| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Three or more independent Poisson sums |
| Difficulty | Moderate -0.8 This is a straightforward S2 Poisson question requiring only recall of conditions (part a) and routine application of standard formulas with simple parameter scaling. The most challenging aspect is part (c) combining two Poisson distributions, but this is a standard textbook technique with clear setup. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | Events occur at a constant rate. Events occur independently or randomly. Events occur singly. (Any two of the three.) | B1, B1 |
| (b)(i) | \(X \sim \text{Po}(6)\). \(P(X \leq 4) - P(X \leq 3) = 0.2851 - 0.1512 = 0.1339\), or \(\frac{e^{-6}6^4}{4!}\) | B1, M1, A1 |
| (b)(ii) | \(1 - P(X \leq 4) = 1 - 0.2851 = 0.7149\), or \(1 - e^{-6}\!\left(\frac{6^4}{4!}+\frac{6^3}{3!}+\frac{6^2}{2!}+\frac{6}{1!}+1\right)\) | M1, A1 |
| (c) | \(P(\text{0 cars and 1 other}) + P(\text{1 car and 0 others})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Alternative: \(\text{Po}(1+2)=\text{Po}(3)\), \(P(X=1)=3e^{-3}=0.149\) | B1, M1, A1, A1 | B1: attempting both possibilities (or B1 for Po(3) in alternative). M1: finding one pair of form \(e^{-\lambda_1}\times\lambda_2 e^{-\lambda_2}\) (or M1 for attempting \(P(X=1)\) with Po(3)). A1: one pair correct (or A1: \(3e^{-3}\)). A1: awrt 0.149. |
## Question 3:
**(a)** | Events occur at a constant rate. Events occur independently or randomly. Events occur singly. (Any two of the three.) | B1, B1 | Need the word "events" at least once. "Independently" and "randomly" are the same reason — award first B1 if only one mark gained. Special case: if 2 of 3 lines given without the word "events", award B0 B1.
**(b)(i)** | $X \sim \text{Po}(6)$. $P(X \leq 4) - P(X \leq 3) = 0.2851 - 0.1512 = 0.1339$, or $\frac{e^{-6}6^4}{4!}$ | B1, M1, A1 | B1: using Po(6). M1: attempting $P(X\leq4)-P(X\leq3)$ or $\frac{e^{-\lambda}\lambda^4}{4!}$. A1: awrt 0.134.
**(b)(ii)** | $1 - P(X \leq 4) = 1 - 0.2851 = 0.7149$, or $1 - e^{-6}\!\left(\frac{6^4}{4!}+\frac{6^3}{3!}+\frac{6^2}{2!}+\frac{6}{1!}+1\right)$ | M1, A1 | M1: attempting $1-P(X\leq4)$. A1: awrt 0.715.
**(c)** | $P(\text{0 cars and 1 other}) + P(\text{1 car and 0 others})$
$= e^{-1}\times 2e^{-2} + 1e^{-1}\times e^{-2}$
$= 0.3679\times0.2707 + 0.3674\times0.1353$
$= 0.0996 + 0.0498 = 0.149$
Alternative: $\text{Po}(1+2)=\text{Po}(3)$, $P(X=1)=3e^{-3}=0.149$ | B1, M1, A1, A1 | B1: attempting both possibilities (or B1 for Po(3) in alternative). M1: finding one pair of form $e^{-\lambda_1}\times\lambda_2 e^{-\lambda_2}$ (or M1 for attempting $P(X=1)$ with Po(3)). A1: one pair correct (or A1: $3e^{-3}$). A1: awrt 0.149.
---3. (a) State two conditions under which a Poisson distribution is a suitable model to use in statistical work.
The number of cars passing an observation point in a 10 minute interval is modelled by a Poisson distribution with mean 1.\\
(b) Find the probability that in a randomly chosen 60 minute period there will be
\begin{enumerate}[label=(\roman*)]
\item exactly 4 cars passing the observation point,
\item at least 5 cars passing the observation point.
The number of other vehicles, other than cars, passing the observation point in a 60 minute interval is modelled by a Poisson distribution with mean 12.\\
(c) Find the probability that exactly 1 vehicle, of any type, passes the observation point in a 10 minute period.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2008 Q3 [11]}}