| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Rescale rate then sum Poissons |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with rate scaling. Part (a) and (b) are direct calculations with given rates adjusted for distance. Part (c) requires recognizing that the sum of independent Poisson variables is Poisson with combined rate, then computing cumulative probabilities. All steps are standard S2 techniques with no novel insight required, making it slightly easier than average. |
| 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 modelling5.02m Poisson: mean = variance = lambda5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X \sim Po(2.8)\) where \(\lambda = 280 \times 0.01 = 2.8\) | B1 | Correct Poisson parameter for 10m |
| \(P(X \leq 5) = 0.9349\) | M1 A1 | M1 for attempting \(P(X \leq 5)\), A1 for correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Y \sim Po(4.4)\) where \(\lambda = 220 \times 0.02 = 4.4\) | B1 | Correct Poisson parameter for 20m |
| \(P(Y = 2) = e^{-4.4} \times \frac{4.4^2}{2!}\) | M1 | Correct Poisson probability formula |
| \(= 0.0738\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T \sim Po(15)\) where \(\lambda = (280+220) \times 0.03 = 15\) | B1 | Correct combined Poisson parameter |
| \(P(12 \leq T \leq 17) = P(T \leq 17) - P(T \leq 11)\) | M1 | Correct probability range attempted |
| \(= 0.8195 - 0.1848\) | A1 | Correct values from tables |
| \(= 0.6347\) | A1 | Correct final answer |
# Question 1:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \sim Po(2.8)$ where $\lambda = 280 \times 0.01 = 2.8$ | B1 | Correct Poisson parameter for 10m |
| $P(X \leq 5) = 0.9349$ | M1 A1 | M1 for attempting $P(X \leq 5)$, A1 for correct answer |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Y \sim Po(4.4)$ where $\lambda = 220 \times 0.02 = 4.4$ | B1 | Correct Poisson parameter for 20m |
| $P(Y = 2) = e^{-4.4} \times \frac{4.4^2}{2!}$ | M1 | Correct Poisson probability formula |
| $= 0.0738$ | A1 | Correct answer |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T \sim Po(15)$ where $\lambda = (280+220) \times 0.03 = 15$ | B1 | Correct combined Poisson parameter |
| $P(12 \leq T \leq 17) = P(T \leq 17) - P(T \leq 11)$ | M1 | Correct probability range attempted |
| $= 0.8195 - 0.1848$ | A1 | Correct values from tables |
| $= 0.6347$ | A1 | Correct final answer |
---1 In a survey of the tideline along a beach, plastic bottles were found at a constant average rate of 280 per kilometre, and drinks cans were found at a constant average rate of 220 per kilometre. It may be assumed that these objects were distributed randomly and independently.
Calculate the probability that:
\begin{enumerate}[label=(\alph*)]
\item a 10 m length of tideline along this beach contains no more than 5 plastic bottles;
\item a 20 m length of tideline along this beach contains exactly 2 drinks cans;
\item a 30 m length of tideline along this beach contains a total of at least 12 but fewer than 18 of these two types of object.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2015 Q1 [9]}}