| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | Specimen |
| 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 standard Poisson calculations (part a), understanding of exponential distribution for waiting times (part b), conditional probability with Poisson (part c), and normal approximation to Poisson for a sum (part d). While it tests multiple concepts and requires careful setup, all techniques are standard S2 material with no novel insights needed. The conditional probability in part (c) and the approximation in part (d) elevate it slightly above average difficulty. |
| 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 | Mark | Guidance |
| Using \(Po(0.8)\) with \(X=1\) or \(X=3\) | M1 | May be implied by 0.359... or 0.0383... |
| \(\left(e^{-\lambda} \times \lambda\right) \times \left(\frac{e^{-\lambda}\lambda^3}{3!}\right)\) | M1 | Consistent lambda, awrt 0.0138 implies first 2 M marks |
| Correct use of conditional probability with denominator \(= \frac{e^{-1.6}1.6^4}{4!}\) | M1 | |
| Fully correct expression | A1 | |
| \(0.25\) | A1 | Allow awrt 0.250 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Writing or using \(N(72, 72)\) | B1 | |
| For exact fraction or awrt 83.3 | M1 | May be implied by 84; Use of \(N(4320, 4320)\) can score B1 and 1st M1 |
| Using \(84 +/- 0.5\) | M1 | |
| Standardising using \(82.5, 83, 83.\dot{3}\), \(83.5, 83.8, 84\) or \(84.5\) | M1 | Using 'their mean' and 'their sd' |
# Question 1:
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $Po(0.8)$ with $X=1$ or $X=3$ | M1 | May be implied by 0.359... or 0.0383... |
| $\left(e^{-\lambda} \times \lambda\right) \times \left(\frac{e^{-\lambda}\lambda^3}{3!}\right)$ | M1 | Consistent lambda, awrt 0.0138 implies first 2 M marks |
| Correct use of conditional probability with denominator $= \frac{e^{-1.6}1.6^4}{4!}$ | M1 | |
| Fully correct expression | A1 | |
| $0.25$ | A1 | Allow awrt 0.250 |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Writing or using $N(72, 72)$ | B1 | |
| For exact fraction **or** awrt 83.3 | M1 | May be implied by 84; Use of $N(4320, 4320)$ can score B1 and 1st M1 |
| Using $84 +/- 0.5$ | M1 | |
| Standardising using $82.5, 83, 83.\dot{3}$, $83.5, 83.8, 84$ or $84.5$ | M1 | Using 'their mean' **and** 'their sd' |
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\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 £60\\
(d) Using a suitable approximation, find the probability that, in 90 days, the total amount of fines issued will be more than $\pounds 5000$\\
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\hfill \mbox{\textit{Edexcel S2 2018 Q1 [16]}}