| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with clear rate adjustments across parts. Part (a) involves direct probability calculations, part (b) combines Poisson with binomial (standard S2 technique), and part (c) requires expected value calculation using probabilities. All steps are routine for S2 students with no novel insights required, making it slightly easier than average. |
| 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 modelling5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(F < 3 \ | F \sim Po(1.5)) = 0.8088\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(F \geq 6) = 1 - P(F \leq 5)\) or \(1 - 0.9955\) | M1 | Writing or using \(1 - P(F \leq 5)\) |
| \(= 0.0045\) | A1 | awrt 0.0045 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R \sim Po(10)\), \(P(R \leq 12) = 0.7916\) | M1 | For 0.792 or better |
| \(X \sim B(15, \text{"0.7916"})\) | M1 | May be implied by a fully correct method for \(P(X=10)\) |
| \(P(X=10) = {}^{15}C_{10}(\text{"0.7916"})^{10}(1-\text{"0.7916"})^5\) | M1 | A correct method to find \(P(X=10)\) using a binomial distribution (implied by awrt 0.114) |
| \(= 0.11405\ldots\) | A1 | awrt 0.114 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H \sim Po(0.4)\) | M1 | Writing or using \(Po(0.4)\) e.g. \(P(H=1) = 0.268\ldots\) |
| \(P(H=0) = e^{-0.4} = 0.6703\ldots\) or \(P(H>0) = 1 - e^{-0.4} = 0.32967\ldots\) | M1 | Correct method to find \(P(H=0)\) or \(P(H>0)\). May be implied by awrt 0.67 or \(1 -\) awrt 0.67 |
| \(\text{Profit} = 2.4 \times \text{"0.6703"} - 3 \times \text{"0.32967"}\) | dM1 | Dependent on previous M. Correct method to find the profit. Allow \(7.4 \times \text{"0.6703"} + 2 \times \text{"0.32967"} - 5\) |
| \(= 0.6197\) | A1 | awrt 0.62. Allow 62p |
# Question 1:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(F < 3 \| F \sim Po(1.5)) = 0.8088$ | B1 | awrt 0.809 |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(F \geq 6) = 1 - P(F \leq 5)$ or $1 - 0.9955$ | M1 | Writing or using $1 - P(F \leq 5)$ |
| $= 0.0045$ | A1 | awrt 0.0045 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R \sim Po(10)$, $P(R \leq 12) = 0.7916$ | M1 | For 0.792 or better |
| $X \sim B(15, \text{"0.7916"})$ | M1 | May be implied by a fully correct method for $P(X=10)$ |
| $P(X=10) = {}^{15}C_{10}(\text{"0.7916"})^{10}(1-\text{"0.7916"})^5$ | M1 | A correct method to find $P(X=10)$ using a binomial distribution (implied by awrt 0.114) |
| $= 0.11405\ldots$ | A1 | awrt 0.114 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H \sim Po(0.4)$ | M1 | Writing or using $Po(0.4)$ e.g. $P(H=1) = 0.268\ldots$ |
| $P(H=0) = e^{-0.4} = 0.6703\ldots$ **or** $P(H>0) = 1 - e^{-0.4} = 0.32967\ldots$ | M1 | Correct method to find $P(H=0)$ or $P(H>0)$. May be implied by awrt 0.67 or $1 -$ awrt 0.67 |
| $\text{Profit} = 2.4 \times \text{"0.6703"} - 3 \times \text{"0.32967"}$ | dM1 | Dependent on previous M. Correct method to find the profit. Allow $7.4 \times \text{"0.6703"} + 2 \times \text{"0.32967"} - 5$ |
| $= 0.6197$ | A1 | awrt 0.62. Allow 62p |
---\begin{enumerate}
\item Bhavna produces rolls of cloth. She knows that faults occur randomly in her cloth at a mean rate of 1.5 every 15 metres.\\
(a) Find the probability that in 15 metres of her cloth there are\\
(i) less than 3 faults,\\
(ii) at least 6 faults.
\end{enumerate}
Each roll contains 100 metres of Bhavna's cloth.\\
She selects 15 rolls at random.\\
(b) Find the probability that exactly 10 of these rolls each have fewer than 13 faults.
Bhavna decides to sell her cloth in pieces.\\
Each piece of her cloth is 4 metres long.\\
The cost to make each piece is $\pounds 5.00$\\
She sells each piece of her cloth that contains no faults for $\pounds 7.40$\\
She sells each piece of her cloth that contains faults for $\pounds 2.00$\\
(c) Find the expected profit that Bhavna will make on each piece of her cloth that she sells.
\hfill \mbox{\textit{Edexcel S2 2022 Q1 [11]}}