| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Mean-variance comparison for Poisson validation |
| Difficulty | Standard +0.3 This is a straightforward application of the Poisson distribution with clearly stated parameters. Parts (a) and (b) require basic probability calculations using tables or formulas, part (c) involves routine mean and variance calculations from summary statistics, part (d) tests understanding of when Poisson is appropriate (mean ≈ variance), and part (e) is another simple Poisson probability calculation. All steps are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02g Calculate mean and standard deviation5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - P(X \leq 2) = 1 - 0.4232 = 0.5768\) or \(1 - e^{-3}(1 + 3 + \frac{3^2}{2!})\) | M1, A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leq 6) - P(X \leq 4) = 0.9665 - 0.8153 = 0.1512\) or \(e^{-3}(\frac{3^5}{5!} + \frac{3^6}{6!})\) | M1, A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu = 3.69\) | B1 | |
| \(\text{Var}(X) = \frac{1386}{80} - (\frac{295}{80})^2 = 3.73/3.72/3.71\) with accept \(s^2 = 3.77\) | M1, A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| For a Poisson model, Mean = Variance; For these data \(3.69 \approx 3.73\) \(\Rightarrow\) Poisson model | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{e^{-3.6875} \cdot 3.6875^4}{4!} = 0.193\) with allow their mean or var; Awrt 0.193 or 0.194 | M1, A1 ft | (2 marks) |
## Part (a)
$1 - P(X \leq 2) = 1 - 0.4232 = 0.5768$ or $1 - e^{-3}(1 + 3 + \frac{3^2}{2!})$ | M1, A1 | (3 marks)
## Part (b)
$P(X \leq 6) - P(X \leq 4) = 0.9665 - 0.8153 = 0.1512$ or $e^{-3}(\frac{3^5}{5!} + \frac{3^6}{6!})$ | M1, A1 | (2 marks)
## Part (c)
$\mu = 3.69$ | B1 |
$\text{Var}(X) = \frac{1386}{80} - (\frac{295}{80})^2 = 3.73/3.72/3.71$ with accept $s^2 = 3.77$ | M1, A1 | (3 marks)
## Part (d)
For a Poisson model, Mean = Variance; For these data $3.69 \approx 3.73$ $\Rightarrow$ Poisson model | B1 | (1 mark)
## Part (e)
$\frac{e^{-3.6875} \cdot 3.6875^4}{4!} = 0.193$ with allow their mean or var; Awrt 0.193 or 0.194 | M1, A1 ft | (2 marks)
---A botanist is studying the distribution of daisies in a field. The field is divided into a number of equal sized squares. The mean number of daisies per square is assumed to be 3. The daisies are distributed randomly throughout the field.
Find the probability that, in a randomly chosen square there will be
\begin{enumerate}[label=(\alph*)]
\item more than 2 daisies, [3]
\item either 5 or 6 daisies. [2]
\end{enumerate}
The botanist decides to count the number of daisies, $x$, in each of 80 randomly selected squares within the field. The results are summarised below
$$\sum x = 295 \quad \sum x^2 = 1386$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Calculate the mean and the variance of the number of daisies per square for the 80 squares. Give your answers to 2 decimal places. [3]
\item Explain how the answers from part (c) support the choice of a Poisson distribution as a model. [1]
\item Using your mean from part (c), estimate the probability that exactly 4 daisies will be found in a randomly selected square. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2009 Q1 [11]}}