| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson parameter from given probability |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on standard Poisson distribution properties. Part (i) uses the additive property of Poisson (sum of Po(1.5) values is Po(4.5)) then requires a simple probability calculation. Part (ii) involves solving e^(-λ) = 0.523 using logarithms. Part (iii) requires setting up an equation using the Poisson probability formula and solving a quadratic. All parts are routine applications of well-known formulas with no novel insight required, making this 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 Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\lambda = 4.5\) | B1 | seen |
| \(1 - e^{-4.5}\left(1 + 4.5 + \frac{4.5^2}{2}\right) = 0.826\) (3 s.f.) | M1 | any \(\lambda\). Allow one end error |
| A1 [3] | ||
| (ii) \(e^{-\lambda} = 0.523\) | B1 | |
| (\(-\lambda = \ln 0.523\)) \(\lambda = 0.648\) (3 s.f.) | B1 [2] | |
| (iii) \(e^{-\mu} \times \frac{\mu^3}{3!} = 24 \times e^{-\mu} \times \mu\) | B1 | For a simplified expression in \(\mu^2\) with \(e^{-\mu}\) and \(\mu\) cancelled and no factorials. |
| \(\frac{\mu^2}{6} = 24\) | M1 | |
| \(\mu = 12\) | A1 [3] |
**(i)** $\lambda = 4.5$ | B1 | seen
$1 - e^{-4.5}\left(1 + 4.5 + \frac{4.5^2}{2}\right) = 0.826$ (3 s.f.) | M1 | any $\lambda$. Allow one end error
| A1 [3]
**(ii)** $e^{-\lambda} = 0.523$ | B1
($-\lambda = \ln 0.523$) $\lambda = 0.648$ (3 s.f.) | B1 [2]
**(iii)** $e^{-\mu} \times \frac{\mu^3}{3!} = 24 \times e^{-\mu} \times \mu$ | B1 | For a simplified expression in $\mu^2$ with $e^{-\mu}$ and $\mu$ cancelled and no factorials.
$\frac{\mu^2}{6} = 24$ | M1
$\mu = 12$ | A1 [3]
---4 (i) The random variable $W$ has the distribution $\operatorname { Po } ( 1.5 )$. Find the probability that the sum of 3 independent values of $W$ is greater than 2 .\\
(ii) The random variable $X$ has the distribution $\operatorname { Po } ( \lambda )$. Given that $\mathrm { P } ( X = 0 ) = 0.523$, find the value of $\lambda$ correct to 3 significant figures.\\
(iii) The random variable $Y$ has the distribution $\operatorname { Po } ( \mu )$, where $\mu \neq 0$. Given that
$$\mathrm { P } ( Y = 3 ) = 24 \times \mathrm { P } ( Y = 1 )$$
find $\mu$.
\hfill \mbox{\textit{CAIE S2 2014 Q4 [8]}}