| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| 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 application of standard Poisson distribution properties: scaling parameters for different time periods, summing independent Poisson variables, and using tables/calculators for probabilities. Part (c) requires recalling that variance equals mean for Poisson distributions. All techniques are routine for Further Maths Statistics students 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!5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\times4=8\), \(X\sim\text{Po}(8)\) | M1 | Selects Poisson model with \(\lambda=2\times4=8\) |
| \(P(X=5) = 0.0916\) AWRT | A1 | Finds \(P(X=5)=0.0916\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2+5=7\), \(X+Y\sim\text{Po}(7)\) | M1 | Selects Poisson model with \(\lambda=2+5=7\) |
| \(P(X+Y>8)=0.27\) (AWRT) | A1 | Uses model to find \(P(X+Y>8)=0.27\) |
| The probability is less than 40% so a new machine will not be purchased | E1F | Concludes correctly whether machine should be purchased; follow through on attempt to combine Poisson distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Variance \(= \text{sd}^2 = 0.25\) or \(\sqrt{\text{mean}}\) for machine A or B or combined machines | M1 | Calculates variance \(=\text{sd}^2\) or \(\sqrt{\text{mean}}\) (PI by clear argument) |
| For a Poisson distribution Mean \(=\) Variance, but means of 2 and 5 are not equal to variance of 0.25 | A1 | Identifies clear contradiction: for Poisson, Mean \(=\) Variance |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\times4=8$, $X\sim\text{Po}(8)$ | M1 | Selects Poisson model with $\lambda=2\times4=8$ |
| $P(X=5) = 0.0916$ AWRT | A1 | Finds $P(X=5)=0.0916$ |
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2+5=7$, $X+Y\sim\text{Po}(7)$ | M1 | Selects Poisson model with $\lambda=2+5=7$ |
| $P(X+Y>8)=0.27$ (AWRT) | A1 | Uses model to find $P(X+Y>8)=0.27$ |
| The probability is less than 40% so a new machine will not be purchased | E1F | Concludes correctly whether machine should be purchased; follow through on attempt to combine Poisson distributions |
## Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Variance $= \text{sd}^2 = 0.25$ or $\sqrt{\text{mean}}$ for machine A or B or combined machines | M1 | Calculates variance $=\text{sd}^2$ or $\sqrt{\text{mean}}$ (PI by clear argument) |
| For a Poisson distribution Mean $=$ Variance, but means of 2 and 5 are not equal to variance of 0.25 | A1 | Identifies clear contradiction: for Poisson, Mean $=$ Variance |6 A company owns two machines, $A$ and $B$, which make toys. Both machines run continuously and independently.
Machine $A$ makes an average of 2 errors per hour.\\
6
\begin{enumerate}[label=(\alph*)]
\item Using a Poisson model, find the probability that the machine makes exactly 5 errors in 4 hours, giving your answer to three significant figures.
6
\item Machine $B$ makes an average of 5 errors per hour. Both machines are switched on and run for 1 hour.
The company finds the probability that the total number of errors made by machines $A$ and $B$ in 1 hour is greater than 8 .
If the probability is greater than 0.4 , a new machine will be purchased.\\
Using a Poisson model, determine whether or not the toy company will purchase a new machine.\\
6
\item After investigation, the standard deviation of errors made by machine $A$ is found to be 0.5 errors per hour and the standard deviation of errors made by machine $B$ is also found to be 0.5 errors per hour.
Explain whether or not the use of Poisson models in parts (a) and (b) is appropriate.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2019 Q6 [7]}}