| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | March |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Scaled time period sums |
| Difficulty | Standard +0.3 This is a straightforward Poisson distribution question requiring standard techniques: stating model assumptions (routine recall), scaling the parameter for different time periods, calculating probabilities using tables/calculator, and applying normal approximation. All steps are textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Orders arrive at constant mean rate (must say mean or rate) | Must be in context (accept 25.2 as context) | |
| Orders arrive at random | B1 | Any one reason correctly stated |
| Orders arrive independently | B1 | A second reason correctly stated |
| Orders arrive singly | SC B1: both correct, not in context | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = \frac{3}{8} \times 25.2 = 9.45\) | B1 | |
| \(e^{-9.45}\left(\frac{9.45^3}{3!} + \frac{9.45^4}{4!} + \frac{9.45^5}{5!}\right)\) or \(e^{-9.45}(140.65 + 332.29 + 628.03)\) or \(0.01107 + 0.02615 + 0.04942\) | M1 | Allow any \(\lambda\). Allow end errors. Expression must be seen. |
| \(= 0.0866\) (3 sf) | A1 | If M0 allow SC B1 for 0.0866 no working seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-3.15} \times 3.15\) or \((1 - e^{-3.15}(1 + 3.15))\) or 0.135 or 0.822 (3 sf) | B1 | |
| \(e^{-3.15} \times 3.15 \times (1 - e^{-3.15}(1 + 3.15))\) | M1 | M1 for product of two Poisson probabilities \(P(1) \times (1 - P(0,1))\) (no end errors accepted). Accept any \(\lambda\). |
| \(\times 2\) or \(0.111 \times 2\) | M1 | M1 for their product of two Poisson probabilities (accept end errors) \(\times 2\). Accept any \(\lambda\). |
| \(0.222\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(N(113.4,\ 113.4)\) | B1 | SOI |
| \(\frac{120.5 - 113.4}{\sqrt{113.4}} = 0.667\) | M1 | Standardise with their values. Allow wrong or no cc. Must have \(\sqrt{\phantom{x}}\). |
| \(1 - \phi(\text{their } 0.667)\) | M1 | For probability area consistent with their values. |
| \(= 0.252\) (3 sf) | A1 |
## Question 2:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Orders arrive at constant mean rate (must say mean or rate) | | Must be in context (accept 25.2 as context) |
| Orders arrive at random | **B1** | Any one reason correctly stated |
| Orders arrive independently | **B1** | A second reason correctly stated |
| Orders arrive singly | | **SC B1**: both correct, not in context |
| | **[2]** | |
## Question 2(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = \frac{3}{8} \times 25.2 = 9.45$ | B1 | |
| $e^{-9.45}\left(\frac{9.45^3}{3!} + \frac{9.45^4}{4!} + \frac{9.45^5}{5!}\right)$ or $e^{-9.45}(140.65 + 332.29 + 628.03)$ or $0.01107 + 0.02615 + 0.04942$ | M1 | Allow any $\lambda$. Allow end errors. Expression must be seen. |
| $= 0.0866$ (3 sf) | A1 | If M0 allow SC B1 for 0.0866 no working seen. |
## Question 2(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-3.15} \times 3.15$ or $(1 - e^{-3.15}(1 + 3.15))$ or 0.135 or 0.822 (3 sf) | B1 | |
| $e^{-3.15} \times 3.15 \times (1 - e^{-3.15}(1 + 3.15))$ | M1 | M1 for product of two Poisson probabilities $P(1) \times (1 - P(0,1))$ (no end errors accepted). Accept any $\lambda$. |
| $\times 2$ or $0.111 \times 2$ | M1 | M1 for their product of two Poisson probabilities (accept end errors) $\times 2$. Accept any $\lambda$. |
| $0.222$ (3 sf) | A1 | |
## Question 2(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $N(113.4,\ 113.4)$ | B1 | SOI |
| $\frac{120.5 - 113.4}{\sqrt{113.4}} = 0.667$ | M1 | Standardise with their values. Allow wrong or no cc. Must have $\sqrt{\phantom{x}}$. |
| $1 - \phi(\text{their } 0.667)$ | M1 | For probability area consistent with their values. |
| $= 0.252$ (3 sf) | A1 | |
---2 The number of orders arriving at a shop during an 8-hour working day is modelled by the random variable $X$ with distribution $\operatorname { Po } ( 25.2 )$.
\begin{enumerate}[label=(\alph*)]
\item State two assumptions that are required for the Poisson model to be valid in this context.
\item \begin{enumerate}[label=(\roman*)]
\item Find the probability that the number of orders that arrive in a randomly chosen 3-hour period is between 3 and 5 inclusive.
\item Find the probability that, in two randomly chosen 1 -hour periods, exactly 1 order will arrive in one of the 1 -hour periods, and at least 2 orders will arrive in the other 1 -hour period. [4]
\end{enumerate}\item The shop can only deal with a maximum of 120 orders during any 36-hour period.
Use a suitable approximating distribution to find the probability that, in a randomly chosen 36-hour period, there will be too many orders for the shop to deal with.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q2 [13]}}