| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson parameter from given probability |
| Difficulty | Moderate -0.3 This is a straightforward Poisson distribution question requiring basic probability calculation and solving a simple equation. Part (i) is direct formula application, parts (ii)-(iii) involve equating two Poisson probabilities and solving algebraically for n, which is a standard textbook exercise requiring no novel insight. Slightly easier than average due to the guided structure and routine nature of the tasks. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks |
|---|---|
| \(0.0842\) (3 sf) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-5} \times \dfrac{5^n}{n!} = e^{-5} \times \dfrac{5^{n+1}}{(n+1)!}\) | B1 | or \(\dfrac{5^n}{n!} = \dfrac{5^{n+1}}{(n+1)!}\) or better; ISW |
| Answer | Marks |
|---|---|
| \(1 = \dfrac{5}{n+1}\), \(n = 4\) | B1 |
**Question 1:**
**Part (i):**
$0.0842$ (3 sf) | B1 |
**Part (ii):**
$e^{-5} \times \dfrac{5^n}{n!} = e^{-5} \times \dfrac{5^{n+1}}{(n+1)!}$ | B1 | or $\dfrac{5^n}{n!} = \dfrac{5^{n+1}}{(n+1)!}$ or better; ISW
**Part (iii):**
$1 = \dfrac{5}{n+1}$, $n = 4$ | B1 |
---1 The random variable $X$ has the distribution $\operatorname { Po } ( 5 )$.\\
(i) Find $\mathrm { P } ( X = 2 )$.\\
It is given that $\mathrm { P } ( X = n ) = \mathrm { P } ( X = n + 1 )$.\\
(ii) Write down an equation in $n$.\\
(iii) Hence or otherwise find the value of $n$.\\
\hfill \mbox{\textit{CAIE S2 2019 Q1 [3]}}