| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single period normal approximation - scaled period (normal approximation only) |
| Difficulty | Moderate -0.3 This is a straightforward S2 Poisson question testing standard techniques: (a) direct Poisson probability calculation, (b) scaling the parameter and using cumulative probabilities, (c) normal approximation to Poisson. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-part structure and need for normal approximation. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=1) = 0.25e^{-0.25} = 0.1947\) | M1A1 | \(0.25e^{-0.25}\) oe; awrt \(0.195\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim Po(1.5)\) | B1 | Stating or using \(Po(1.5)\) |
| \(P(X>2) = 1 - P(X\leq 2)\) | M1 | Stating or using \(1 - P(X\leq 2)\) |
| \(= 1 - 0.8088\) | ||
| \(= 0.1912\) | A1 | awrt \(0.191\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([\lambda = 300\times 0.25 = 75]\) | ||
| \(X \sim N(75, 75)\) | B1 B1 | 1st B1: normal approximation and correct mean; 2nd B1: \(\text{Var}(X) = 75\) or \(\text{sd} = \sqrt{75}\) or awrt \(8.66\) (may be given if correct in standardisation formula) |
| \(P(X < 90) = P\!\left(X \leq \frac{89.5-75}{\sqrt{75}}\right)\) | M1M1 | 1st M1: using either \(89.5\) or \(88.5\); 2nd M1: standardising using their mean and sd, using \([89.5, 88.5\) or \(89]\) and finding correct area |
| \(= P(Z \leq 1.6743..)\) | ||
| \(=\) awrt \(0.953\) or \(0.952\) | A1 | NB use of Poisson gives \(0.9498\) and gains no marks |
# Question 2:
## Part 2(a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=1) = 0.25e^{-0.25} = 0.1947$ | M1A1 | $0.25e^{-0.25}$ oe; awrt $0.195$ |
## Part 2(b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim Po(1.5)$ | B1 | Stating or using $Po(1.5)$ |
| $P(X>2) = 1 - P(X\leq 2)$ | M1 | Stating or using $1 - P(X\leq 2)$ |
| $= 1 - 0.8088$ | | |
| $= 0.1912$ | A1 | awrt $0.191$ |
## Part 2(c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[\lambda = 300\times 0.25 = 75]$ | | |
| $X \sim N(75, 75)$ | B1 B1 | 1st B1: normal approximation and correct mean; 2nd B1: $\text{Var}(X) = 75$ or $\text{sd} = \sqrt{75}$ or awrt $8.66$ (may be given if correct in standardisation formula) |
| $P(X < 90) = P\!\left(X \leq \frac{89.5-75}{\sqrt{75}}\right)$ | M1M1 | 1st M1: using either $89.5$ or $88.5$; 2nd M1: standardising using their mean and sd, using $[89.5, 88.5$ or $89]$ and finding correct area |
| $= P(Z \leq 1.6743..)$ | | |
| $=$ awrt $0.953$ or $0.952$ | A1 | **NB** use of Poisson gives $0.9498$ and gains no marks |\begin{enumerate}
\item The number of defects per metre in a roll of cloth has a Poisson distribution with mean 0.25
\end{enumerate}
Find the probability that\\
(a) a randomly chosen metre of cloth has 1 defect,\\
(b) the total number of defects in a randomly chosen 6 metre length of cloth is more than 2
A tailor buys 300 metres of cloth.\\
(c) Using a suitable approximation find the probability that the tailor's cloth will contain less than 90 defects.
\hfill \mbox{\textit{Edexcel S2 2013 Q2 [10]}}