| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Consecutive non-overlapping periods |
| Difficulty | Standard +0.8 This is a multi-part S2 question requiring understanding of Poisson properties (scaling, memoryless property, conditional probability, and normal approximation). Part (a) is routine, but parts (b)-(d) require conceptual understanding beyond standard recall—particularly the memoryless property in (b), conditional probability with independent Poisson processes in (c), and normal approximation setup in (d). The question tests multiple sophisticated concepts but remains within standard S2 scope without requiring novel insight. |
| Spec | 2.04d Normal approximation to binomial5.02i 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/Working | Marks | Guidance |
| \(X \sim Po(3.2)\) | B1 | Writing or using \(Po(3.2)\) |
| \(P(X=3) = \frac{e^{-3.2}3.2^3}{3!}\) | M1 | \(\frac{e^{-\lambda}\lambda^3}{3!}\) |
| \(= 0.2226\) awrt \(0.223\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y \sim Po(1.6)\) | B1 | Writing or using \(Po(1.6)\) |
| \(P(Y \geq 1) = 1 - P(Y=0)\) | M1 | \(1-P(Y=0)\) or \(1-e^{-\lambda}\) |
| \(= 1 - e^{-1.6}\) | ||
| \(= 0.7981\) awrt \(0.798\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim Po(0.8)\) | ||
| \(\frac{P(X=1) \times P(X=3)}{P(Y=4)} = \frac{(e^{-0.8} \times 0.8) \times \left(\frac{e^{-0.8}0.8^3}{3!}\right)}{\frac{e^{-1.6}1.6^4}{4!}}\) | M1 M1 | 1st M1 using \(Po(0.8)\) with \(X=1\) or \(X=3\) (may be implied by \(0.359...\) or \(0.0383...\)); 2nd M1 \((e^{-\lambda}\times\lambda)\times\left(\frac{e^{-\lambda}\lambda^3}{3!}\right)\) consistent lambda, awrt \(0.0138\) implies 1st 2 M marks |
| M1 A1 | 3rd M1 correct use of conditional probability with denominator \(= \frac{e^{-1.6}1.6^4}{4!}\); 1st A1 fully correct expression | |
| \(= \frac{0.3594 \times 0.0383}{0.05513}\) | ||
| \(= 0.25\) | A1 | 2nd A1 \(0.25\) (allow awrt \(0.250\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A \sim Po(72)\) approximated by \(N(72,72)\) | B1 | Writing or using \(N(72,72)\) |
| \(\frac{5000}{60} = 83.33\) | M1 | 1st M1 for exact fraction or awrt \(83.3\) (may be implied by \(84\)); Note: Use of \(N(4320,4320)\) can score B1 and 1st M1 |
| \(P(A \geq 84) = P\left(Z \geq \frac{83.5-72}{\sqrt{72}}\right)\) | M1 M1 | 2nd M1 Using \(84 +/- 0.5\); 3rd M1 standardising using \(82.5, 83, 83.\dot{3}\) (awrt \(83.3\)), \(83.5, 83.8, 84\) or \(84.5\), 'their mean' and 'their sd' |
| \(= P(Z \geq 1.355...)\) | ||
| \(= 0.0869\) awrt \(0.087/0.088\) | A1 |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim Po(3.2)$ | B1 | Writing or using $Po(3.2)$ |
| $P(X=3) = \frac{e^{-3.2}3.2^3}{3!}$ | M1 | $\frac{e^{-\lambda}\lambda^3}{3!}$ |
| $= 0.2226$ awrt $0.223$ | A1 | |
**(3 marks)**
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y \sim Po(1.6)$ | B1 | Writing or using $Po(1.6)$ |
| $P(Y \geq 1) = 1 - P(Y=0)$ | M1 | $1-P(Y=0)$ or $1-e^{-\lambda}$ |
| $= 1 - e^{-1.6}$ | | |
| $= 0.7981$ awrt $0.798$ | A1 | |
**(3 marks)**
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim Po(0.8)$ | | |
| $\frac{P(X=1) \times P(X=3)}{P(Y=4)} = \frac{(e^{-0.8} \times 0.8) \times \left(\frac{e^{-0.8}0.8^3}{3!}\right)}{\frac{e^{-1.6}1.6^4}{4!}}$ | M1 M1 | 1st M1 using $Po(0.8)$ with $X=1$ or $X=3$ (may be implied by $0.359...$ or $0.0383...$); 2nd M1 $(e^{-\lambda}\times\lambda)\times\left(\frac{e^{-\lambda}\lambda^3}{3!}\right)$ consistent lambda, awrt $0.0138$ implies 1st 2 M marks |
| | M1 A1 | 3rd M1 correct use of conditional probability with denominator $= \frac{e^{-1.6}1.6^4}{4!}$; 1st A1 fully correct expression |
| $= \frac{0.3594 \times 0.0383}{0.05513}$ | | |
| $= 0.25$ | A1 | 2nd A1 $0.25$ (allow awrt $0.250$) |
**(5 marks)**
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A \sim Po(72)$ approximated by $N(72,72)$ | B1 | Writing or using $N(72,72)$ |
| $\frac{5000}{60} = 83.33$ | M1 | 1st M1 for exact fraction **or** awrt $83.3$ (may be implied by $84$); Note: Use of $N(4320,4320)$ can score B1 and 1st M1 |
| $P(A \geq 84) = P\left(Z \geq \frac{83.5-72}{\sqrt{72}}\right)$ | M1 M1 | 2nd M1 Using $84 +/- 0.5$; 3rd M1 standardising using $82.5, 83, 83.\dot{3}$ (awrt $83.3$), $83.5, 83.8, 84$ or $84.5$, 'their mean' **and** 'their sd' |
| $= P(Z \geq 1.355...)$ | | |
| $= 0.0869$ awrt $0.087/0.088$ | A1 | |
**(5 marks)**
---\begin{enumerate}
\item The number of cars caught speeding per day, by a particular camera, has a Poisson distribution with mean 0.8\\
(a) Find the probability that in a given 4 day period exactly 3 cars will be caught speeding by this camera.
\end{enumerate}
A car has been caught speeding by this camera.\\
(b) Find the probability that the period of time that elapses before the next car is caught speeding by this camera is less than 48 hours.
Given that 4 cars were caught speeding by this camera in a two day period,\\
(c) find the probability that 1 was caught on the first day and 3 were caught on the second day.
Each car that is caught speeding by this camera is fined $\pounds 60$\\
(d) Using a suitable approximation, find the probability that, in 90 days, the total amount of fines issued will be more than $\pounds 5000$\\
\hfill \mbox{\textit{Edexcel S2 2015 Q1 [16]}}