| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Scaled time period sums |
| Difficulty | Moderate -0.3 This is a straightforward application of Poisson distribution with clear parameter scaling across different time intervals. Parts (a) and (b) are direct calculator/formula work, while part (c) requires recognizing when normal approximation is appropriate—a standard S2 technique. The multi-part structure and normal approximation elevate it slightly above pure recall, but it remains easier than average as it follows textbook patterns without requiring problem-solving insight. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-4.2} \times \frac{4.2^4}{4!}\) | M1 | P(4), allow any \(\lambda\) |
| \(0.194\) (3 sf) | A1 | As final answer. SC Unsupported correct answer scores B1 only. |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - e^{-8.4}\left(1 + 8.4 + \frac{8.4^2}{2} + \frac{8.4^3}{3!}\right)\) | M1 | Allow M1 with incorrect \(\lambda\). Accept one end error. |
| \(0.968\) (3 sf) | A1 | As final answer. SC Unsupported correct answer scores B1 only. |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(N(50.4,\ 50.4)\) | M1 | SOI |
| \(\frac{39.5 - 50.4}{\sqrt{50.4}}\ [= -1.535]\) | M1 | Allow wrong or no continuity correction. Must have \(\sqrt{\phantom{0}}\) |
| \(\Phi(-1.535) = 1 - \Phi(1.535)\) | M1 | For correct probability area consistent with *their* working. |
| \(0.0624\) (3 sf) or \(0.0623\) | A1 | |
| Total: 4 |
## Question 7:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-4.2} \times \frac{4.2^4}{4!}$ | M1 | P(4), allow any $\lambda$ |
| $0.194$ (3 sf) | A1 | As final answer. **SC** Unsupported correct answer scores B1 only. |
| **Total: 2** | | |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - e^{-8.4}\left(1 + 8.4 + \frac{8.4^2}{2} + \frac{8.4^3}{3!}\right)$ | M1 | Allow M1 with incorrect $\lambda$. Accept one end error. |
| $0.968$ (3 sf) | A1 | As final answer. **SC** Unsupported correct answer scores B1 only. |
| **Total: 2** | | |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $N(50.4,\ 50.4)$ | M1 | SOI |
| $\frac{39.5 - 50.4}{\sqrt{50.4}}\ [= -1.535]$ | M1 | Allow wrong or no continuity correction. Must have $\sqrt{\phantom{0}}$ |
| $\Phi(-1.535) = 1 - \Phi(1.535)$ | M1 | For correct probability area consistent with *their* working. |
| $0.0624$ (3 sf) or $0.0623$ | A1 | |
| **Total: 4** | | |7 Customers arrive at a particular shop at random times. It has been found that the mean number of customers who arrive during a 5 -minute interval is 2.1 .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that exactly 4 customers arrive during a 10 -minute interval.
\item Find the probability that at least 4 customers arrive during a 20 -minute interval.
\item Use a suitable approximating distribution to find the probability that fewer than 40 customers arrive during a 2-hour interval.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q7 [8]}}