| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | October |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Consecutive non-overlapping periods |
| Difficulty | Challenging +1.2 This is a comprehensive multi-part Poisson distribution question requiring various techniques: basic probability calculation, finding critical values using tables/cumulative probabilities, scaling parameters for different time periods, exponential distribution (memoryless property), conditional probability, and hypothesis testing. While it covers many S2 topics and requires careful work, each individual part uses standard techniques without requiring novel insight or particularly complex reasoning. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.02i 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 modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X=8) = \frac{e^{-6}6^8}{8!}\) or \(0.8472 - 0.7440\) | M1 | Correct formula or correct use of tables |
| \(= 0.10325\ldots\) | A1 | awrt 0.103 |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([X \sim Po(6)\ldots] P(X \ldots n) < 0.05\) for \(P(X \text{ "} n-1) > 0.95\) | M1 | A correct probability statement. Implied by correct answer |
| \(n = 11\) | A1cao | cao |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(K \sim Po(0.6m)\) and \(P(K=0) < 0.05\) i.e. \(e^{-0.6m} < 0.05\) / \(-0.6m < \ln 0.05\) oe or \(\lambda = 3\) | M1 | Forming an equation or inequality or identifying \(\lambda = 3\) |
| \(m = 5\) | A1cao | cao |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(Y \sim Po(3)\) | B1 | Writing Po(3) [implied by 0.0498... or correct answer] |
| \(P(Y \ldots 1) = 1 - P(Y=0)\) | M1 | Writing or using \(1 - P(Y=0)\) |
| \(= 0.9502\) | A1 | Allow 0.95 or better |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([W \sim Po(18)]\ P(W=15) = \frac{e^{-18}18^{15}}{15!} = 0.078575\ldots\) | M1 | Using Po(18) to find \(P(W=15)\) |
| \(\frac{P(Y=1\ [Y \sim Po(3)]) \times P(T=14\ [T \sim Po(15)])}{"0.078575..."}\) | dM1 | Dep on 1st M1. Attempt at conditional probability with \(P(Y=1) \times P(T=14)\) (any value of \(\lambda\)) on numerator and their \(P(W=15)\) on denominator (may be implied) |
| \(= \frac{(e^{-3} \times 3)[=0.149\ldots] \times \left(\frac{e^{-15}15^{14}}{14!}\right)[=0.102\ldots]}{" 0.078575..."}\) | dM1 | Dep on 2nd M1. Correct ratio of probabilities. ALT: \(= 15\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{14}\) |
| \(= 0.1947\ldots\) | A1 | awrt 0.195 |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(J \sim Po(9)\); \(P(J \leq 13) = 0.9261\); \(P(J \leq 14) = 0.9585\) | M1 | Writing or using Po(9). Implied by correct CR |
| So critical region is \(J \geq 15\) | A1 | cao. Allow \(J > 14\). Do not allow as part of a probability statement |
| (2) | ||
| Total 15 |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=8) = \frac{e^{-6}6^8}{8!}$ or $0.8472 - 0.7440$ | M1 | Correct formula or correct use of tables |
| $= 0.10325\ldots$ | A1 | awrt 0.103 |
| | **(2)** | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[X \sim Po(6)\ldots] P(X \ldots n) < 0.05$ for $P(X \text{ "} n-1) > 0.95$ | M1 | A correct probability statement. Implied by correct answer |
| $n = 11$ | A1cao | cao |
| | **(2)** | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $K \sim Po(0.6m)$ and $P(K=0) < 0.05$ i.e. $e^{-0.6m} < 0.05$ / $-0.6m < \ln 0.05$ oe or $\lambda = 3$ | M1 | Forming an equation or inequality or identifying $\lambda = 3$ |
| $m = 5$ | A1cao | cao |
| | **(2)** | |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $Y \sim Po(3)$ | B1 | Writing Po(3) [implied by 0.0498... or correct answer] |
| $P(Y \ldots 1) = 1 - P(Y=0)$ | M1 | Writing or using $1 - P(Y=0)$ |
| $= 0.9502$ | A1 | Allow 0.95 or better |
| | **(3)** | |
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[W \sim Po(18)]\ P(W=15) = \frac{e^{-18}18^{15}}{15!} = 0.078575\ldots$ | M1 | Using Po(18) to find $P(W=15)$ |
| $\frac{P(Y=1\ [Y \sim Po(3)]) \times P(T=14\ [T \sim Po(15)])}{"0.078575..."}$ | dM1 | Dep on 1st M1. Attempt at conditional probability with $P(Y=1) \times P(T=14)$ (any value of $\lambda$) on numerator and their $P(W=15)$ on denominator (may be implied) |
| $= \frac{(e^{-3} \times 3)[=0.149\ldots] \times \left(\frac{e^{-15}15^{14}}{14!}\right)[=0.102\ldots]}{" 0.078575..."}$ | dM1 | Dep on 2nd M1. Correct ratio of probabilities. ALT: $= 15\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{14}$ |
| $= 0.1947\ldots$ | A1 | awrt 0.195 |
| | **(4)** | |
## Part (f)
| Answer | Mark | Guidance |
|--------|------|----------|
| $J \sim Po(9)$; $P(J \leq 13) = 0.9261$; $P(J \leq 14) = 0.9585$ | M1 | Writing or using Po(9). Implied by correct CR |
| So critical region is $J \geq 15$ | A1 | cao. Allow $J > 14$. Do not allow as part of a probability statement |
| | **(2)** | |
| | **Total 15** | |
---\begin{enumerate}
\item The number of cars entering a safari park per 10 -minute period can be modelled by a Poisson distribution with mean 6\\
(a) Find the probability that in a given 10 -minute period exactly 8 cars will enter the safari park.\\
(b) Find the smallest value of $n$ such that the probability that at least $n$ cars enter the safari park in 10 minutes is less than 0.05
\end{enumerate}
The probability that no cars enter the safari park in $m$ minutes, where $m$ is an integer, is less than 0.05\\
(c) Find the smallest value of $m$
A car enters the safari park.\\
(d) Find the probability that there is less than 5 minutes before the next car enters the safari park.
Given that exactly 15 cars entered the safari park in a 30-minute period,\\
(e) find the probability that exactly 1 car entered the safari park in the first 5 minutes of the 30-minute period.
Aston claims that the mean number of cars entering the safari park per 10-minute period is more than 6 He selects a 15-minute period at random in order to test whether there is evidence to support his claim.\\
(f) Determine the critical region for the test at the $5 \%$ level of significance.
\hfill \mbox{\textit{Edexcel S2 2021 Q4 [15]}}