| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with geometric or waiting time |
| Difficulty | Moderate -0.8 This is a straightforward Poisson distribution question requiring only standard formula application: scaling the rate parameter (6 per hour to 9 per 90 minutes), calculating P(X=7) and P(X≥10) using tables or calculator, and recognizing part (b) as P(X≥1) over 15 minutes. All steps are routine textbook exercises with no problem-solving insight required. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | \(\text{Po}(9)\) written or used in either (i) or (ii) | Must be stated or implied |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leq 7) - P(X \leq 6) = 0.3239 - 0.2068\) | M1 | Writing \(P(X \leq 7) - P(X \leq 6)\) or \(\frac{e^{-\lambda}\lambda^7}{7!}\); may be implied by \(0.3239 - 0.2068\) |
| \(= 0.1171\) | A1 | awrt \(0.117\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \geq 10) = 1 - P(X \leq 9)\) | M1 | Writing \(1 - P(X \leq 9)\); may be implied by \(1 - 0.5874\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.4126\) | A1 | awrt \(0.413\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Po}(1.5)\) | B1 | \(\text{Po}(1.5)\) written or used |
| \(P(\text{next patient before } 11{:}45) = 1 - P(0)\) | M1 | Writing or using \(1 - P(0)\) or \(1 - e^{-\lambda}\); may be implied by \(1 - 0.2231\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.7769\) | A1 | awrt \(0.777\) |
## Question 1:
### Part (a)
**B1** | $\text{Po}(9)$ written or used in either (i) or (ii) | Must be stated or implied
---
### Part (a)(i)
$P(X \leq 7) - P(X \leq 6) = 0.3239 - 0.2068$ | **M1** | Writing $P(X \leq 7) - P(X \leq 6)$ or $\frac{e^{-\lambda}\lambda^7}{7!}$; may be implied by $0.3239 - 0.2068$
$= 0.1171$ | **A1** | awrt $0.117$
---
### Part (a)(ii)
$P(X \geq 10) = 1 - P(X \leq 9)$ | **M1** | Writing $1 - P(X \leq 9)$; may be implied by $1 - 0.5874$
$= 1 - 0.5874$
$= 0.4126$ | **A1** | awrt $0.413$
**(5 marks)**
---
### Part (b)
$\text{Po}(1.5)$ | **B1** | $\text{Po}(1.5)$ written or used
$P(\text{next patient before } 11{:}45) = 1 - P(0)$ | **M1** | Writing or using $1 - P(0)$ or $1 - e^{-\lambda}$; may be implied by $1 - 0.2231$
$= 1 - e^{-1.5}$
$= 0.7769$ | **A1** | awrt $0.777$
**(3 marks) [8 marks total]**\begin{enumerate}
\item Patients arrive at a hospital accident and emergency department at random at a rate of 6 per hour.\\
(a) Find the probability that, during any 90 minute period, the number of patients arriving at the hospital accident and emergency department is\\
(i) exactly 7\\
(ii) at least 10
\end{enumerate}
A patient arrives at 11.30 a.m.\\
(b) Find the probability that the next patient arrives before 11.45 a.m.\\
\hfill \mbox{\textit{Edexcel S2 2014 Q1 [8]}}