| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Mean-variance comparison for Poisson validation |
| Difficulty | Moderate -0.3 This is a straightforward Further Statistics question testing standard Poisson distribution properties. Parts (a)-(c) require textbook recall of assumptions and mean-variance equality. Parts (d)-(e) involve routine probability calculations. Part (f) requires basic understanding of how increased clustering affects probabilities. While it's Further Maths content, the question demands no novel insight or complex multi-step reasoning—just systematic application of well-rehearsed techniques. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Value of \(F\) | 0 | 1 | \(\geqslant 2\) |
| Probability | 0.200 |
| Answer | Marks | Guidance |
|---|---|---|
| Failures occur to no fixed pattern/are not predictable | B1 [1] | OE. *NOT* "independent" |
| Answer | Marks | Guidance |
|---|---|---|
| Failures occur independently of one another | B1 | Not recoverable from (a) if independence not restated here; must be contextualised |
| and at constant average rate | B1 [2] | Ignore "singly". Allow "uniform" rate, not "constant rate" or "constant probability"; must be contextualised |
| Answer | Marks | Guidance |
|---|---|---|
| Variance \((1.6384) \approx\) mean; So suggests that it is likely to be well modelled | M1, A1 [2] | Compare variance (or SD). Allow square/square-root confusion; Correct comparison and conclusion, \(1.64\) or better seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-1.61}\) | B1 [1] | Exact needed, allow even if \(0!\) or \(1.61^0\) or both left in |
| Answer | Marks |
|---|---|
| \(1\) | \(\geq 2\) |
| \(0.322\) | \(0.478\) |
| B1, B1 [2] | One correct e.g. \(0.3218\); Other correct e.g. \(0.4783\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(F=1)\) will be smaller as single failures are less likely | B1\*, depB1 [2] | OE. Partial answer: B1 |
# Question 6:
## Part (a)
Failures occur to no fixed pattern/are not predictable | **B1** [1] | OE. *NOT* "independent"
## Part (b)
Failures occur independently of one another | **B1** | Not recoverable from **(a)** if independence not restated here; must be contextualised
and at constant average rate | **B1** [2] | Ignore "singly". Allow "uniform" rate, not "constant rate" or "constant probability"; must be contextualised
## Part (c)
Variance $(1.6384) \approx$ mean; So suggests that it is likely to be well modelled | **M1, A1** [2] | Compare variance (or SD). Allow square/square-root confusion; Correct comparison and conclusion, $1.64$ or better seen
## Part (d)
$e^{-1.61}$ | **B1** [1] | Exact needed, allow even if $0!$ or $1.61^0$ or both left in
## Part (e)
| $1$ | $\geq 2$ |
|---|---|
| $0.322$ | $0.478$ |
**B1, B1** [2] | One correct e.g. $0.3218$; Other correct e.g. $0.4783$
## Part (f)
$P(F=1)$ will be smaller as single failures are less likely | **B1\*, depB1** [2] | OE. Partial answer: B1
---
6 A statistician investigates the number, $F$, of signal failures per week on a railway network.
\begin{enumerate}[label=(\alph*)]
\item The statistician assumes that signal failures occur randomly.
Explain what this statement means.
\item State two further assumptions needed for $F$ to be well modelled by a Poisson distribution.
In a random sample of 50 weeks, the statistician finds that the mean number of failures per week is 1.61, with standard deviation 1.28.
\item Explain whether this suggests that $F$ is likely to be well modelled by a Poisson distribution.
Assume first that $F \sim \operatorname { Po } ( 1.61 )$.
\item Write down an exact expression for $\mathrm { P } ( F = 0 )$.
\item Complete the table in the Printed Answer Booklet to show the probabilities of different values of $F$, correct to three significant figures.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
Value of $F$ & 0 & 1 & $\geqslant 2$ \\
\hline
Probability & 0.200 & & \\
\hline
\end{tabular}
\end{center}
After further investigation, the statistician decides to use a different model for the distribution of $F$. In this model it is now assumed that $\mathrm { P } ( F = 0 )$ is still 0.200 , but that if one failure occurs, there is an increased probability that further failures occur.
\item Explain the effect of this assumption on the value of $\mathrm { P } ( F = 1 )$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2020 Q6 [10]}}