| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Explain or apply conditions in context |
| Difficulty | Standard +0.3 This is a straightforward application of standard Poisson distribution techniques: basic probability calculations, normal approximation for large λ, and sum of independent Poisson variables. Part (iii) tests conceptual understanding of Poisson assumptions. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=0) = \frac{e^{-2.1} \cdot 2.1^0}{0!} = 0.1225\) | M1 | For calculation |
| \(= 0.1225\) | A1 | CAO; Allow 0.122 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.3796\) | M1 | M1 for use of correct structure; M0 for use of \(1 - P(X \leq 2)\) or \(1 - 0.6796\); Using \(\lambda = 2.0\) leading to \(1 - 0.4060\) gets M1 |
| \(= 0.6204\) | A1 | CAO; Allow 0.6203, 0.620 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New \(\lambda = 5 \times 2.1 = 10.5\) | B1 | For mean (SOI) |
| \(P(\text{Between 5 and 10 in 5 mins}) = 0.5207 - 0.0211\) | M1 | For \(P(X \leq 10) - P(X \leq 4)\) used; e.g. \(1 - 0.9379\) leads to B0M1A0 |
| \(= 0.4996\) | A1 | CAO; Allow 0.500, 0.50; Condone 0.5 www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean number in 60 minutes \(= 60 \times 2.1 = 126\) | ||
| Using Normal approx. to the Poisson, \(X \sim N(126, 126)\) | B1 | For Normal approximation |
| B1 | For correct parameters (SOI) | |
| \(P(X \geq 130) = P\!\left(Z \geq \frac{129.5 - 126}{\sqrt{126}}\right)\) | B1 | For correct continuity correction |
| \(= P(Z > 0.3118) = 1 - \Phi(0.3118)\) | M1 | For correct probability structure |
| \(= 1 - 0.6224 = 0.3776\) | A1 | CAO; Do not FT wrong or omitted CC; Allow 0.378www & 0.3775 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (Because if butterflies are blown in pairs,) the events will no longer be occurring singly | E1 | Accept 'not independent' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(3 \text{ or fewer}) = P(3 \text{ or fewer individuals and no pairs}) + P(0 \text{ or } 1 \text{ individual and } 1 \text{ pair})\) | M1 | For \(P(0 \text{ pairs}) (= 0.8187)\) |
| \(= (0.9068 \times 0.8187) + (0.4932 \times (0.9825 - 0.8187))\) | M1 | For \(P(1 \text{ pair}) (= 0.1638 \text{ or } 0.1637)\); First two M1s can be awarded for 0.9825 |
| \(= (0.9068 \times 0.8187) + (0.4932 \times 0.1638)\) | M2 | For structure M2 for correct 6 combinations identified and their probabilities added; M1 for 5 combinations identified and their probabilities added |
| \(= 0.7424 + 0.0808 = 0.8232\) | A1 | CAO; Allow awrt 0.823 |
# Question 2:
## Part (i)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=0) = \frac{e^{-2.1} \cdot 2.1^0}{0!} = 0.1225$ | M1 | For calculation |
| $= 0.1225$ | A1 | CAO; Allow 0.122 |
## Part (i)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.3796$ | M1 | M1 for use of correct structure; M0 for use of $1 - P(X \leq 2)$ or $1 - 0.6796$; Using $\lambda = 2.0$ leading to $1 - 0.4060$ gets M1 |
| $= 0.6204$ | A1 | CAO; Allow 0.6203, 0.620 |
## Part (i)(C)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New $\lambda = 5 \times 2.1 = 10.5$ | B1 | For mean (SOI) |
| $P(\text{Between 5 and 10 in 5 mins}) = 0.5207 - 0.0211$ | M1 | For $P(X \leq 10) - P(X \leq 4)$ used; e.g. $1 - 0.9379$ leads to B0M1A0 |
| $= 0.4996$ | A1 | CAO; Allow 0.500, 0.50; Condone 0.5 www |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean number in 60 minutes $= 60 \times 2.1 = 126$ | | |
| Using Normal approx. to the Poisson, $X \sim N(126, 126)$ | B1 | For Normal approximation |
| | B1 | For correct parameters (SOI) |
| $P(X \geq 130) = P\!\left(Z \geq \frac{129.5 - 126}{\sqrt{126}}\right)$ | B1 | For correct continuity correction |
| $= P(Z > 0.3118) = 1 - \Phi(0.3118)$ | M1 | For correct probability structure |
| $= 1 - 0.6224 = 0.3776$ | A1 | CAO; Do not FT wrong or omitted CC; Allow 0.378www & 0.3775 |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| (Because if butterflies are blown in pairs,) the events will no longer be occurring singly | E1 | Accept 'not independent' |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(3 \text{ or fewer}) = P(3 \text{ or fewer individuals and no pairs}) + P(0 \text{ or } 1 \text{ individual and } 1 \text{ pair})$ | M1 | For $P(0 \text{ pairs}) (= 0.8187)$ |
| $= (0.9068 \times 0.8187) + (0.4932 \times (0.9825 - 0.8187))$ | M1 | For $P(1 \text{ pair}) (= 0.1638 \text{ or } 0.1637)$; First two M1s can be awarded for 0.9825 |
| $= (0.9068 \times 0.8187) + (0.4932 \times 0.1638)$ | M2 | For structure M2 for correct 6 combinations identified and their probabilities added; M1 for 5 combinations identified and their probabilities added |
| $= 0.7424 + 0.0808 = 0.8232$ | A1 | CAO; Allow awrt 0.823 |
---2 John is observing butterflies being blown across a fence in a strong wind. He uses the Poisson distribution with mean 2.1 to model the number of butterflies he observes in one minute.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that John observes\\
(A) no butterflies in a minute,\\
(B) at least 2 butterflies in a minute,\\
(C) between 5 and 10 butterflies inclusive in a period of 5 minutes.
\item Use a suitable approximating distribution to find the probability that John observes at least 130 butterflies in a period of 1 hour.
In fact some of the butterflies John observes being blown across the fence are being blown in pairs.
\item Explain why this invalidates one of the assumptions required for a Poisson distribution to be a suitable model.
John decides to revise his model for the number of butterflies he observes in one minute. In this new model, the number of pairs of butterflies is modelled by the Poisson distribution with mean 0.2 , and the number of single butterflies is modelled by an independent Poisson distribution with mean 1.7.
\item Find the probability that John observes no more than 3 butterflies altogether in a period of one minute.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2013 Q2 [18]}}