| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Moderate -0.3 This is a straightforward S2 question testing standard applications of Poisson and uniform distributions with some basic binomial probability. Parts (a)-(d) are direct formula applications requiring no problem-solving, while part (e) requires recognizing independence and multiplying probabilities—a small step beyond pure recall but still routine for S2 level. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02e Discrete uniform distribution5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=4) = P(X \leq 4) - P(X \leq 3)\) or \(\frac{e^{-3}3^4}{4!}\) | M1 | For a correct method; may use incorrect \(\lambda\) |
| \(= 0.8153 - 0.6472 = 0.1681\) or \(0.1680313...\) (awrt \(0.168\)) | A1 | For awrt 0.168 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y\) [= number of customers joining queue in next 20 mins] \(\sim Po(6)\) | B1 | For writing or using \(Po(6)\) |
| \(P(Y>10) = 1 - P(Y \leq 10)\) | M1 | For writing or using \(1-P(Y \leq 10)\) |
| \(= 1 - 0.9574 = 0.0426(209...)\) (awrt \(0.0426\)) | A1 | For awrt 0.0426 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T > 3.5) = 0.3\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(C \sim B(5, 0.3)\) | M1 | For identifying \(C \sim B(5, 0.3)\); follow through their 0.3 |
| \(P(C \geq 3) = 1 - P(C \leq 2)\) | M1 | For writing or using \(1-P(C \leq 2)\) |
| \(= 1 - 0.8369 = 0.1631\) (or \(0.16308...\)) (awrt \(0.163\)) | A1 | For awrt 0.163. SC: normal distribution may get M0 M1 A0 if \(P(C \geq 2.5)\) found |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{Bethan served in} < 4 \text{ minutes}) = 0.8\) | B1 | For 0.8 |
| \(J\) = number joining queue in 4 mins, \(J \sim Po(1.2)\) | M1 | For identifying \(Po(1.2)\) |
| \(P(J=0) = e^{-1.2} = 0.30119...\) | A1 | For \(e^{-1.2}\) or awrt 0.301 |
| \(P(\text{Bethan served and } J=0) = 0.8 \times e^{-1.2} = 0.240955...\) (awrt \(0.241\)) | A1 | For awrt 0.241 |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=4) = P(X \leq 4) - P(X \leq 3)$ or $\frac{e^{-3}3^4}{4!}$ | M1 | For a correct method; may use incorrect $\lambda$ |
| $= 0.8153 - 0.6472 = 0.1681$ or $0.1680313...$ (awrt $0.168$) | A1 | For awrt 0.168 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y$ [= number of customers joining queue in next 20 mins] $\sim Po(6)$ | B1 | For writing or using $Po(6)$ |
| $P(Y>10) = 1 - P(Y \leq 10)$ | M1 | For writing or using $1-P(Y \leq 10)$ |
| $= 1 - 0.9574 = 0.0426(209...)$ (awrt $0.0426$) | A1 | For awrt 0.0426 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T > 3.5) = 0.3$ | B1 | |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $C \sim B(5, 0.3)$ | M1 | For identifying $C \sim B(5, 0.3)$; follow through their 0.3 |
| $P(C \geq 3) = 1 - P(C \leq 2)$ | M1 | For writing or using $1-P(C \leq 2)$ |
| $= 1 - 0.8369 = 0.1631$ (or $0.16308...$) (awrt $0.163$) | A1 | For awrt 0.163. SC: normal distribution may get M0 M1 A0 if $P(C \geq 2.5)$ found |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{Bethan served in} < 4 \text{ minutes}) = 0.8$ | B1 | For 0.8 |
| $J$ = number joining queue in 4 mins, $J \sim Po(1.2)$ | M1 | For identifying $Po(1.2)$ |
| $P(J=0) = e^{-1.2} = 0.30119...$ | A1 | For $e^{-1.2}$ or awrt 0.301 |
| $P(\text{Bethan served and } J=0) = 0.8 \times e^{-1.2} = 0.240955...$ (awrt $0.241$) | A1 | For awrt 0.241 |
---\begin{enumerate}
\item In a village shop the customers must join a queue to pay. The number of customers joining the queue in a 10 minute interval is modelled by a Poisson distribution with mean 3
\end{enumerate}
Find the probability that\\
(a) exactly 4 customers join the queue in the next 10 minutes,\\
(b) more than 10 customers join the queue in the next 20 minutes.
When a customer reaches the front of the queue the customer pays the assistant. The time each customer takes paying the assistant, $T$ minutes, has a continuous uniform distribution over the interval $[ 0,5 ]$. The random variable $T$ is independent of the number of people joining the queue.\\
(c) Find $\mathrm { P } ( T > 3.5 )$
In a random sample of 5 customers, the random variable $C$ represents the number of customers who took more than 3.5 minutes paying the assistant.\\
(d) Find $\mathrm { P } ( C \geqslant 3 )$
Bethan has just reached the front of the queue and starts paying the assistant.\\
(e) Find the probability that in the next 4 minutes Bethan finishes paying the assistant and no other customers join the queue.\\
\hfill \mbox{\textit{Edexcel S2 2013 Q5 [13]}}