OCR Further Statistics AS 2020 November — Question 6 10 marks

Exam BoardOCR
ModuleFurther Statistics AS (Further Statistics AS)
Year2020
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeMean-variance comparison for Poisson validation
DifficultyModerate -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.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities

6 A statistician investigates the number, \(F\), of signal failures per week on a railway network.
  1. The statistician assumes that signal failures occur randomly. Explain what this statement means.
  2. 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.
  3. Explain whether this suggests that \(F\) is likely to be well modelled by a Poisson distribution. Assume first that \(F \sim \operatorname { Po } ( 1.61 )\).
  4. Write down an exact expression for \(\mathrm { P } ( F = 0 )\).
  5. Complete the table in the Printed Answer Booklet to show the probabilities of different values of \(F\), correct to three significant figures.
    Value of \(F\)01\(\geqslant 2\)
    Probability0.200
    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.
  6. Explain the effect of this assumption on the value of \(\mathrm { P } ( F = 1 )\).

Question 6:
Part (a)
AnswerMarks Guidance
Failures occur to no fixed pattern/are not predictableB1 [1] OE. *NOT* "independent"
Part (b)
AnswerMarks Guidance
Failures occur independently of one anotherB1 Not recoverable from (a) if independence not restated here; must be contextualised
and at constant average rateB1 [2] Ignore "singly". Allow "uniform" rate, not "constant rate" or "constant probability"; must be contextualised
Part (c)
AnswerMarks Guidance
Variance \((1.6384) \approx\) mean; So suggests that it is likely to be well modelledM1, A1 [2] Compare variance (or SD). Allow square/square-root confusion; Correct comparison and conclusion, \(1.64\) or better seen
Part (d)
AnswerMarks Guidance
\(e^{-1.61}\)B1 [1] Exact needed, allow even if \(0!\) or \(1.61^0\) or both left in
Part (e)
AnswerMarks
\(1\)\(\geq 2\)
\(0.322\)\(0.478\)
B1, B1 [2]One correct e.g. \(0.3218\); Other correct e.g. \(0.4783\)
Part (f)
AnswerMarks Guidance
\(P(F=1)\) will be smaller as single failures are less likelyB1\*, 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

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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]}}