| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding minimum n for P(X≥k) threshold |
| Difficulty | Standard +0.3 This is a standard S2 Poisson distribution question with straightforward applications: direct probability calculations, scaling the parameter, normal approximation, and solving an equation numerically. All techniques are routine for this module, though part (iv)(b) requires some careful numerical work. Slightly above average due to the multi-part nature and the numerical verification requirement, but no novel insight needed. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: 1 − P(≤ 7) = 1 − 0.9881 | Marks: M1 | Guidance: Allow for 0.0038 or 0.0335 |
| Answer/Working: = 0.0119 | Marks: A1, 2 | Guidance: Answer, a.r.t. 0.0119 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Po(12) | Marks: M1 | Guidance: Po(12) stated or implied |
| Answer/Working: P(≤ 14) − P(≤ 12) [0.7720 − 0.5760] | Marks: M1 | Guidance: Formula, 2 consecutive correct terms, or tables, e.g. .0905 or .3104 or .1629 |
| Answer/Working: = 0.196 | Marks: A1, 3 | Guidance: Answer, a.r.t. 0.196 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Po(60) ≈ N(60, 60) | Marks: M1 | Guidance: |
| Answer/Working: \(\Phi\left(\frac{69.5-60}{\sqrt{60}}\right) = \Phi(1.226)\) | Marks: A1 | Guidance: Variance or SD 60; Standardise, λ & \(\sqrt{\lambda}\), allow λ or wrong or no cc |
| Answer/Working: = 0.8899 | Marks: A1, 5 | Guidance: \(\sqrt{\lambda}\) and cc both correct; Answer 0.89 or a.r.t. 0.890 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: 1 − e^−λ(1 + 3m) | Marks: M1 | Guidance: M1 for one error, e.g. no "1 −" or extra term, or \(0^{th}\) term missing; answer_aesf |
| Answer/Working: | Marks: A1, 2 | Guidance: |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: m = 1.29, p = 0.89842 | Marks: A1 | Guidance: Substitute 1.29 or 1.3 into appropriate fn |
| Answer/Working: m = 1.3, p = 0.9008 | Marks: A1 | Guidance: |
| Comp | 0.9 | 0.1 |
| 1.29 | 0.898 | 0.10158 |
| 1.3 | 0.901 | 0.09918 |
| Answer/Working: Straddles 0.9, therefore solution between 1.29 and 1.3 | Marks: A1, 4 | Guidance: Explicit comparison with relevant value, & conclusion, needs both ps correct |
| Answer/Working: Method for iteration; 1.296... 1.2965 or better; conclusion stated | Marks: M1A1 | Guidance: Can be implied by at least 1.296...; Need at least 4 dp for M1A2 |
### (i)
**Answer/Working:** 1 − P(≤ 7) = 1 − 0.9881 | **Marks:** M1 | **Guidance:** Allow for 0.0038 or 0.0335
**Answer/Working:** = 0.0119 | **Marks:** A1, 2 | **Guidance:** Answer, a.r.t. 0.0119
### (ii)
**Answer/Working:** Po(12) | **Marks:** M1 | **Guidance:** Po(12) stated or implied
**Answer/Working:** P(≤ 14) − P(≤ 12) [0.7720 − 0.5760] | **Marks:** M1 | **Guidance:** Formula, 2 consecutive correct terms, or tables, e.g. .0905 or .3104 or .1629
**Answer/Working:** = 0.196 | **Marks:** A1, 3 | **Guidance:** Answer, a.r.t. 0.196
### (iii)
**Answer/Working:** Po(60) ≈ N(60, 60) | **Marks:** M1 | **Guidance:**
**Answer/Working:** $\Phi\left(\frac{69.5-60}{\sqrt{60}}\right) = \Phi(1.226)$ | **Marks:** A1 | **Guidance:** Variance or SD 60; Standardise, λ & $\sqrt{\lambda}$, allow λ or wrong or no cc
**Answer/Working:** = 0.8899 | **Marks:** A1, 5 | **Guidance:** $\sqrt{\lambda}$ and cc both correct; Answer 0.89 or a.r.t. 0.890
### (iv)
#### (a)
**Answer/Working:** 1 − e^−λ(1 + 3m) | **Marks:** M1 | **Guidance:** M1 for one error, e.g. no "1 −" or extra term, or $0^{th}$ term missing; answer_aesf
**Answer/Working:** | **Marks:** A1, 2 | **Guidance:**
#### (b)
**Answer/Working:** m = 1.29, p = 0.89842 | **Marks:** A1 | **Guidance:** Substitute 1.29 or 1.3 into appropriate fn
**Answer/Working:** m = 1.3, p = 0.9008 | **Marks:** A1 | **Guidance:**
| Comp | 0.9 | 0.1 | 0 |
|------|-----|-----|---------|
| 1.29 | 0.898 | 0.10158 | −0.0158 |
| 1.3 | 0.901 | 0.09918 | .0008146 |
**Answer/Working:** Straddles 0.9, therefore solution between 1.29 and 1.3 | **Marks:** A1, 4 | **Guidance:** Explicit comparison with relevant value, & conclusion, needs both ps correct
**Answer/Working:** Method for iteration; 1.296... 1.2965 or better; conclusion stated | **Marks:** M1A1 | **Guidance:** Can be implied by at least 1.296...; Need at least 4 dp for M1A2Buttercups in a meadow are distributed independently of one another and at a constant average incidence of 3 buttercups per square metre.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that in 1 square metre there are more than 7 buttercups. [2]
\item Find the probability that in 4 square metres there are either 13 or 14 buttercups. [3]
\item Use a suitable approximation to find the probability that there are no more than 69 buttercups in 20 square metres. [5]
\item \begin{enumerate}[label=(\alph*)]
\item Without using an approximation, find an expression for the probability that in $m$ square metres there are at least 2 buttercups. [2]
\item It is given that the probability that there are at least 2 buttercups in $m$ square metres is 0.9. Using your answer to part (a), show numerically that $m$ lies between 1.29 and 1.3. [4]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2010 Q9 [16]}}