| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson approximation justification or comparison |
| Difficulty | Standard +0.3 This is a straightforward application of standard Poisson approximation to binomial with clear parameters (n=10000, p=1/1296). Part (a) is trivial probability multiplication, (b) requires stating standard conditions for Poisson approximation, (c) involves direct calculator use of Poisson probabilities, and (d) requires recognizing a binomial-Poisson structure but follows standard methods. Slightly easier than average due to clear setup and routine application of well-practiced techniques. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02f Geometric distribution: conditions5.02i 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 modelling |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 1296 | B1 | |
| [1] | 1.1 | AG |
| 1 | (b) | Random or independent |
| Answer | Marks |
|---|---|
| Poisson distribution is also appropriate | B1 |
| Answer | Marks |
|---|---|
| [3] | 2.4 |
| Answer | Marks |
|---|---|
| 2.4 | For two correct statements |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (c) | 10000 625 |
| Answer | Marks |
|---|---|
| P(X > 10) = 0.157 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | BC 0.091864… |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (d) | P(A person does not throw 4 sixes in 20 tries) = |
| Answer | Marks |
|---|---|
| P(No more than 2) = 0.9586 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | Or Poisson approx. mean 0.765975 (from 50×0.015319) |
Question 1:
1 | (a) | 4
1 1
P(4 sixes) = ( ) =
6 1296 | B1
[1] | 1.1 | AG
1 | (b) | Random or independent
Fixed probability of success
Fixed number of trials
The random variable is the number of successes oe
There are only 2 outcomes
1
Because n (= 10 000) is large and p (= ) is small a
1296
Poisson distribution is also appropriate | B1
B1
B1
[3] | 2.4
2.4
2.4 | For two correct statements
For at least 4 correct comments
For explanation of Poisson. Must mention Poisson.
.
1 | (c) | 10000 625
Poisson( ) or Poisson(7.716…) or Poisson( )
1296 81
P(X = 10) = 0.0919
P(X > 10) = 0.157 | M1
A1
A1
[3] | 3.3
1.1
1.1 | BC 0.091864…
BC 0.156963…
1 | (d) | P(A person does not throw 4 sixes in 20 tries) =
20
1295
( ) = [0.98468…]
1296
B(50, 1 – 0.98468…) [=B(50, 0.015319…)]
P(No more than 2) = 0.9586 | M1
M1
A1
[3] | 3.1b
3.3
1.1 | Or Poisson approx. mean 0.765975 (from 50×0.015319)
BC 0.958647… Poisson approx. leads to 0.9573…
Allow awrt 0.96
NB Using B(50, 20/1296 or 5/324) gets answer 0.9579 but
scores zero1 A website simulates the outcome of throwing four fair dice. Ten thousand people take part in a challenge using the website in which they have one attempt at getting four sixes in the four throws of the dice. The number of people who succeed in getting four sixes is denoted by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item Show that, for each person, the probability that the person gets four sixes is equal to $\frac { 1 } { 1296 }$.
\item Explain why you could use either a binomial distribution or a Poisson distribution to model the distribution of $X$.
\item Use a Poisson distribution to calculate each of the following probabilities.
\begin{itemize}
\item $\mathrm { P } ( X = 10 )$
\item $\mathrm { P } ( X > 10 )$
\item In another challenge on the website, 50 people are each given 20 independent attempts to try to get four sixes as often as they can.
\end{itemize}
Determine the probability that no more than 2 people succeed in getting four sixes at least once in their 20 attempts.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2023 Q1 [10]}}