| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| 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 Poisson approximation to binomial with standard calculations. Part (a) requires stating standard conditions (large n, small p, np moderate), while parts (b) and (c) involve routine probability calculations using tables or calculators. The multi-day extension in (c) is predictable and requires only basic understanding of independence. Slightly above average due to the multi-part structure and need to recognize np for the week, but no novel insight required. |
| 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 |
|---|---|---|
| 4 | (a) | A binomial distribution (to model the number of |
| Answer | Marks |
|---|---|
| Poisson distribution is also appropriate | B2,1,0 |
| Answer | Marks |
|---|---|
| [3] | 2.4 |
| 2.4 | B1 for two of these conditions stated |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (b) | Poisson(𝟏𝟓𝟎𝟎𝟎×𝟎.𝟎𝟎𝟎𝟓) or Poisson(7.5) |
| Answer | Marks |
|---|---|
| P(X ≥ 8) = 0.4754 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Poisson soi. e.g. by 7.5 seen |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (c) | (i) |
| P(≥ 40) = awrt 0.363 | B1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | BC (0.362862… from Poisson or 0.362839… from |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (c) | (ii) |
| = 0.0243 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | FT [their P(X ≥ 8)]5 Allow M1 for Y ~ B(5, their P(X ≥ 8)) |
Question 4:
4 | (a) | A binomial distribution (to model the number of
successes) could be suitable as
• There is a fixed number of parcels delivered each
day. [n = 15000]
• Each delivery has two possible outcomes (wrong
address or not)
• The deliveries could be assumed to be
independent.
• There is constant probability of delivery to a
wrong address [p = 0.0005]
Because n = 15 000 is large and p = 0.0005 is small a
Poisson distribution is also appropriate | B2,1,0
B1
[3] | 2.4
2.4 | B1 for two of these conditions stated
B2 for correct comments with at least three conditions
stated, and at least one given in context with reference to
binomial.
SCB1 if first two B1 marks not awarded but B(15000,
0.0005) stated
Allow correctly worded explanations that Poisson could be
suitable due to the mean being close to the variance
4 | (b) | Poisson(𝟏𝟓𝟎𝟎𝟎×𝟎.𝟎𝟎𝟎𝟓) or Poisson(7.5)
P(X = 5) = 0.1094
P(X ≥ 8) = 0.4754 | B1
B1
B1
[3] | 3.3
1.1
1.1 | Poisson soi. e.g. by 7.5 seen
BC (0.109374…)
BC (0.475361…)
4 | (c) | (i) | Poisson (𝟓×𝟕.𝟓) or B(75000, 0.0005) oe seen
P(≥ 40) = awrt 0.363 | B1
B1
[2] | 3.3
1.1 | BC (0.362862… from Poisson or 0.362839… from
binomial)
4 | (c) | (ii) | P(𝑋 ≥ 8)5 = 0.4753614…5
= 0.0243 | M1
A1
[2] | 3.3
1.1 | FT [their P(X ≥ 8)]5 Allow M1 for Y ~ B(5, their P(X ≥ 8))
and P(Y = 5)
BC (0.024272…) cao SCB1 for correct answer without
working shown.4 At a parcel delivery company it is known that the probability that a parcel is delivered to the wrong address is 0.0005 . On a particular day, 15000 parcels are delivered. The number of parcels delivered to the wrong address is denoted by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item Explain why the binomial distribution and the Poisson distribution could both be suitable models for the distribution of $X$.
\item Use a Poisson distribution to find each of the following.
\begin{itemize}
\item $\mathrm { P } ( X = 5 )$
\item $\mathrm { P } ( X \geqslant 8 )$
\end{itemize}
You are given that 15000 parcels are delivered each day in a 5-day working week.
\item \begin{enumerate}[label=(\roman*)]
\item Determine the probability that at least 40 parcels are delivered to the wrong address during the week.
\item Determine the probability that at least 8 parcels are delivered to the wrong address on each of the 5 days in the week.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2023 Q4 [10]}}