| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Standard Poisson approximation to binomial |
| Difficulty | Moderate -0.3 This is a straightforward application of recognizing when Poisson approximation to binomial is appropriate and performing standard probability calculations. Part (a) requires identifying binomial B(2000, 1/1500) and Poisson(4/3), which is textbook knowledge. Parts (b) and (c) involve routine calculator work with Poisson probabilities. No novel insight or complex multi-step reasoning is required, making it slightly easier than average but not trivial since it requires understanding of approximation conditions. |
| Spec | 5.02d Binomial: mean np and variance np(1-p)5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | Binomial. |
| Answer | Marks |
|---|---|
| λ = 4/3 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | For Binomial. Allow B(…) |
| Answer | Marks |
|---|---|
| (b) | P(X = 2) = 0.2343 (Poisson) or 0.2344 (Binomial) |
| Answer | Marks |
|---|---|
| = 0.1506 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | BC Allow answers from binomial or Poisson distributions |
| Answer | Marks |
|---|---|
| (c) | Use of B 20000, 1 |
| Answer | Marks |
|---|---|
| = 0.8552 allow 0.855www | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | 40 |
Question 2:
2 | (a) | Binomial.
n = 2000, p = 1/1500
Poisson
λ = 4/3 | M1
A1
M1
A1
[4] | 3.3
1.1
3.3
1.1 | For Binomial. Allow B(…)
For parameters correct to at least 3 s.f.
For Poisson. Allow Po(…). Do not allow P(…)
For parameter correct to at least 3 s.f.
(b) | P(X = 2) = 0.2343 (Poisson) or 0.2344 (Binomial)
P(X > 2) = 1 – 0.8494
= 0.1506 | B1
M1
A1
[3] | 1.1
1.1
1.1 | BC Allow answers from binomial or Poisson distributions
Allow 0.234www
BC
Allow 0.151www
(c) | Use of B 20000, 1
1500
Or P( ≥ 10 ) = 1 – 0.1448
= 0.8552 allow 0.855www | M1
A1
[2] | 1.1
1.1 | 40
Or use ofPo
3
P( ≥ 10 ) = 1 – 0.1449
= 0.8551 allow 0.855www2 Almost all plants of a particular species have red flowers. However on average 1 in every 1500 plants of this species have white flowers. A random sample of 2000 plants of this species is selected. The random variable $X$ represents the number of plants in the sample that have white flowers.
\begin{enumerate}[label=(\alph*)]
\item Name two distributions which could be used to model the distribution of $X$, stating the parameters of each of these distributions.
You may use either of the distributions you have named in the rest of this question.
\item Calculate each of the following.
\begin{itemize}
\item $\mathrm { P } ( X = 2 )$
\item $\mathrm { P } ( X > 2 )$
\item A random sample of 20000 plants of this species is selected.
\end{itemize}
Calculate the probability that there are at least 10 plants in the sample that have white flowers.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2019 Q2 [9]}}