| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard parameter scaling (part a), followed by a routine two-stage Poisson problem (part b), and a normal approximation to Poisson (part c). All techniques are standard S2 bookwork with clear signposting of methods required. Slightly above average only due to the multi-part structure and need to recognize when approximation is appropriate. |
| Spec | 5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(X = \text{Number of items of litter found in a 2m}^2 \text{ area of the beach}\) So \(X \sim \text{Po}(8)\) | M1 |
| \(P(X=5) = \frac{e^{-8} \times 8^5}{5!}\) or \(P(X_r, 5) - P(X_r, 4) = 0.1912 - 0.0996\) | M1 | |
| \(= 0.09160...\) awrt 0.0916 | A1 | (2) |
| (b) | \(Y = \text{Number of face masks found in a 5m}^2 \text{ area of the beach}\) | M1 |
| \(Y \sim \text{Po}(6)\) or \(P(Y ..5) = 1 - P(Y_r, 4) = 1 - 0.2851\) | M1 | |
| \(= 0.7149\) awrt 0.715 | A1 | (2) |
| (c) | \(W = \text{Number of items of litter that are not face masks found in a 20m}^2 \text{ area of the beach}\) | M1 |
| \(W \sim N(56, 56)\) | M1 | |
| \(P(W < 60) = P\left(Z < \frac{59.5-56}{\sqrt{56}}\right)\) | M1 M1 | |
| Tables \([=P(Z < 0.47)] = 0.6808\) calculator 0.68000... awrt 0.68 | A1 | (4) |
| Total | 8 |
| (a) | $X = \text{Number of items of litter found in a 2m}^2 \text{ area of the beach}$ So $X \sim \text{Po}(8)$ | M1 | |
|---|---|---|---|
| | $P(X=5) = \frac{e^{-8} \times 8^5}{5!}$ or $P(X_r, 5) - P(X_r, 4) = 0.1912 - 0.0996$ | M1 | |
| | $= 0.09160...$ awrt 0.0916 | A1 | (2) |
| (b) | $Y = \text{Number of face masks found in a 5m}^2 \text{ area of the beach}$ | M1 | |
| | $Y \sim \text{Po}(6)$ or $P(Y ..5) = 1 - P(Y_r, 4) = 1 - 0.2851$ | M1 | |
| | $= 0.7149$ awrt 0.715 | A1 | (2) |
| (c) | $W = \text{Number of items of litter that are not face masks found in a 20m}^2 \text{ area of the beach}$ | M1 | |
| | $W \sim N(56, 56)$ | M1 | |
| | $P(W < 60) = P\left(Z < \frac{59.5-56}{\sqrt{56}}\right)$ | M1 M1 | |
| | Tables $[=P(Z < 0.47)] = 0.6808$ calculator 0.68000... awrt 0.68 | A1 | (4) |
| | **Total** | **8** | |
**Notes for Question 1:**
- **(a) M1:** for use of $\frac{e^{-\lambda}\lambda^5}{5!}$ or $P(X_r, 5) - P(X_r, 4)$
- **(a) A1:** awrt 0.0916 (correct answer scores 2 out of 2) for writing or using Po(6)
- **(b) M1:** or for a correct probability statement $1 - P(Y_r, 4)$ or $P(Y ..5)$; e.g. $P(Y ..5) = 1 - P(Y_r, 5)$ is M0
- **(b) A1:** awrt 0.715 (correct answer scores 2 out of 2) for writing or using N(56, 56) may be seen in standardisation
- **(c) M1:** (may be implied by the standardisation $\frac{x-56}{\sqrt{56}}$)
- **(c) M1:** standardising with 59.5/60/60.5, their mean and their standard deviation
- **(c) M1:** using a continuity correction $60 + 0.5 = [60.5]$ or $60 - 0.5 = [59.5]$
- **(c) A1:** awrt 0.68 (NB Use of exact Poisson gives 0.68617...and scores 0 out of 4)
---\begin{enumerate}
\item During an annual beach-clean, the people doing the clean are asked to conduct a litter survey.\\
At a particular beach-clean, litter was found at a rate of 4 items per square metre.\\
(a) Find the probability that, in a randomly selected area of 2 square metres on this beach, exactly 5 items of litter were found.
\end{enumerate}
Of the litter found on the beach, 30\% of the items were face masks.\\
(b) Find the probability that, in a randomly selected area of 5 square metres on this beach, more than 4 face masks were found.\\
(c) Using a suitable approximation, find the probability that, in a randomly selected area of 20 square metres on this beach, less than 60 items of litter were found that were not face masks.
\hfill \mbox{\textit{Edexcel S2 2024 Q1}}