| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with geometric or waiting time |
| Difficulty | Standard +0.3 This is a standard Further Maths statistics question on Poisson processes and exponential distributions. Part (a) is routine Poisson probability calculation. Part (b) involves showing the exponential distribution arises from Poisson (a well-known result often seen in textbooks), differentiating to find the pdf, and recognizing the exponential distribution's mean/SD. While it requires understanding the connection between Poisson and exponential distributions, the steps are methodical and this topic is core A-level Further Maths content with no novel problem-solving required. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.03a Continuous random variables: pdf and cdf |
| Answer | Marks | Guidance |
|---|---|---|
| (a) The number of arrivals \(X\) is \(\text{Poi}(7.5)\); \(P(X = 5) = \frac{e^{-7.5} \times 7.5^3}{5!} = 0.109(3745...)\) | B1, M1, A1 | AO3, AO1, AO1 |
| (b)(i) \(P(T > t) = P(\text{No customers arrive between 11am and } t \text{ mins after 11am}) = e^{-0.5t}\) | B1 | AO2 |
| (ii) The cumulative distribution function of \(T\) is \(F(t) = P(T \leq t) = P(T > t) = 1 - e^{-0.5t}\) | M1, A1 | AO3, AO2 |
| Let \(f(t)\) denote the probability density function of \(T\): \(f(t) = F'(t) = 0.5e^{-0.5t}\) | M1, A1 | AO2, AO1 |
| (iii) This is the exponential distribution. Therefore mean = standard deviation = 1/0.5 = 2 | B1, B1 | AO2, AO2 |
**(a)** The number of arrivals $X$ is $\text{Poi}(7.5)$; $P(X = 5) = \frac{e^{-7.5} \times 7.5^3}{5!} = 0.109(3745...)$ | B1, M1, A1 | AO3, AO1, AO1 | Or straight from the calculator
**(b)(i)** $P(T > t) = P(\text{No customers arrive between 11am and } t \text{ mins after 11am}) = e^{-0.5t}$ | B1 | AO2
**(ii)** The cumulative distribution function of $T$ is $F(t) = P(T \leq t) = P(T > t) = 1 - e^{-0.5t}$ | M1, A1 | AO3, AO2
Let $f(t)$ denote the probability density function of $T$: $f(t) = F'(t) = 0.5e^{-0.5t}$ | M1, A1 | AO2, AO1
**(iii)** This is the exponential distribution. Therefore mean = standard deviation = 1/0.5 = 2 | B1, B1 | AO2, AO2 | [10]
---Customers arrive at a shop such that the number of arrivals in a time interval of $t$ minutes follows a Poisson distribution with mean $0.5t$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that exactly 5 customers arrive between 11 a.m. and 11.15 a.m. [3]
\item A customer arrives at exactly 11 a.m.
\begin{enumerate}[label=(\roman*)]
\item Let the next customer arrive at $T$ minutes past 11 a.m. Show that
$$P(T > t) = e^{-0.5t}.$$
\item Hence find the probability density function, $f(t)$, of $T$.
\item Hence, giving a reason, write down the mean and the standard deviation of the time between the arrivals of successive customers. [7]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 Q6 [10]}}