| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Justify Poisson approximation only |
| Difficulty | Moderate -0.3 This is a straightforward application of standard Poisson approximation criteria (large n, small p, np moderate). Part (a)(i) requires stating textbook conditions, part (a)(ii) involves routine probability calculations, and part (b) is a standard negative binomial problem. While it requires understanding of when approximations are valid, it demands no novel insight or complex multi-step reasoning—slightly easier than average for Further Maths Statistics. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks |
|---|---|
| 2 | (a) |
| (i) | People with the antigen occur randomly and |
| Answer | Marks |
|---|---|
| a Poisson distribution is also appropriate | E1 |
| Answer | Marks |
|---|---|
| [3] | 2.4 |
| Answer | Marks |
|---|---|
| 2.4 | For partial explanation of binomial |
| Answer | Marks |
|---|---|
| 2 | (a) |
| (ii) | Binomial (1200, 0.00025) or Poisson (0.3) |
| Answer | Marks |
|---|---|
| P(X > 3) = 0.0003 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | NB both distributions give same |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | P(≤5000 needed) = 1 – P(0 or 1 in 5000 have antigen) |
| Answer | Marks |
|---|---|
| = 1 – 0.6446 = 0.3554 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
Question 2:
2 | (a)
(i) | People with the antigen occur randomly and
occurrences are independent with constant probability
0.00025.
The number of people with the antigen out of 1200 is
being counted, so a binomial distribution is
appropriate.
Because n (= 1200) is large and p (= 0.00025)is small
a Poisson distribution is also appropriate | E1
E1
E1
[3] | 2.4
2.4
2.4 | For partial explanation of binomial
For full explanation
For explanation of Poisson.
2 | (a)
(ii) | Binomial (1200, 0.00025) or Poisson (0.3)
P(X = 3) = 0.0033
P(X > 3) = 0.0003 | M1
A1
A1
[3] | 3.3
1.1
1.1 | NB both distributions give same
answer to 4 dp
2 | (b) | P(≤5000 needed) = 1 – P(0 or 1 in 5000 have antigen)
Use B(5000, 0.00025) or Po(1.25)
= 1 – 0.6446 = 0.3554 | M1
B1
A1
[3] | 3.3
3.4
1.12 On average 1 in 4000 people have a particular antigen in their blood (an antigen is a molecule which may cause an adverse reaction).
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item A random sample of 1200 people is selected. The random variable $X$ represents the number of people in the sample who have this antigen in their blood. Explain why you could use either a binomial distribution or a Poisson distribution to model the distribution of $X$.
\item Use either a binomial or a Poisson distribution to calculate each of the following probabilities.
\begin{itemize}
\end{enumerate}\item $\mathrm { P } ( X = 3 )$
\item $\mathrm { P } ( X > 3 )$
\item A researcher needs to find 2 people with the antigen. Find the probability that at most 5000 people have to be tested in order to achieve this.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2020 Q2 [9]}}