| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Validity of Poisson model |
| Difficulty | Standard +0.3 This is a straightforward multi-part Poisson question requiring standard techniques: direct probability calculations, scaling the rate parameter for different time periods, normal approximation for large λ, and contextual discussion of assumptions. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02i 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 = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (A) \(P(X=1) = 0.1712 - 0.0408 = 0.1304\) OR \(= e^{-3.2}\frac{3.2^1}{1!} = 0.1304\) | M1 A1 | M1 for tables; A1 (2 s.f. WWW) |
| (B) \(P(X \geq 6) = 1 - P(X \leq 5) = 1 - 0.8946 = 0.1054\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (A) \(\lambda = 3.2 \div 5 = 0.64\) | B1 | For mean (SOI) |
| \(P(X=1) = e^{-0.64}\frac{0.64^1}{1!} = 0.3375\) | M1 A1 | |
| (B) P(exactly one in each of 5 mins) \(= 0.3375^5 = 0.004379\) | B1 | FT to at least 2 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mean in 1 hour \(= 12 \times 3.2 = 38.4\); \(X \sim N(38.4, 38.4)\) | B1 | Normal approx. with correct parameters (SOI) |
| \(P(X \leq 45.5) = P\!\left(Z \leq \frac{45.5 - 38.4}{\sqrt{38.4}}\right)\) | B1 | For continuity correction |
| \(= P(Z \leq 1.146) = \Phi(1.146) = 0.874\) (3 s.f.) | M1 A1 | M1 for probability using correct tail; A1 CAO (but FT 44.5 or omitted CC) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (A) Suitable arguments for/against each assumption | E1, E1 | |
| (B) Suitable arguments for/against each assumption | E1, E1 |
# Question 3:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| (A) $P(X=1) = 0.1712 - 0.0408 = 0.1304$ OR $= e^{-3.2}\frac{3.2^1}{1!} = 0.1304$ | M1 A1 | M1 for tables; A1 (2 s.f. WWW) |
| (B) $P(X \geq 6) = 1 - P(X \leq 5) = 1 - 0.8946 = 0.1054$ | M1 A1 | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| (A) $\lambda = 3.2 \div 5 = 0.64$ | B1 | For mean (SOI) |
| $P(X=1) = e^{-0.64}\frac{0.64^1}{1!} = 0.3375$ | M1 A1 | |
| (B) P(exactly one in each of 5 mins) $= 0.3375^5 = 0.004379$ | B1 | FT to at least 2 s.f. |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean in 1 hour $= 12 \times 3.2 = 38.4$; $X \sim N(38.4, 38.4)$ | B1 | Normal approx. with correct parameters (SOI) |
| $P(X \leq 45.5) = P\!\left(Z \leq \frac{45.5 - 38.4}{\sqrt{38.4}}\right)$ | B1 | For continuity correction |
| $= P(Z \leq 1.146) = \Phi(1.146) = 0.874$ (3 s.f.) | M1 A1 | M1 for probability using correct tail; A1 CAO (but FT 44.5 or omitted CC) |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| (A) Suitable arguments for/against each assumption | E1, E1 | |
| (B) Suitable arguments for/against each assumption | E1, E1 | |
---3 The number of calls received at an office per 5 minutes is modelled by a Poisson distribution with mean 3.2.
\begin{enumerate}[label=(\roman*)]
\item Find the probability of\\
(A) exactly one call in a 5 -minute period,\\
(B) at least 6 calls in a 5 -minute period.
\item Find the probability of\\
(A) exactly one call in a 1 -minute period,\\
(B) exactly one call in each of five successive 1-minute periods.
\item Use a suitable approximating distribution to find the probability of at most 45 calls in a period of 1 hour.
Two assumptions required for a Poisson distribution to be a suitable model are that calls arrive
\begin{itemize}
\item at a uniform average rate,
\item independently of each other.
\item Comment briefly on the validity of each of these assumptions if the office is\\
(A) the enquiry department of a bank,\\
(B) a police emergency control room.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2007 Q3 [16]}}