| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Moderate -0.8 This is a straightforward Poisson distribution question requiring only standard recall and routine calculations: stating a textbook condition, computing probabilities using given parameters (λ=1.2×12=14.4), and applying the normal approximation for large λ (λ=168). No problem-solving insight or novel reasoning is required—all techniques are standard S2 procedures. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Constant average rate | B1 | OE. Accept constant rate. Allow without context. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 14.4\) | B1 | |
| \(e^{-14.4}\!\left(\frac{14.4^{13}}{13!} + \frac{14.4^{14}}{14!} + \frac{14.4^{15}}{15!}\right)\) or \(e^{-14.4}(183837 + 189089 + 181526)\) or \((0.102469 + 0.105396 + 0.101181)\) | M1 | Poisson \(P(13, 14, 15)\). Expression must be seen. Allow one end error; allow any \(\lambda\). Allow fully correct sigma notation. |
| \(= 0.309\) (3sf) | A1 | SC: \(0.309\) with no working scores B1 B1. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(N(140 \times 1.2,\ 140 \times 1.2)\) or \(N(168, 168)\) | B1 | Stated or implied. |
| \(\frac{149.5 - 168}{\sqrt{168}}\ [= -1.427]\) | M1 | Standardising using their mean and variance. Allow with wrong or no continuity correction. |
| \(\Phi(\text{"}{-1.427}\text{"}) = 1 - \Phi(\text{"}{1.427}\text{"})\) | M1 | For area consistent with their working. |
| \(= 0.0768\) or \(0.0767\) (3sf) | A1 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Constant average rate | B1 | OE. Accept constant rate. Allow without context. |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 14.4$ | B1 | |
| $e^{-14.4}\!\left(\frac{14.4^{13}}{13!} + \frac{14.4^{14}}{14!} + \frac{14.4^{15}}{15!}\right)$ or $e^{-14.4}(183837 + 189089 + 181526)$ or $(0.102469 + 0.105396 + 0.101181)$ | M1 | Poisson $P(13, 14, 15)$. Expression must be seen. Allow one end error; allow any $\lambda$. Allow fully correct sigma notation. |
| $= 0.309$ (3sf) | A1 | SC: $0.309$ with no working scores **B1 B1**. |
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $N(140 \times 1.2,\ 140 \times 1.2)$ or $N(168, 168)$ | B1 | Stated or implied. |
| $\frac{149.5 - 168}{\sqrt{168}}\ [= -1.427]$ | M1 | Standardising using their mean and variance. Allow with wrong or no continuity correction. |
| $\Phi(\text{"}{-1.427}\text{"}) = 1 - \Phi(\text{"}{1.427}\text{"})$ | M1 | For area consistent with their working. |
| $= 0.0768$ or $0.0767$ (3sf) | A1 | |5 Sales of cell phones at a certain shop occur singly, randomly and independently.
\begin{enumerate}[label=(\alph*)]
\item State one further condition that must be satisfied for the number of sales in a certain time period to be well modelled by a Poisson distribution.\\
The average number of sales per hour is 1.2 .\\
Assume now that a Poisson distribution is a suitable model.
\item Find the probability that the number of sales during a randomly chosen 12 -hour period will be more than 12 and less than 16 .
\item Use a suitable approximating distribution to find the probability that the number of sales during a randomly chosen 1-month period (140 hours) will be less than 150 .
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q5 [8]}}