| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - large lambda direct |
| Difficulty | Standard +0.3 This is a straightforward application of standard Poisson distribution techniques with clear signposting. Parts (a)-(d) involve basic recognition and calculation with λ=2.25, while part (e) requires a routine normal approximation to Poisson with λ=20. All steps are textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim P_0(2.25)\) | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = 0) = e^{-2.25} = 0.105399\ldots\) | awrt 0.105 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 2) = 1 - P(X \leq 2), = 1 - e^{-2.25}\left[1 + 2.25 + \frac{(2.25)^2}{2!}\right]\) | M1, M1 | |
| \(1 - 0.60933\ldots = 0.39066\) | awrt 0.391 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Sheep would tend to cluster – no longer randomly scattered | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Y < 15) = P(Y \leq 14.5), = P\left(Z \leq \frac{14.5 - 20}{\sqrt{20}}\right)\) | \(\pm \frac{1}{2}\) | M1, M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 1 - 0.8907 = 0.1093\) | AWRT 0.109 | M1 A1 |
## Part (a)
$X \sim P_0(2.25)$ | B1 | (1 mark)
## Part (b)
$P(X = 0) = e^{-2.25} = 0.105399\ldots$ | awrt 0.105 | B1 | (1 mark)
## Part (c)
$P(X > 2) = 1 - P(X \leq 2), = 1 - e^{-2.25}\left[1 + 2.25 + \frac{(2.25)^2}{2!}\right]$ | M1, M1
$1 - 0.60933\ldots = 0.39066$ | awrt 0.391 | A1 | (4 marks)
## Part (d)
Sheep would tend to cluster – no longer randomly scattered | B1 | (1 mark)
## Part (e)
$Y \sim P_0(20) \Rightarrow$ normal approx., $\mu = 20, \sigma = \sqrt{20}$
$P(Y < 15) = P(Y \leq 14.5), = P\left(Z \leq \frac{14.5 - 20}{\sqrt{20}}\right)$ | $\pm \frac{1}{2}$ | M1, M1
$= P(Z \leq -1.2298\ldots)$
$= 1 - 0.8907 = 0.1093$ | AWRT 0.109 | M1 A1 | (7 marks)
**Total: 14 marks**
---A biologist is studying the behaviour of sheep in a large field. The field is divided up into a number of equally sized squares and the average number of sheep per square is 2.25. The sheep are randomly spread throughout the field.
\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable model for the number of sheep in a square and give a value for any parameter or parameters required. [1]
\end{enumerate}
Calculate the probability that a randomly selected sample square contains
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item no sheep, [1]
\item more than 2 sheep. [4]
\end{enumerate}
A sheepdog has been sent into the field to round up the sheep.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Explain why the model may no longer be applicable. [1]
\end{enumerate}
In another field, the average number of sheep per square is 20 and the sheep are randomly scattered throughout the field.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Using a suitable approximation, find the probability that a randomly selected square contains fewer than 15 sheep. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [14]}}