| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Basic sum of two Poissons |
| Difficulty | Standard +0.3 This is a straightforward application of standard Poisson distribution techniques from Further Statistics 1. Parts (a)-(c) involve routine calculations with independent Poisson processes (direct probability, sum of Poissons, scaling parameter). Parts (d)-(e) test the standard Poisson approximation to binomial with textbook conditions (large n, small p). All parts follow predictable patterns with no novel problem-solving required, making this slightly easier than an average A-level question overall. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02i 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 modelling5.02m Poisson: mean = variance = lambda5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim \text{Po}(2.6)\), \(Y \sim \text{Po}(1.2)\) | ||
| \(\text{P(each hire 2 in 1 hour)} = \text{P}(X=2) \times \text{P}(Y=2) = 0.25104\ldots \times 0.21685\ldots\) | M1 | For \(\text{P}(X=2) \times \text{P}(Y=2)\) from correct models (may be implied by correct answer) |
| \(= 0.05444\ldots\) awrt \(\mathbf{0.0544}\) | A1 | awrt 0.0544 |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(W = X + Y \rightarrow W \sim \text{Po}(3.8)\) | M1 | For combining Poisson distributions and use of Po(3.8) (may be implied by correct answer) |
| \(\text{P}(W=3) = 0.20458\ldots\) awrt \(\mathbf{0.205}\) | A1 | awrt 0.205 |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T \sim \text{Po}((2.6+1.2)\times 2)\) | M1 | For setting up new model and attempting mean of Poisson distribution (may be implied by correct answer) |
| \(\text{P}(T < 9) = 0.64819\ldots\) awrt \(\mathbf{0.648}\) | A1 | awrt 0.648 |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) Mean \(= np = \mathbf{2.4}\) | B1 | For 2.4 |
| (ii) Variance \(= np(1-p) = 2.3904\) awrt \(\mathbf{2.39}\) | B1 | For awrt 2.39 |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \([D \sim \text{Po}(2.4)\), \(\text{P}(D \leqslant 4)]= 0.9041\ldots\) awrt \(\mathbf{0.904}\) | B1 | For awrt 0.904 |
| (ii) Since \(n\) is large and \(p\) is small/mean is approximately equal to variance | B1 | For a correct explanation to support use of Poisson approximation in this case |
| (2 marks) |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim \text{Po}(2.6)$, $Y \sim \text{Po}(1.2)$ | | |
| $\text{P(each hire 2 in 1 hour)} = \text{P}(X=2) \times \text{P}(Y=2) = 0.25104\ldots \times 0.21685\ldots$ | M1 | For $\text{P}(X=2) \times \text{P}(Y=2)$ from correct models (may be implied by correct answer) |
| $= 0.05444\ldots$ awrt $\mathbf{0.0544}$ | A1 | awrt 0.0544 |
| **(2 marks)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $W = X + Y \rightarrow W \sim \text{Po}(3.8)$ | M1 | For combining Poisson distributions and use of Po(3.8) (may be implied by correct answer) |
| $\text{P}(W=3) = 0.20458\ldots$ awrt $\mathbf{0.205}$ | A1 | awrt 0.205 |
| **(2 marks)** | | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T \sim \text{Po}((2.6+1.2)\times 2)$ | M1 | For setting up new model and attempting mean of Poisson distribution (may be implied by correct answer) |
| $\text{P}(T < 9) = 0.64819\ldots$ awrt $\mathbf{0.648}$ | A1 | awrt 0.648 |
| **(2 marks)** | | |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** Mean $= np = \mathbf{2.4}$ | B1 | For 2.4 |
| **(ii)** Variance $= np(1-p) = 2.3904$ awrt $\mathbf{2.39}$ | B1 | For awrt 2.39 |
| **(2 marks)** | | |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $[D \sim \text{Po}(2.4)$, $\text{P}(D \leqslant 4)]= 0.9041\ldots$ awrt $\mathbf{0.904}$ | B1 | For awrt 0.904 |
| **(ii)** Since $n$ is large and $p$ is small/mean is approximately equal to variance | B1 | For a correct explanation to support use of Poisson approximation in this case |
| **(2 marks)** | | |
---\begin{enumerate}
\item Two car hire companies hire cars independently of each other.
\end{enumerate}
Car Hire A hires cars at a rate of 2.6 cars per hour.\\
Car Hire B hires cars at a rate of 1.2 cars per hour.\\
(a) In a 1 hour period, find the probability that each company hires exactly 2 cars.\\
(b) In a 1 hour period, find the probability that the total number of cars hired by the two companies is 3\\
(c) In a 2 hour period, find the probability that the total number of cars hired by the two companies is less than 9
On average, 1 in 250 new cars produced at a factory has a defect.\\
In a random sample of 600 new cars produced at the factory,\\
(d) (i) find the mean of the number of cars with a defect,\\
(ii) find the variance of the number of cars with a defect.\\
(e) (i) Use a Poisson approximation to find the probability that no more than 4 of the cars in the sample have a defect.\\
(ii) Give a reason to support the use of a Poisson approximation.
\section*{Q uestion 3 continued}
\hfill \mbox{\textit{Edexcel FS1 AS Q3 [10]}}