| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single scaled time period |
| Difficulty | Standard +0.3 This is a standard S2 Poisson distribution question requiring routine scaling of the rate parameter and straightforward probability calculations. Part (c) tests understanding of independence, (d) involves expected value with a simple inequality, and (e) is a standard hypothesis test. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02i 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 modelling5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([X = \text{number of faults in } 4\text{ m}^2 \text{ so } X \sim \text{Po}(3)]\) | ||
| \(P(X=5) = P(X \leqslant 5) - P(X \leqslant 4) = 0.9161 - 0.8153\) or \(\frac{e^{-3}3^5}{5!}\) | M1 | Allow \(\lambda\) instead of 3; accept letter \(\lambda\) or any value of \(\lambda\) |
| \(= 0.1008\) or \(0.100818...\) | A1 | awrt 0.101 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([Y = \text{number of faults in } 6\text{ m}^2]\), \(Y \sim \text{Po}(4.5)\) and \(P(Y>5) = 1 - P(Y \leqslant 5) = 1 - 0.7029\) | M1 | Writing or using Po(4.5) and sight of \([P(Y>5)] = 1 - P(Y \leqslant 5)\); implied by sight of \(1 - 0.7029\) |
| \(= 0.2971\) or (calc) \(0.29706956...\) | A1 | awrt 0.297 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 0.101 (or ft their answer to (a)) | B1ft | |
| Faults occur independently/randomly | B1 | 2nd B1: comment about faults occurring randomly/independently or Poisson has "no memory" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([F = \text{number of faults in a small rug}]\), \(F \sim \text{Po}(0.9)\) | B1 | Writing or using Po(0.9); may be implied by sight of 0.407 or 0.593 |
| \(e^{-0.9}n \times 80 + (1 - e^{-0.9})n \times 60 \geqslant 4000\) or (awrt 0.407)\(n \times 80\) + (awrt 0.593)\(n \times 60 \geqslant 4000\) | M1 | 1st M1: for \(e^{-\lambda}n \times 80 + (1-e^{-\lambda})n \times 60 \geqslant 4000\), any value for \(\lambda\); allow = 4000 |
| \(n \geqslant \frac{4000}{20e^{-0.9}+60} = 58.71...\) | M1 | 2nd M1: for solving leading to a positive value of \(n\); allow any \(\lambda\) and \(n = ...\) |
| \(n = \) 59 | A1 | For answer of 59 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \lambda = 9\), \(H_1: \lambda > 9\) | B1 | Both hypotheses correct with \(\lambda\) or \(\mu\); allow 3 or 0.75 or 0.9 instead of 9 |
| \(R \sim \text{Po}(``0.9"\times 10)\) and \([P(R \geqslant 13)] = 1 - P(R \leqslant 12) = 1 - 0.8758\) | M1 | 1st M1: writing or using Po("9") and using \(1 - P(R \leqslant 12)\); implied by \(1 - 0.8758\) or one of the listed values |
| \(P(R \leqslant 13) = 0.9261\) or \(P(R \geqslant 14) = 0.0739\) or \(P(R \leqslant 14) = 0.9585\) or \(P(R \geqslant 15) = 0.0415\) | ||
| \([P(R \geqslant 13)] = 0.1242\) awrt 0.124 or CR \(R \geqslant 15\) (oe) | A1 | 1st A1: probability = awrt 0.124 or CR of \(R \geqslant 15\) e.g. \(R > 14\) |
| So insufficient evidence to reject \(H_0\)/not significant/not in critical region | M1 | 2nd M1: correct conclusion based on their prob & 0.05 or their CR & 13; no contradicting conclusions |
| There is insufficient evidence that the rate at which faults occur is higher for Rhiannon | A1 | 2nd A1: dep on both Ms; correct contextual comment including words in bold |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[X = \text{number of faults in } 4\text{ m}^2 \text{ so } X \sim \text{Po}(3)]$ | | |
| $P(X=5) = P(X \leqslant 5) - P(X \leqslant 4) = 0.9161 - 0.8153$ or $\frac{e^{-3}3^5}{5!}$ | M1 | Allow $\lambda$ instead of 3; accept letter $\lambda$ or any value of $\lambda$ |
| $= 0.1008$ or $0.100818...$ | A1 | awrt **0.101** |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[Y = \text{number of faults in } 6\text{ m}^2]$, $Y \sim \text{Po}(4.5)$ and $P(Y>5) = 1 - P(Y \leqslant 5) = 1 - 0.7029$ | M1 | Writing or using Po(4.5) and sight of $[P(Y>5)] = 1 - P(Y \leqslant 5)$; implied by sight of $1 - 0.7029$ |
