| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Three or more independent Poisson sums |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard Poisson distribution properties. Part (a) involves routine recall (naming distribution, applying E(aX+b) and Var(aX+b) formulas) and a simple algebraic manipulation to show a recurrence relation. Part (b) applies the additive property of Poisson distributions with basic rate conversions. All techniques are standard textbook exercises requiring minimal problem-solving insight, making this slightly easier than average for A-level. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02m Poisson: mean = variance = lambda5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks |
|---|---|
| Poisson | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(3X - 1) = 3\lambda - 1\) | B1 | |
| \(\text{Var}(3X - 1) = 9\lambda\) | B1 | oe (allow \(3^2\lambda\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = x+1) = \frac{e^{-\lambda} \times \lambda^{x+1}}{(x+1)!}\) | B1 | |
| \(P(X = x+1) = \frac{e^{-\lambda} \times \lambda^{x+1}}{(x+1)!} = \frac{e^{-\lambda} \times \lambda^x \times \lambda}{(x+1)x!} = \frac{\lambda}{x+1} \times \frac{e^{-\lambda} \times \lambda^x}{x!} = \frac{\lambda}{x+1}P(X = x)\) | Mdep1 Adep1 | dep(B1) AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda_{\text{car}} = 500/\text{hour}\) \(\lambda_{\text{coach}} = 10/\text{hour}\) \(\Rightarrow \lambda_{\text{vehicle}} = 510/\text{hour} = 8.5/\min\) | B1 | for 8.5 stated / used special case: \(\lambda = 10 \Rightarrow B1M0A0\) B1 \(\Rightarrow 1 - 0.458\) or 0.542 |
| \(P(V \geq 10) = 1 - 0.6530 = 0.347\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu_{\text{car}} = 836/\text{hour}\) \(\mu_{\text{coach}} = 22/\text{hour}\) \(\Rightarrow \mu_{\text{vehicle}} = 858/\text{hour} = 14.3/\min\) | B1 | for 14.3 stated /used |
| \(P(V \leq 3) = P(V = 0,1,2,3) = \begin{cases} e^{-14.3}\left[1 + \frac{14.3}{1} + \frac{14.3^2}{2} + \frac{14.3^3}{6}\right] \\ = e^{-14.3} \times 604.91283 \\ \\ 0.0003726 \text{ to } 0.000373 \end{cases}\) | M1 | All 4 terms required for any \(\lambda > 0\) M0 for use of normal approximation |
| \(= 0.00037\) (2sf) | Adep1 | dep M1 |
## (a)(i)
| Poisson | B1 | |
## (a)(ii)
| $E(3X - 1) = 3\lambda - 1$ | B1 | |
| $\text{Var}(3X - 1) = 9\lambda$ | B1 | oe (allow $3^2\lambda$) |
## (a)(iii)
| $P(X = x+1) = \frac{e^{-\lambda} \times \lambda^{x+1}}{(x+1)!}$ | B1 | |
| $P(X = x+1) = \frac{e^{-\lambda} \times \lambda^{x+1}}{(x+1)!} = \frac{e^{-\lambda} \times \lambda^x \times \lambda}{(x+1)x!} = \frac{\lambda}{x+1} \times \frac{e^{-\lambda} \times \lambda^x}{x!} = \frac{\lambda}{x+1}P(X = x)$ | Mdep1 Adep1 | dep(B1) AG |
## (b)(i)
| $\lambda_{\text{car}} = 500/\text{hour}$ $\lambda_{\text{coach}} = 10/\text{hour}$ $\Rightarrow \lambda_{\text{vehicle}} = 510/\text{hour} = 8.5/\min$ | B1 | for 8.5 stated / used special case: $\lambda = 10 \Rightarrow B1M0A0$ B1 $\Rightarrow 1 - 0.458$ or 0.542 |
| $P(V \geq 10) = 1 - 0.6530 = 0.347$ | M1 A1 | |
## (b)(ii)
| $\mu_{\text{car}} = 836/\text{hour}$ $\mu_{\text{coach}} = 22/\text{hour}$ $\Rightarrow \mu_{\text{vehicle}} = 858/\text{hour} = 14.3/\min$ | B1 | for 14.3 stated /used |
| $P(V \leq 3) = P(V = 0,1,2,3) = \begin{cases} e^{-14.3}\left[1 + \frac{14.3}{1} + \frac{14.3^2}{2} + \frac{14.3^3}{6}\right] \\ = e^{-14.3} \times 604.91283 \\ \\ 0.0003726 \text{ to } 0.000373 \end{cases}$ | M1 | All 4 terms required for any $\lambda > 0$ M0 for use of normal approximation |
| $= 0.00037$ (2sf) | Adep1 | dep M1 |
**Total: 12 marks**
---4
\begin{enumerate}[label=(\alph*)]
\item A discrete random variable $X$ has a probability function defined by
$$\mathrm { P } ( X = x ) = \frac { \mathrm { e } ^ { - \lambda } \lambda ^ { x } } { x ! } \quad \text { for } x = 0,1,2,3,4 , \ldots \ldots$$
\begin{enumerate}[label=(\roman*)]
\item State the name of the distribution of $X$.
\item Write down, in terms of $\lambda$, expressions for $\mathrm { E } ( 3 X - 1 )$ and $\operatorname { Var } ( 3 X - 1 )$.
\item Write down an expression for $\mathrm { P } ( X = x + 1 )$, and hence show that
$$\mathrm { P } ( X = x + 1 ) = \frac { \lambda } { x + 1 } \mathrm { P } ( X = x )$$
\end{enumerate}\item The number of cars and the number of coaches passing a certain road junction may be modelled by independent Poisson distributions.
\begin{enumerate}[label=(\roman*)]
\item On a winter morning, an average of 500 cars per hour and an average of 10 coaches per hour pass this junction.
Determine the probability that a total of at least 10 such vehicles pass this junction during a particular 1 -minute interval on a winter morning.
\item On a summer morning, an average of 836 cars per hour and an average of 22 coaches per hour pass this junction.
Calculate the probability that a total of at most 3 such vehicles pass this junction during a particular 1 -minute interval on a summer morning. Give your answer to two significant figures.\\
(3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q4 [12]}}