| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Two independent Poisson sums |
| Difficulty | Moderate -0.8 Part (i) requires straightforward recall of Poisson conditions (independence and constant rate). Parts (ii) and (iii) involve standard Poisson calculations with no novel problem-solving. This is a routine textbook exercise testing basic knowledge and application of the Poisson distribution, making it easier than average. |
| Spec | 5.02i 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 modelling5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | Goals are scored independently p |
| Answer | Marks |
|---|---|
| S | e |
| Answer | Marks |
|---|---|
| [2] | 1.2 |
| 1.2 | Not “singly” |
| Must be in context | Allow “constant average rate” |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (ii) | (a) |
| Answer | Marks | Guidance |
|---|---|---|
| r! | B1 | |
| [1] | 1.1 | Must be seen |
| 5 | (ii) | (b) |
| [1] | 1.1 | |
| 5 | (iii) | Total ~P (cid:11)1.9(cid:14)(cid:79)(cid:12) |
| Answer | Marks |
|---|---|
| hence a reasonable estimate is 0.4 | M1 |
| Answer | Marks |
|---|---|
| [4] | 2.2a |
| Answer | Marks |
|---|---|
| 3.2a | n |
| Answer | Marks |
|---|---|
| m | BC |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | 2 | 0 |
| 5(ii)(a) | 0 | 0 |
| 5(ii)(b) | 1 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(iii) | 1 | 1 |
| 6(i)(a) | 0 | 1 |
| 6(i)(b) | 2 | 0 |
Question 5:
5 | (i) | Goals are scored independently p
Goals are scored at uniform rate
S | e
B1
B1
[2] | 1.2
1.2 | Not “singly”
Must be in context | Allow “constant average rate”
but not “constant rate”.
B0 for any answer that implies
fixed numbers in given time.
B0 for “events must occur
randomly”, “independently”,
“singly” or “at constant rate”
oe
5 | (ii) | (a) | 1.9r
P(X (cid:32)r)(cid:32)e(cid:16)1.9
r! | B1
[1] | 1.1 | Must be seen
5 | (ii) | (b) | P(X (cid:32)3)(cid:32)0.171 | B1
[1] | 1.1
5 | (iii) | Total ~P (cid:11)1.9(cid:14)(cid:79)(cid:12)
o
(cid:79)(cid:32)(cid:11)1.9(cid:14)1.31(cid:12), P(cid:11)(cid:33)3(cid:12)(cid:32)0.399...
(cid:79)(cid:32)(cid:11)1.9(cid:14)1.32(cid:12), P(cid:11)(cid:33)3(cid:12)(cid:32)0.401...
0.399… < 0.4 and 0.401… > 0.4,
hence a reasonable estimate is 0.4 | M1
M1
A1
E1
[4] | 2.2a
3.1b
1.1
3.2a | n
Use 1.9 + (cid:79)
Evaluate RH tail probability for 1.31
e
and 1.32
Both evaluations correct
m | BC
BC
--- 5(i) ---
5(i) | 2 | 0 | 0 | 0 | 2
5(ii)(a) | 0 | 0 | 01 | 1
5(ii)(b) | 1 | 0 | 0 | 0 | 1
--- 5(iii) ---
5(iii) | 1 | 1 | 20 | 4
6(i)(a) | 0 | 1 | 0 | 01
6(i)(b) | 2 | 0 | 0 | 0 | 25 The number of goals scored by the home team in a randomly chosen hockey match is denoted by $X$.
\begin{enumerate}[label=(\roman*)]
\item In order for $X$ to be modelled by a Poisson distribution it is assumed that goals scored are random events. State two other conditions needed for $X$ to be modelled by a Poisson distribution in this context.
Assume now that $X$ can be modelled by the distribution $\operatorname { Po } ( 1.9 )$.
\item (a) Write down an expression for $\mathrm { P } ( X = r )$.\\
(b) Hence find $\mathrm { P } ( X = 3 )$.
\item Assume also that the number of goals scored by the away team in a randomly chosen hockey match has an independent Poisson distribution with mean $\lambda$ between 1.31 and 1.32. Find an estimate for the probability that more than 3 goals are scored altogether in a randomly chosen match.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics Q5 [8]}}