| $= 0.2971$ or (calc) $0.29706956...$ | A1 | awrt **0.297** |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| **0.101** (or ft their answer to (a)) | B1ft | |
| Faults occur **independently/randomly** | B1 | 2nd B1: comment about faults occurring **randomly/independently** or Poisson has "**no memory**" |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[F = \text{number of faults in a small rug}]$, $F \sim \text{Po}(0.9)$ | B1 | Writing or using Po(0.9); may be implied by sight of 0.407 or 0.593 |
| $e^{-0.9}n \times 80 + (1 - e^{-0.9})n \times 60 \geqslant 4000$ or (awrt 0.407)$n \times 80$ + (awrt 0.593)$n \times 60 \geqslant 4000$ | M1 | 1st M1: for $e^{-\lambda}n \times 80 + (1-e^{-\lambda})n \times 60 \geqslant 4000$, any value for $\lambda$; allow = 4000 |
| $n \geqslant \frac{4000}{20e^{-0.9}+60} = 58.71...$ | M1 | 2nd M1: for solving leading to a positive value of $n$; allow any $\lambda$ and $n = ...$ |
| $n = $ **59** | A1 | For answer of 59 only |
## Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \lambda = 9$, $H_1: \lambda > 9$ | B1 | Both hypotheses correct with $\lambda$ or $\mu$; allow 3 or 0.75 or 0.9 instead of 9 |
| $R \sim \text{Po}(``0.9"\times 10)$ and $[P(R \geqslant 13)] = 1 - P(R \leqslant 12) = 1 - 0.8758$ | M1 | 1st M1: writing or using Po("9") and using $1 - P(R \leqslant 12)$; implied by $1 - 0.8758$ or one of the listed values |
| $P(R \leqslant 13) = 0.9261$ or $P(R \geqslant 14) = 0.0739$ or $P(R \leqslant 14) = 0.9585$ or $P(R \geqslant 15) = 0.0415$ | | |
| $[P(R \geqslant 13)] = 0.1242$ awrt 0.124 or CR $R \geqslant 15$ (oe) | A1 | 1st A1: probability = awrt 0.124 or CR of $R \geqslant 15$ e.g. $R > 14$ |
| So insufficient evidence to reject $H_0$/not significant/not in critical region | M1 | 2nd M1: correct conclusion based on their prob & 0.05 or their CR & 13; no contradicting conclusions |
| There is insufficient evidence that the **rate** at which **faults** occur is higher for **Rhiannon** | A1 | 2nd A1: dep on both Ms; correct contextual comment including words in bold |
---\begin{enumerate}
\item Luis makes and sells rugs. He knows that faults occur randomly in his rugs at a rate of 3 every $4 \mathrm {~m} ^ { 2 }$\\
(a) Find the probability of there being exactly 5 faults in one of his rugs that is $4 \mathrm {~m} ^ { 2 }$ in size.\\
(b) Find the probability that there are more than 5 faults in one of his rugs that is $6 \mathrm {~m} ^ { 2 }$ in size.
\end{enumerate}
Luis makes a rug that is $4 \mathrm {~m} ^ { 2 }$ in size and finds it has exactly 5 faults in it.\\
(c) Write down the probability that the next rug that Luis makes, which is $4 \mathrm {~m} ^ { 2 }$ in size, will have exactly 5 faults. Give a reason for your answer.
A small rug has dimensions 80 cm by 150 cm . Faults still occur randomly at a rate of 3 every $4 \mathrm {~m} ^ { 2 }$\\
Luis makes a profit of $\pounds 80$ on each small rug he sells that contains no faults but a profit of $\pounds 60$ on any small rug he sells that contains faults.
Luis sells $n$ small rugs and expects to make a profit of at least $\pounds 4000$\\
(d) Calculate the minimum value of $n$
Luis wishes to increase the productivity of his business and employs Rhiannon. Faults also occur randomly in Rhiannon's rugs and independently to faults made by Luis. Luis randomly selects 10 small rugs made by Rhiannon and finds 13 faults.\\
(e) Test, at the $5 \%$ level of significance, whether or not there is evidence to support the suggestion that the rate at which faults occur is higher for Rhiannon than for Luis. State your hypotheses clearly.
\hfill \mbox{\textit{Edexcel S2 2021 Q2 [15]}}