| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson parameter from given probability |
| Difficulty | Standard +0.3 This is a straightforward Poisson distribution question requiring standard techniques: stating the range (trivial), solving for λ using the probability formula (algebraic manipulation of a simple equation), calculating probabilities from tables, and using normal approximation for large μ (standard S2 technique). All parts are routine applications 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 probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks |
|---|---|
| \(0, 1, 2, 3, \ldots\) | B1 |
| Answer | Marks |
|---|---|
| \(3\) | B1 |
| Answer | Marks |
|---|---|
| \(e^{-3}\left(\frac{3^4}{4!} + \frac{3^5}{5!} + \frac{3^6}{6!}\right)\) | M1 |
| \(0.319\) (3 sf) | A1 |
| Answer | Marks |
|---|---|
| \(\Phi^{-1}(0.0668) = -1.500\) | M1 |
| \(N(\mu, \mu)\) | M1 |
| \(\frac{45.5 - \mu}{\sqrt{\mu}} = \text{"-1.500"}\) | M1 |
| \(\mu - \text{"1.500"}\sqrt{\mu} - 45.5 = 0\); \(\sqrt{\mu} = \frac{\text{"1.5"} \pm \sqrt{(-1.5)^2 + 4 \times 45.5}}{2} (= 7.5369)\) | M1 |
| \(\mu = 56.8\) (3 sf) | A1 |
## Question 5:
**Part 5(a)(i):**
| $0, 1, 2, 3, \ldots$ | B1 | |
**Part 5(a)(ii):**
| $3$ | B1 | |
**Part 5(a)(iii):**
| $e^{-3}\left(\frac{3^4}{4!} + \frac{3^5}{5!} + \frac{3^6}{6!}\right)$ | M1 | |
| $0.319$ (3 sf) | A1 | |
**Part 5(b):**
| $\Phi^{-1}(0.0668) = -1.500$ | M1 | |
| $N(\mu, \mu)$ | M1 | |
| $\frac{45.5 - \mu}{\sqrt{\mu}} = \text{"-1.500"}$ | M1 | |
| $\mu - \text{"1.500"}\sqrt{\mu} - 45.5 = 0$; $\sqrt{\mu} = \frac{\text{"1.5"} \pm \sqrt{(-1.5)^2 + 4 \times 45.5}}{2} (= 7.5369)$ | M1 | |
| $\mu = 56.8$ (3 sf) | A1 | |
---5
\begin{enumerate}[label=(\alph*)]
\item The random variable $X$ has the distribution $\operatorname { Po } ( \lambda )$.
\begin{enumerate}[label=(\roman*)]
\item State the values that $X$ can take.\\
It is given that $\mathrm { P } ( X = 1 ) = 3 \times \mathrm { P } ( X = 0 )$.
\item Find $\lambda$.
\item Find $\mathrm { P } ( 4 \leqslant X \leqslant 6 )$.
\end{enumerate}\item The random variable $Y$ has the distribution $\operatorname { Po } ( \mu )$ where $\mu$ is large. Using a suitable approximating distribution, it is found that $\mathrm { P } ( Y < 46 ) = 0.0668$, correct to 4 decimal places.
Find $\mu$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q5 [9]}}