| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Minor (Further Statistics Minor) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson parameter from given probability |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics question testing standard Poisson distribution techniques: calculating probabilities with given parameters, using geometric distribution for repeated trials, and applying the variance formula E(X²)=Var(X)+[E(X)]² to find μ. All parts are routine applications of formulas with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^25.02i 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 |
|---|---|---|
| 5 | (a) | (i) |
| = 0.0839 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 1.1 | Or P(X ≥ 10) = 1 – P(X ≤ 9) |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | (ii) |
| P(X < 50) = 0.0844 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | soi |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | 0.916120 |
| = 0.1733 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | soi |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | Use of Var(X) = E(X) |
| Answer | Marks |
|---|---|
| P(X < 5) = 0.8153 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | For equation |
Question 5:
5 | (a) | (i) | P(X ≥ 10) = 1 – 0.9161
= 0.0839 | M1
A1
[2] | 3.1b
1.1 | Or P(X ≥ 10) = 1 – P(X ≤ 9)
BC (0.083924…)
5 | (a) | (ii) | Poisson (60)
P(X < 50) = 0.0844 | M1
A1
[2] | 3.3
1.1 | soi
BC (0.084406…)
5 | (b) | 0.916120
= 0.1733 | M1
A1FT
[2] | 3.1a
1.1 | soi
or 0.1732 from calculator value in (a)
5 | (c) | Use of Var(X) = E(X)
Var(X) = E(X 2) – (E(X))2 µ = 12 – µ2
µ = 3
P(X < 5) = 0.8153 | M1
M1
A1
A1
[4] | 3.1b
2.2a
1.1
1.1 | For equation
BC (0.815263…)5 Biological cell membranes have receptor molecules which perform various functions. It is known that the number of receptor molecules of a particular type can be modelled by a Poisson distribution with mean 6 per area of 1 square unit.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Determine the probability that there are at least 10 of these receptor molecules in an area of 1 square unit.
\item Determine the probability that there are fewer than 50 of these receptor molecules in an area of 10 square units.
\end{enumerate}\item A scientist is looking at areas of 1 square unit of cell membrane in order to find one which has at least 10 receptor molecules.
Find the probability that she has to look at more than 20 to find such an area.
It is known that the number of receptor molecules of another type in an area of 1 square unit can be modelled by the random variable $X$ which has a Poisson distribution with mean $\mu$. It is given that $\mathrm { E } \left( X ^ { 2 } \right) = 12$.
\item Determine $\mathrm { P } ( X < 5 )$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2021 Q5 [10]}}