| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Moderate -0.3 This is a straightforward multi-part Poisson question requiring standard techniques: scaling the parameter for different time periods, calculating P(X>4) using complement rule, and applying normal approximation with continuity correction. All steps are routine applications of textbook methods with no novel problem-solving required, making it slightly easier than average. |
| 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 | Marks | Guidance |
| \(\lambda = 6.6\) | B1 | |
| \(e^{-6.6} \times \dfrac{6.6^6}{6!}\) | M1 | Any \(\lambda\) |
| 0.156 (3 s.f.) | A1 | If M0 awarded SC B1 for 0.156 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - e^{-2.2}\!\left(1 + 2.2 + \dfrac{2.2^2}{2} + \dfrac{2.2^3}{3!} + \dfrac{2.2^4}{4!}\right)\) | M1 | Allow one end error; need \(1 - \ldots\); any \(\lambda\) |
| 0.0725 (3 s.f.) | A1 | If M0 awarded SC B1 for 0.0725 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(N(26.4,\ 26.4)\) | B1 | Give at early stage \(2.2 \times 12\) |
| \(\dfrac{19.5 - \text{'26.4'}}{\sqrt{\text{'26.4'}}}\) \([= -1.343]\) | M1 | Standardising with *their* values; allow wrong or no continuity correction |
| \(\phi(-1.343) = 1 - \phi(\text{'1.343'})\) | M1 | Area consistent with *their* working |
| 0.0897 or 0.0896 (3 s.f.) | A1 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = 6.6$ | B1 | |
| $e^{-6.6} \times \dfrac{6.6^6}{6!}$ | M1 | Any $\lambda$ |
| 0.156 (3 s.f.) | A1 | If M0 awarded SC B1 for 0.156 |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - e^{-2.2}\!\left(1 + 2.2 + \dfrac{2.2^2}{2} + \dfrac{2.2^3}{3!} + \dfrac{2.2^4}{4!}\right)$ | M1 | Allow one end error; need $1 - \ldots$; any $\lambda$ |
| 0.0725 (3 s.f.) | A1 | If M0 awarded SC B1 for 0.0725 |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $N(26.4,\ 26.4)$ | B1 | Give at early stage $2.2 \times 12$ |
| $\dfrac{19.5 - \text{'26.4'}}{\sqrt{\text{'26.4'}}}$ $[= -1.343]$ | M1 | Standardising with *their* values; allow wrong or no continuity correction |
| $\phi(-1.343) = 1 - \phi(\text{'1.343'})$ | M1 | Area consistent with *their* working |
| 0.0897 or 0.0896 (3 s.f.) | A1 | |
---5 The number of clients who arrive at an information desk has a Poisson distribution with mean 2.2 per 5-minute period.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that, in a randomly chosen 15 -minute period, exactly 6 clients arrive at the desk.
\item If more than 4 clients arrive during a 5 -minute period, they cannot all be served.
Find the probability that, during a randomly chosen 5 -minute period, not all the clients who arrive at the desk can be served.
\item Use a suitable approximating distribution to find the probability that, during a randomly chosen 1-hour period, fewer than 20 clients arrive at the desk.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q5 [9]}}