| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Validity of Poisson model |
| Difficulty | Standard +0.3 Part (a) is a standard application of summing independent Poisson distributions (λ=26) and using tables/calculator for P(X≥30). Part (b) requires recognizing that for Poisson, variance equals mean, so SD should be √26≈5.1, not 10, indicating the model is invalid. This is straightforward Further Maths Statistics content with no novel insight required, slightly above average due to the two-part structure and the conceptual check in part (b). |
| Spec | 5.02i Poisson distribution: random events model5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim \text{Po}(26)\) | M1 (AO 3.3) | Translate situation into correct Poisson model; PI |
| \(P(X \geq 30) = 1 - P(X \leq 29)\) | M1 (AO 3.4) | Uses Poisson model to calculate \(P(X \geq 30)\), \(P(X > 30)\), \(P(X \leq 29)\) or \(P(X \leq 30)\) |
| \(= 1 - 0.759 = 0.241\) | A1 (AO 1.1b) | AWRT \(0.241\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Compares either \(26\) with \(100\) or \(\sqrt{26}\) with \(10\) | M1 (AO 3.5b) | |
| The Poisson model is not valid. Variance of total of daisies and dandelions, \(10^2\), is not approximately equal to the mean, \(26\) | A1F (AO 2.4) | Concludes mean/variance not approximately equal to \(100\), or standard deviation not approximately equal to \(10\); must not use or imply \(\neq\) |
## Question 4(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim \text{Po}(26)$ | M1 (AO 3.3) | Translate situation into correct Poisson model; PI |
| $P(X \geq 30) = 1 - P(X \leq 29)$ | M1 (AO 3.4) | Uses Poisson model to calculate $P(X \geq 30)$, $P(X > 30)$, $P(X \leq 29)$ or $P(X \leq 30)$ |
| $= 1 - 0.759 = 0.241$ | A1 (AO 1.1b) | **AWRT** $0.241$ |
## Question 4(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Compares either $26$ with $100$ or $\sqrt{26}$ with $10$ | M1 (AO 3.5b) | |
| The Poisson model is not valid. Variance of total of daisies and dandelions, $10^2$, is not approximately equal to the mean, $26$ | A1F (AO 2.4) | Concludes mean/variance not approximately equal to $100$, or standard deviation not approximately equal to $10$; must not use or imply $\neq$ |4 Daisies and dandelions are the only flowers growing in a field.
The number of daisies per square metre in the field has a mean of 16\\
The number of dandelions per square metre in the field has a mean of 10\\
The number of daisies per square metre and the number of dandelions per square metre are independent.
4
\begin{enumerate}[label=(\alph*)]
\item Using a Poisson model, find the probability that a randomly selected square metre from the field has a total of at least 30 flowers, giving your answer to three decimal places.\\
4
\item A survey of the entire field is taken.\\
The standard deviation of the total number of flowers per square metre is 10 State, with a reason, whether the model used in part (a) is valid.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2022 Q4 [5]}}