understanding but is a common exam question. Slightly above average difficulty due to the multi-step nature and the binomial-Poisson combination in part (b), but all techniques are standard textbook exercises requiring no novel insight. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(M \sim \text{Po}(2)\); \(P(M \leqslant 3) = 0.8571\) | B1 | awrt \(0.857\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(M \geqslant 6) = 1 - P(M \leqslant 5)\) | M1 | for \(1 - P(M \leqslant 5)\) or \(1 - 0.9834\); do not allow \(1 - P(M < 6)\) unless \(1 - P(M \leqslant 5)\) is used |
| \(= 0.0166\) | A1 | awrt \(0.0166\); correct answer scores 2 out of 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(Q \sim \text{Po}(6)\) | M1 | for writing or using \(\text{Po}(6)\) |
| \(P(4 \leqslant Q \leqslant 7) = P(Q \leqslant 7) - P(Q \leqslant 3) = 0.5928\) | M1 | for \(P(Q \leqslant 7) - P(Q \leqslant 3)\) or \(0.7440 - 0.1512\) or awrt \(0.593\) |
| \(X \sim \text{B}(20,\ ``0.5928\text{''})\) and \(P(X = 15)\) or \(^{20}C_{15}(``0.5928\text{''})^{15}(1 - ``0.5928\text{''})^{5}\) | M1 | implied by awrt \(0.068\) |
| \(= 0.06809\ldots\) | A1 | awrt \(0.068\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \lambda = 2\), \(H_1: \lambda > 2\) | B1 | for correct hypotheses in terms of \(\lambda\) or \(\mu\); allow 5 instead of 2; must be attached to \(H_0\) and \(H_1\) correctly |
| \(R \sim \text{Po}(5)\): \(P(R \geqslant 10)\) or \(1 - P(R \leqslant 9)\) | M1 | for writing or using \(\text{Po}(5)\) and \(P(R \geqslant 10)\) or \(1 - P(R \leqslant 9)\); may be implied by awrt \(0.0318\) or correct CR |
| \(= 0.0318\) or CR: \(R \geqslant 10\) | A1 | awrt \(0.0318\); allow CR: \([R] \geqslant 10\); allow any letter or no letter for CR |
| Reject \(H_0\) or Significant or in the critical region | M1 | for a correct ft statement consistent with their \(p\)-value and \(0.05\) or with \(10\) and their CR; need not be contextual but there must be no contradicting non-contextual comments |
| There is evidence to support the manager's belief / rate of customers arriving at the garage has increased | A1 | dep on 1st and 2nd M1 for a correct conclusion in context which must be rejecting \(H_0\); must use bold words (oe) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The number of tyres bought is likely not to occur singly / tyres are not sold independently | B1 | for the idea that tyres may be bought in e.g. pairs oe / the idea that tyre sales are not independent |
# Question 1:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $M \sim \text{Po}(2)$; $P(M \leqslant 3) = 0.8571$ | B1 | awrt $0.857$ |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(M \geqslant 6) = 1 - P(M \leqslant 5)$ | M1 | for $1 - P(M \leqslant 5)$ or $1 - 0.9834$; do not allow $1 - P(M < 6)$ unless $1 - P(M \leqslant 5)$ is used |
| $= 0.0166$ | A1 | awrt $0.0166$; correct answer scores 2 out of 2 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $Q \sim \text{Po}(6)$ | M1 | for writing or using $\text{Po}(6)$ |
| $P(4 \leqslant Q \leqslant 7) = P(Q \leqslant 7) - P(Q \leqslant 3) = 0.5928$ | M1 | for $P(Q \leqslant 7) - P(Q \leqslant 3)$ or $0.7440 - 0.1512$ or awrt $0.593$ |
| $X \sim \text{B}(20,\ ``0.5928\text{''})$ and $P(X = 15)$ or $^{20}C_{15}(``0.5928\text{''})^{15}(1 - ``0.5928\text{''})^{5}$ | M1 | implied by awrt $0.068$ |
| $= 0.06809\ldots$ | A1 | awrt $0.068$ |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \lambda = 2$, $H_1: \lambda > 2$ | B1 | for correct hypotheses in terms of $\lambda$ or $\mu$; allow 5 instead of 2; must be attached to $H_0$ and $H_1$ correctly |
| $R \sim \text{Po}(5)$: $P(R \geqslant 10)$ or $1 - P(R \leqslant 9)$ | M1 | for writing or using $\text{Po}(5)$ **and** $P(R \geqslant 10)$ or $1 - P(R \leqslant 9)$; may be implied by awrt $0.0318$ or correct CR |
| $= 0.0318$ or CR: $R \geqslant 10$ | A1 | awrt $0.0318$; allow CR: $[R] \geqslant 10$; allow any letter or no letter for CR |
| Reject $H_0$ or Significant or in the critical region | M1 | for a correct ft statement consistent with their $p$-value and $0.05$ or with $10$ and their CR; need not be contextual but there must be no contradicting non-contextual comments |
| There is evidence to support the **manager's belief** / rate of **customers** arriving at the **garage** has **increased** | A1 | dep on 1st and 2nd M1 for a correct conclusion in context which must be rejecting $H_0$; must use bold words (oe) |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| The number of tyres bought is likely not to occur singly / tyres are not sold independently | B1 | for the idea that tyres may be bought in e.g. pairs oe / the idea that tyre sales are not independent |1 A garage sells tyres. The number of customers arriving at the garage to buy tyres in a 10-minute period is modelled by a Poisson distribution with mean 2
\begin{enumerate}[label=(\alph*)]
\item Find the probability that
\begin{enumerate}[label=(\roman*)]
\item fewer than 4 customers arrive to buy tyres in the next 10 minutes,
\item more than 5 customers arrive to buy tyres in the next 10 minutes.
The manager randomly selects 20 non-overlapping, 30-minute periods.
\end{enumerate}\item Find the probability that there are between 4 and 7 (inclusive) customers arriving to buy tyres in exactly 15 of these 30-minute periods.
The manager believes that placing an advert in the local paper will lead to a significant increase in the number of customers arriving at the garage.\\
A week after the advert is placed, the manager randomly selects a 25 -minute period and finds that 10 customers arrive at the garage to buy tyres.
\item Test, at the $5 \%$ level of significance, whether or not there is evidence to support the manager's belief.\\
State your hypotheses clearly.
\item Explain why the Poisson distribution is unlikely to be valid for the number of tyres sold during a 10-minute period.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2024 Q1 [13]}}