| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Basic sum of two Poissons |
| Difficulty | Standard +0.3 This is a straightforward S2 Poisson distribution question requiring standard cumulative probability calculations and knowledge that independent Poisson variables sum to another Poisson. Part (c)(iii) involves binomial probability with the result from (c)(ii), but all steps are routine applications of formulas with no novel problem-solving required. Slightly above average difficulty due to the multi-part nature and calculator work, but well within typical S2 scope. |
| Spec | 2.04b Binomial distribution: as model B(n,p)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 |
|---|---|---|
| \(\mathbb{P}(X \geq 9) = 1 - \mathbb{P}(X \leq 8) = 1 - 0.5231 = 0.4769\) | B2,1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbb{P}(5 < X < 10) = \mathbb{P}(X \leq 9) - \mathbb{P}(X \leq 5) = 0.653 - 0.1496 = 0.5034\) | B3,2,1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbb{P}(Y < 2) = \mathbb{P}(Y \leq 1) = \mathbb{P}(Y = 0 \text{ or } Y = 1) = e^{-1.5} + e^{-1.5} \times 1.5\) | M1 | (both) |
| \([0.2231 + 0.3347] = 0.5578254 = 0.558\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda = 8.5 + 1.5 = 10\) | B1 | 1 |
| Answer | Marks |
|---|---|
| \(\mathbb{P}(T > 16) = 1 - \mathbb{P}(T \leq 16) = 1 - 0.9730 = 0.027\) | M1 |
| A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = ^3C_5 0.027^3 \times 0.973\) | M1 | for either term correct |
| \(+ 0.027^3\) | M1 | for addition of the two correct terms |
| \(= 0.002128 + 0.00001968 = 0.0021[dp]\) | A1 | 3 |
| Alternative: | ||
| \(p = 1 - \mathbb{P}(X \leq 1)\) | (M1) | for either term correct |
| \(\mathbb{P}(X = 0) + \mathbb{P}(X = 1)\) | ||
| \(= 0.973^3 + 3 \times 0.973^2 \times 0.027\) | (M1) | for 1 – [sum of two correct terms] |
| \(= 0.921167 + 0.076685\) | ||
| \(p = 1 - 0.99785\) | (A1) | 0.0021 or 0.0022 [iff M1M1 (+ 4dp)] |
| \(= 0.0021\) | ||
| 13 |
### 5(a)(i)
| $\mathbb{P}(X \geq 9) = 1 - \mathbb{P}(X \leq 8) = 1 - 0.5231 = 0.4769$ | B2,1 | 2 | $1 - 0.6530 = 0.347$ (B1) awfw 0.476 and 0.477 |
### 5(a)(ii)
| $\mathbb{P}(5 < X < 10) = \mathbb{P}(X \leq 9) - \mathbb{P}(X \leq 5) = 0.653 - 0.1496 = 0.5034$ | B3,2,1 | 3 | awfw 0.503 to 0.504 0.7634 – 0.1496 = 0.613 to 0.614 (B2) 0.6530 – 0.2562 = 0.397 to 0.398 (B2) 0.7634 – 0.2562 = 0.507 to 0.508 (B1) $\alpha - 0.1496$ or $0.653 - \alpha$ (B1) iff $0 < p < 1$ |
### 5(b)
| $\mathbb{P}(Y < 2) = \mathbb{P}(Y \leq 1) = \mathbb{P}(Y = 0 \text{ or } Y = 1) = e^{-1.5} + e^{-1.5} \times 1.5$ | M1 | (both) |
| --- | --- | --- |
| $[0.2231 + 0.3347] = 0.5578254 = 0.558$ | A1 | 2 | awfw 0.557 to 0.56 |
### 5(c)(i)
| $\lambda = 8.5 + 1.5 = 10$ | B1 | 1 | Allow $\mathbb{P}(10)$ or $\text{Po}(10)$ |
### 5(c)(ii)
| $\mathbb{P}(T > 16) = 1 - \mathbb{P}(T \leq 16) = 1 - 0.9730 = 0.027$ | M1 | |
| --- | --- | --- |
| | A1 | 2 | |
### 5(c)(iii)
| $p = ^3C_5 0.027^3 \times 0.973$ | M1 | for either term correct |
| --- | --- | --- |
| $+ 0.027^3$ | M1 | for addition of the two correct terms |
| $= 0.002128 + 0.00001968 = 0.0021[dp]$ | A1 | 3 | 0.0021 or 0.0022 [iff M1M1 (+ 4dp)] |
| **Alternative:** | | |
| $p = 1 - \mathbb{P}(X \leq 1)$ | (M1) | for either term correct |
| $\mathbb{P}(X = 0) + \mathbb{P}(X = 1)$ | | |
| $= 0.973^3 + 3 \times 0.973^2 \times 0.027$ | (M1) | for 1 – [sum of two correct terms] |
| $= 0.921167 + 0.076685$ | | |
| $p = 1 - 0.99785$ | (A1) | 0.0021 or 0.0022 [iff M1M1 (+ 4dp)] |
| $= 0.0021$ | | |
| | | **13** | |5
\begin{enumerate}[label=(\alph*)]
\item The number of minor accidents occurring each year at RapidNut engineering company may be modelled by the random variable $X$ having a Poisson distribution with mean 8.5.
Determine the probability that, in any particular year, there are:
\begin{enumerate}[label=(\roman*)]
\item at least 9 minor accidents;
\item more than 5 but fewer than 10 minor accidents.
\end{enumerate}\item The number of major accidents occurring each year at RapidNut engineering company may be modelled by the random variable $Y$ having a Poisson distribution with mean 1.5.
Calculate the probability that, in any particular year, there are fewer than 2 major accidents.
\item The total number of minor and major accidents occurring each year at RapidNut engineering company may be modelled by the random variable $T$ having the probability distribution
$$\mathrm { P } ( T = t ) = \left\{ \begin{array} { c l }
\frac { \mathrm { e } ^ { - \lambda } \lambda ^ { t } } { t ! } & t = 0,1,2,3 , \ldots \\
0 & \text { otherwise }
\end{array} \right.$$
Assuming that the number of minor accidents is independent of the number of major accidents:
\begin{enumerate}[label=(\roman*)]
\item state the value of $\lambda$;
\item determine $\mathrm { P } ( T > 16 )$;
\item calculate the probability that there will be a total of more than 16 accidents in each of at least two out of three years, giving your answer to four decimal places.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q5 [13]}}