| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| 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 testing standard techniques: cumulative probability calculations, sum of independent Poisson variables, normal approximation, and finding a parameter from equal probabilities. Part (b) requires solving e^(-λ)λ³/3! = e^(-λ)λ⁵/5!, which simplifies to a basic equation. All parts are routine applications of well-practiced methods 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 probabilities5.02n Sum of Poisson variables: is Poisson5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-2.3}\left(\frac{2.3^2}{2} + \frac{2.3^3}{3!} + \frac{2.3^4}{4!}\right)\) | M1 | Allow one end error |
| \(= 0.585\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((\lambda) = 4.6\) | B1 | |
| \(1 - e^{-4.6}\left(1 + 4.6 + \frac{4.6^2}{2}\right)\) | M1 | any \(\lambda\), Allow one end error |
| \(= 0.837\) (3 sf) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(S \sim N(115, 115)\) | B1 | May be implied |
| \(\frac{110.5 - 115}{\sqrt{115}}\) \((= -0.420)\) | M1 | Allow with wrong or no cc OR no \(\sqrt{\phantom{x}}\) |
| \(1 - \Phi('0.420')\) \((= 1 - 0.663)\) | M1 | |
| \(= 0.337\) | A1 | Accept alternative method using \(N(2.3, 2.3)\) no mixed methods |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-\lambda} \times \frac{\lambda^3}{3!} = e^{-\lambda} \times \frac{\lambda^5}{5!}\) | M1 | |
| \(\lambda^3 = \frac{\lambda^5}{4\times5}\) or \(\lambda^2 = 20\) oe | A1 | any correct simplification without \(e^{-\lambda}\) or \(!\) |
| \(\lambda = \sqrt{20}\) or \(2\sqrt{5}\) or \(4.47\) (3 sf) | A1 | |
| Total: 3 |
## Question 5(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-2.3}\left(\frac{2.3^2}{2} + \frac{2.3^3}{3!} + \frac{2.3^4}{4!}\right)$ | M1 | Allow one end error |
| $= 0.585$ | A1 | |
| **Total: 2** | | |
---
## Question 5(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(\lambda) = 4.6$ | B1 | |
| $1 - e^{-4.6}\left(1 + 4.6 + \frac{4.6^2}{2}\right)$ | M1 | any $\lambda$, Allow one end error |
| $= 0.837$ (3 sf) | A1 | |
| **Total: 3** | | |
---
## Question 5(a)(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $S \sim N(115, 115)$ | B1 | May be implied |
| $\frac{110.5 - 115}{\sqrt{115}}$ $(= -0.420)$ | M1 | Allow with wrong or no cc OR no $\sqrt{\phantom{x}}$ |
| $1 - \Phi('0.420')$ $(= 1 - 0.663)$ | M1 | |
| $= 0.337$ | A1 | Accept alternative method using $N(2.3, 2.3)$ no mixed methods |
| **Total: 4** | | |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-\lambda} \times \frac{\lambda^3}{3!} = e^{-\lambda} \times \frac{\lambda^5}{5!}$ | M1 | |
| $\lambda^3 = \frac{\lambda^5}{4\times5}$ or $\lambda^2 = 20$ oe | A1 | any correct simplification without $e^{-\lambda}$ or $!$ |
| $\lambda = \sqrt{20}$ or $2\sqrt{5}$ or $4.47$ (3 sf) | A1 | |
| **Total: 3** | | |
---5
\begin{enumerate}[label=(\alph*)]
\item The random variable $X$ has the distribution $\operatorname { Po } ( 2.3 )$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { P } ( 2 \leqslant X \leqslant 4 )$.
\item Find the probability that the sum of two independent values of $X$ is greater than 2 .
\item The random variable $S$ is the sum of 50 independent values of $X$. Use a suitable approximating distribution to find $\mathrm { P } ( S \leqslant 110 )$.
\end{enumerate}\item The random variable $Y$ has the distribution $\mathrm { Po } ( \lambda )$. Given that $\mathrm { P } ( Y = 3 ) = \mathrm { P } ( Y = 5 )$, find $\lambda$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2019 Q5 [12]}}