| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Poisson to the Normal distribution |
| Type | Combined Poisson approximation and exact calculation |
| Difficulty | Standard +0.3 This is a standard S2 question on Poisson distribution with normal approximation. Parts (a)-(d) involve routine identification of Poisson model, parameter conversion (150/hour → 2.5/minute), and basic probability calculations. Part (e) requires normal approximation with continuity correction, which is a textbook procedure. The multi-step nature and 16 total marks make it slightly above average difficulty, but all techniques are standard S2 content with no novel problem-solving required. |
| Spec | 5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(X \sim \text{Po}(2.5)\) | M1A1 | (2 marks) |
| (b) Cars arrive at the toll booth independently/randomly; Cars arrive one at a time; The rate of arrival at a toll booth remains constant at 2.5 per minute | B1, B1 | (2 marks) |
| (c)(i) \(P(X = 0) = e^{-2.5} = 0.0821\) | B1 | (1 mark) |
| (c)(ii) \(P(X > 3) = 1 - P(X \leq 3) = 0.2424\) | M1, A1 | (2 marks) |
| (d) Use of Po(10); \(1 - 0.0487 = 0.9513\); \(m = 15\) | M1, M1, A1cao | (3 marks) |
| (e) \(Y \sim N(25, 25)\); \(P(X < 15) = P(Y \leq 14.5) = P\left(Z \leq \frac{14.5-25}{5}\right) = P(Z \leq -2.1) = 0.01786\) | B1B1, M1, M1, A1, A1 | (6 marks) |
**(a)** $X \sim \text{Po}(2.5)$ | M1A1 | (2 marks)
**(b)** Cars arrive at the toll booth independently/randomly; Cars arrive one at a time; The rate of arrival at a toll booth remains constant at 2.5 per minute | B1, B1 | (2 marks)
**(c)(i)** $P(X = 0) = e^{-2.5} = 0.0821$ | B1 | (1 mark)
**(c)(ii)** $P(X > 3) = 1 - P(X \leq 3) = 0.2424$ | M1, A1 | (2 marks)
**(d)** Use of Po(10); $1 - 0.0487 = 0.9513$; $m = 15$ | M1, M1, A1cao | (3 marks)
**(e)** $Y \sim N(25, 25)$; $P(X < 15) = P(Y \leq 14.5) = P\left(Z \leq \frac{14.5-25}{5}\right) = P(Z \leq -2.1) = 0.01786$ | B1B1, M1, M1, A1, A1 | (6 marks)
**Notes:**
- (a) M1 Poisson; A1 2.5.
- (b) Any two of the statements or equivalent. At least one must be in context. Need words that imply "cars arrive" or "rate of arrival." SC no context but 2 correct reasons B1B0. No context but 1 correct reason B0B0.
- (c)(i) B1 awrt 0.0821.
- (c)(ii) M1 for writing or finding $1 - P(X \leq 3)$; A1 awrt 0.242.
- (d) M1 writing or using Po(10); M1 for 1- 0.0487 or 0.9513 seen or implied by correct value for $m$; A1 cao.
- (e) B1 use of normal; B1 using or seeing mean and variance of 25. These first two marks may be given if the following are seen in the correct places in the standardisation formula : 25 and $\sqrt{25}$ or 5. M1 for attempting a continuity correction $(14 \pm 0.5)$ or $(15 \pm 0.5)$; M1 for standardising using their mean and their standard deviation and using [14.5, 14, 13.5, 15 or 15.5] accept ± z.; A1 correct z value ±2.1 or ± $\frac{14.5-25}{5}$; A1 awrt 0.0179. NB use of calculator gets full marks if the answer is awrt 0.0179.
**[16 marks total]**
---Cars arrive at a motorway toll booth at an average rate of 150 per hour.
\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable distribution to model the number of cars arriving at the toll booth, $X$, per minute.
[2]
\item State clearly any assumptions you have made by suggesting this model.
[2]
\end{enumerate}
Using your model,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the probability that in any given minute
\begin{enumerate}[label=(\roman*)]
\item no cars arrive,
\item more than 3 cars arrive.
\end{enumerate}
[3]
\item In any given 4 minute period, find $m$ such that P($X > m$) = 0.0487
[3]
\item Using a suitable approximation find the probability that fewer than 15 cars arrive in any given 10 minute period.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2011 Q6 [16]}}