| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Minimum time or stock level |
| Difficulty | Standard +0.3 This question tests standard Poisson distribution techniques: scaling the rate parameter for different time periods, summing independent Poisson variables, and using cumulative probabilities. All parts are routine applications of textbook methods with no novel problem-solving required, making it slightly easier than average for A-level statistics. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\lambda = 2.7]\ 1 - e^{-2.7}(1 + 2.7 + \frac{2.7^2}{2})\) or \(1 - e^{-2.7}(1 + 2.7 + 3.645)\) or \(1-(0.06721 + 0.1815 + 0.2450)\) | M1 | Any \(\lambda\). Allow one end error. Must see expression |
| \(= 0.506\) (3 sf) | A1 | SC unsupported answer 0.506 scores B1 |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 1.95\) | B1 | |
| \(e^{-1.95}(1 + 1.95 + \frac{1.95^2}{2} + \frac{1.95^3}{3!})\) or \(e^{-1.95}(1 + 1.95 + 1.90125 + 1.2358)\) or \(0.1423 + 0.2774 + 0.2705 + 0.1758\) | M1 | Any \(\lambda\). Allow one end error. Must see expression |
| \(= 0.866\) | A1 | SC unsupported answer 0.866 scores B1B1 |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - e^{-2.1x} \geqslant 0.90\) or \(1 - e^{-\lambda} \geqslant 0.90\) | M1 | OE Condone use of '=' throughout |
| \([e^{-2.1x} \leq 0.1]\) or \(e^{-\lambda} \leq 0.1\); \(-2.1x \leq \ln 0.1\) or \(-\lambda \leq \ln 0.1\) \([\lambda \geq 2.3026, 2.3026/2.1]\) | M1 | Rearrange and attempt take logs of relevant form |
| 1.096 or 10.96 accept 1.097 or 10.97 | *A1 | Seen |
| She must wait for at least 11 minutes | A1 dep | SC Use of trial and improvement: Use of \(1-e^{-\lambda}\) any numerical \(\lambda\) (not 2.1) i.e. one trial M1. Use of enough trials to give answer of 0.90 (2sf) M1. \(\lambda=2.30\) i.e. 3sf accuracy AND 1.09... or 10.9... A1. Then 11 A1 dep |
| Total: 4 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\lambda = 2.7]\ 1 - e^{-2.7}(1 + 2.7 + \frac{2.7^2}{2})$ or $1 - e^{-2.7}(1 + 2.7 + 3.645)$ or $1-(0.06721 + 0.1815 + 0.2450)$ | M1 | Any $\lambda$. Allow one end error. Must see expression |
| $= 0.506$ (3 sf) | A1 | SC unsupported answer 0.506 scores **B1** |
| **Total: 2** | | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 1.95$ | B1 | |
| $e^{-1.95}(1 + 1.95 + \frac{1.95^2}{2} + \frac{1.95^3}{3!})$ or $e^{-1.95}(1 + 1.95 + 1.90125 + 1.2358)$ or $0.1423 + 0.2774 + 0.2705 + 0.1758$ | M1 | Any $\lambda$. Allow one end error. Must see expression |
| $= 0.866$ | A1 | SC unsupported answer 0.866 scores **B1B1** |
| **Total: 3** | | |
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - e^{-2.1x} \geqslant 0.90$ or $1 - e^{-\lambda} \geqslant 0.90$ | M1 | OE Condone use of '=' throughout |
| $[e^{-2.1x} \leq 0.1]$ or $e^{-\lambda} \leq 0.1$; $-2.1x \leq \ln 0.1$ or $-\lambda \leq \ln 0.1$ $[\lambda \geq 2.3026, 2.3026/2.1]$ | M1 | Rearrange and attempt take logs of relevant form |
| 1.096 or 10.96 accept 1.097 or 10.97 | *A1 | Seen |
| She must wait for at least 11 minutes | A1 dep | SC Use of trial and improvement: Use of $1-e^{-\lambda}$ any numerical $\lambda$ (not 2.1) i.e. one trial **M1**. Use of enough trials to give answer of 0.90 (2sf) **M1**. $\lambda=2.30$ i.e. 3sf accuracy AND 1.09... or 10.9... **A1**. Then 11 **A1** dep |
| **Total: 4** | | |6 The numbers of customers arriving at service desks $A$ and $B$ during a 10 -minute period have the independent distributions $\operatorname { Po } ( 1.8 )$ and $\operatorname { Po } ( 2.1 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that during a randomly chosen 15 -minute period more than 2 customers will arrive at $\operatorname { desk } A$.
\item Find the probability that during a randomly chosen 5-minute period the total number of customers arriving at both desks is less than 4 .\\
\includegraphics[max width=\textwidth, alt={}, center]{acd6f1c9-bbaf-40ca-b5cb-8466ddb9f596-08_2720_35_109_2012}
\item An inspector waits at desk $B$. She wants to wait long enough to be $90 \%$ certain of seeing at least one customer arrive at the desk.
Find the minimum time for which she should wait, giving your answer correct to the nearest minute.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q6 [9]}}