| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson parameter from given probability |
| Difficulty | Standard +0.3 This is a straightforward Poisson distribution question requiring standard techniques: calculating cumulative probabilities, finding mode using P(X=n)=P(X=n+1), and applying binomial distribution to repeated intervals. Part (b) uses the standard λ/(n+1)=1 formula, and part (d) combines Poisson with binomial in a routine way. Slightly above average due to multiple parts and the need to recognize the binomial application in (d), but all techniques are standard S2 material with no novel insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02i 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 |
|---|---|---|
| Working | Mark | Guidance |
| \(P(X < 5) = P(X \leq 4) = 0.947346...\) | M1 | For writing or using \(P(X \leq 4)\); if answer incorrect this must be shown |
| awrt 0.947 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(Y \sim \text{Po}(6)\) | B1 | For writing or using Po(6) |
| \(e^{-6}\frac{6^n}{(n)!} = e^{-6}\frac{6^{n+1}}{(n+1)!} \rightarrow (n+1)=6\) or \(P(Y=4)=0.13385...\), \(P(Y=5)=0.16062...\), \(P(Y=6)=0.16062...\) | M1 | For correct expressions for \(P(Y=n)\) and \(P(Y=n+1)\) using their \(\lambda\), or finding at least 2 of given probabilities for Po(6) correct to 3sf |
| \(\underline{n=5}\) | A1 | Must be only one value of \(n\) given |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(c \leq D \leq 12) = P(D \leq 12) - P(D \leq c-1)\) or \(P(X \leq d) = 0.8758 - 0.8546\) | M1 | 1st M1 for the expression \(P(D \leq 12) - P(D \leq c-1)\); condone \(P(D \leq 12) - P(D \leq c)\) |
| \(0.8546 = 0.8758 - P(D \leq c-1)\), \(P(D \leq c-1) = 0.0212\), \(P(X \leq 3) = 0.0212\) | dM1 | 2nd dM1 for correct substitution of 0.8758 (awrt 0.876) and 0.8546 leading to \(P(D \leq c-1)\) = a probability |
| \(c - 1 = 3\), \(\underline{c=4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(X=2) = 0.27067...\) | B1 | 1st B1 for awrt 0.271 seen |
| \(W \sim \text{B}(6, \text{ awrt } 0.27067...)\) | B1ft | 2nd B1ft follow through their \(P(X=2)\); must see B(6, "0.27067...") or working of form \(("0.27067...")^4(1-"0.27067...")^2\) |
| \(P(W=4) = {}^6C_4("0.27067...")^4(1-"0.27067...")^2 = 0.0428...\) awrt 0.043 | M1, A1 | M1: \({}^6C_4\, p^4(1-p)^2\) or \(P(W \leq 4) - P(W \leq 3)\) or \(0.9932...- 0.9504...\); NB if \(P(X=2)\) not stated and 0.271 not seen, do not ft |
# Question 2:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $P(X < 5) = P(X \leq 4) = 0.947346...$ | M1 | For writing or using $P(X \leq 4)$; if answer incorrect this must be shown |
| awrt **0.947** | A1 | |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $Y \sim \text{Po}(6)$ | B1 | For writing or using Po(6) |
| $e^{-6}\frac{6^n}{(n)!} = e^{-6}\frac{6^{n+1}}{(n+1)!} \rightarrow (n+1)=6$ **or** $P(Y=4)=0.13385...$, $P(Y=5)=0.16062...$, $P(Y=6)=0.16062...$ | M1 | For correct expressions for $P(Y=n)$ and $P(Y=n+1)$ using their $\lambda$, or finding at least 2 of given probabilities for Po(6) correct to 3sf |
| $\underline{n=5}$ | A1 | Must be only one value of $n$ given |
## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| $P(c \leq D \leq 12) = P(D \leq 12) - P(D \leq c-1)$ or $P(X \leq d) = 0.8758 - 0.8546$ | M1 | 1st M1 for the expression $P(D \leq 12) - P(D \leq c-1)$; condone $P(D \leq 12) - P(D \leq c)$ |
| $0.8546 = 0.8758 - P(D \leq c-1)$, $P(D \leq c-1) = 0.0212$, $P(X \leq 3) = 0.0212$ | dM1 | 2nd dM1 for correct substitution of 0.8758 (awrt 0.876) and 0.8546 leading to $P(D \leq c-1)$ = a probability |
| $c - 1 = 3$, $\underline{c=4}$ | A1 | |
## Part (d)
| Working | Mark | Guidance |
|---------|------|----------|
| $P(X=2) = 0.27067...$ | B1 | 1st B1 for awrt 0.271 seen |
| $W \sim \text{B}(6, \text{ awrt } 0.27067...)$ | B1ft | 2nd B1ft follow through their $P(X=2)$; must see B(6, "0.27067...") or working of form $("0.27067...")^4(1-"0.27067...")^2$ |
| $P(W=4) = {}^6C_4("0.27067...")^4(1-"0.27067...")^2 = 0.0428...$ awrt **0.043** | M1, A1 | M1: ${}^6C_4\, p^4(1-p)^2$ or $P(W \leq 4) - P(W \leq 3)$ or $0.9932...- 0.9504...$; NB if $P(X=2)$ not stated and 0.271 not seen, do not ft |
---\begin{enumerate}
\item During morning hours, employees arrive randomly at an office drinks dispenser at a rate of 2 every 10 minutes.
\end{enumerate}
The number of employees arriving at the drinks dispenser is assumed to follow a Poisson distribution.\\
(a) Find the probability that fewer than 5 employees arrive at the drinks dispenser during a 10-minute period one morning.
During a 30 -minute period one morning, the probability that $n$ employees arrive at the drinks dispenser is the same as the probability that $n + 1$ employees arrive at the drinks dispenser.\\
(b) Find the value of $n$
During a 45-minute period one morning, the probability that between $c$ and 12, inclusive, employees arrive at the drinks dispenser is 0.8546\\
(c) Find the value of $C$\\
(d) Find the probability that exactly 2 employees arrive at the drinks dispenser in exactly 4 of the 6 non-overlapping 10-minute intervals between 10 am and 11am one morning.
\hfill \mbox{\textit{Edexcel S2 2019 Q2 [12]}}