| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Multiple independent time periods |
| Difficulty | Moderate -0.8 This is a straightforward Poisson distribution question requiring only standard recall and routine calculations. Part (a) asks for a textbook condition (independence), parts (b)-(d) involve direct application of Poisson probability formulas with simple parameter scaling and binomial combination. No problem-solving insight or complex multi-step reasoning required. |
| Spec | 2.04c Calculate binomial probabilities5.02i 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 |
| Customers arrive independently or singly or at random | B1 | Any one of these in context |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-2.3} \times \frac{2.3^3}{3!}\) | M1 | Attempt correct expression seen |
| \(= 0.203\) (3sf) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Po}(4.6)\) | B1 | SOI |
| \(1 - e^{-4.6}\left(1 + 4.6 + \frac{4.6^2}{2!} + \frac{4.6^3}{3!}\right)\) | M1 | Correct expression, with any \(\lambda\), allow one end error |
| \(= 0.674\) (3sf) | A1 | As final answer |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{none arrive}) = e^{-2.3}\) \((= 0.10026)\) | M1 | Must be clearly *their* P(none arrive) |
| \(^5C_2(e^{-2.3})^2(1 - e^{-2.3})^3\) | M1 | FT *their* \(e^{-2.3}\) |
| \(= 0.0732\) or \(0.0733\) (3sf) | A1 | |
| Total: 3 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Customers arrive independently or singly or at random | B1 | Any one of these in context |
| Total: 1 | | |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-2.3} \times \frac{2.3^3}{3!}$ | M1 | Attempt correct expression seen |
| $= 0.203$ (3sf) | A1 | |
| Total: 2 | | |
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Po}(4.6)$ | B1 | SOI |
| $1 - e^{-4.6}\left(1 + 4.6 + \frac{4.6^2}{2!} + \frac{4.6^3}{3!}\right)$ | M1 | Correct expression, with any $\lambda$, allow one end error |
| $= 0.674$ (3sf) | A1 | As final answer |
| Total: 3 | | |
## Question 5(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{none arrive}) = e^{-2.3}$ $(= 0.10026)$ | M1 | Must be clearly *their* P(none arrive) |
| $^5C_2(e^{-2.3})^2(1 - e^{-2.3})^3$ | M1 | FT *their* $e^{-2.3}$ |
| $= 0.0732$ or $0.0733$ (3sf) | A1 | |
| Total: 3 | | |5 Customers arrive at a shop at a constant average rate of 2.3 per minute.
\begin{enumerate}[label=(\alph*)]
\item State another condition for the number of customers arriving per minute to have a Poisson distribution.\\
It is now given that the number of customers arriving per minute has the distribution $\mathrm { Po } ( 2.3 )$.
\item Find the probability that exactly 3 customers arrive during a 1 -minute period.
\item Find the probability that more than 3 customers arrive during a 2 -minute period.
\item Five 1-minute periods are chosen at random. Find the probability that no customers arrive during exactly 2 of these 5 periods.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q5 [9]}}