AQA S2 2005 June — Question 1 7 marks

Exam BoardAQA
ModuleS2 (Statistics 2)
Year2005
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSum of Poisson processes
TypeMulti-period repeated application
DifficultyStandard +0.3 This is a straightforward Poisson distribution question requiring basic probability calculations and understanding that Poisson parameters scale with time. Part (a) is direct formula application, part (b)(i) requires knowing that λ scales linearly (2.6×5=13), and part (b)(ii) involves calculating P(Y≥15) then raising to the fourth power for independence. While multi-step, all techniques are standard S2 material with no novel insight required, making it slightly easier than average.
Spec5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson

1 The number of cars, \(X\), passing along a road each minute can be modelled by a Poisson distribution with a mean of 2.6.
  1. Calculate \(\mathrm { P } ( X = 2 )\).
    1. Write down the distribution of \(Y\), the number of cars passing along this road in a 5-minute interval.
    2. Hence calculate the probability that at least 15 cars pass along this road in each of four successive 5 -minute intervals.

Question 1:
Part (a)
AnswerMarks Guidance
\(P(X=2) = \frac{e^{-2.6}(2.6)^2}{2!} = 0.251\)M1, A1 (2 marks) \(0.5184 - 0.2674 = 0.251\)
Part (b)(i)
AnswerMarks Guidance
\(Y \sim P_o(13)\)B1 (1 mark) Must state Poisson and 13
Part (b)(ii)
AnswerMarks Guidance
\(P(Y \geq 15) = 1 - P(Y < 14) = 1 - 0.6751 = 0.3249 \approx 0.325\)M1, A1\(\checkmark\) (follow through on their \(\lambda\))
\(\therefore p = (0.3249)^4\)M\(\hat{1}\) On their \(p(Y \geq 15)\)
\(p = 0.0111\) to \(0.0112\)(4 marks total)
## Question 1:

### Part (a)
$P(X=2) = \frac{e^{-2.6}(2.6)^2}{2!} = 0.251$ | M1, A1 (2 marks) | $0.5184 - 0.2674 = 0.251$

### Part (b)(i)
$Y \sim P_o(13)$ | B1 (1 mark) | Must state Poisson **and** 13

### Part (b)(ii)
$P(Y \geq 15) = 1 - P(Y < 14) = 1 - 0.6751 = 0.3249 \approx 0.325$ | M1, A1$\checkmark$ (follow through on their $\lambda$)

$\therefore p = (0.3249)^4$ | M$\hat{1}$ | On their $p(Y \geq 15)$

$p = 0.0111$ to $0.0112$ | (4 marks total)

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1 The number of cars, $X$, passing along a road each minute can be modelled by a Poisson distribution with a mean of 2.6.
\begin{enumerate}[label=(\alph*)]
\item Calculate $\mathrm { P } ( X = 2 )$.
\item \begin{enumerate}[label=(\roman*)]
\item Write down the distribution of $Y$, the number of cars passing along this road in a 5-minute interval.
\item Hence calculate the probability that at least 15 cars pass along this road in each of four successive 5 -minute intervals.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA S2 2005 Q1 [7]}}