| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Explain or apply conditions in context |
| Difficulty | Moderate -0.3 This question tests standard Poisson distribution knowledge: stating assumptions (textbook recall), writing the probability formula, and performing routine calculations including scaling parameters and adding independent Poisson variables. While it has multiple parts (12 marks total), each component is straightforward application of theory without requiring problem-solving insight or novel approaches. The most demanding part (v) involves recognizing that independent Poisson variables sum to another Poisson distribution, which is a standard result taught in Further Stats AS. |
| 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 modelling5.02m Poisson: mean = variance = lambda5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) | p |
| Answer | Marks |
|---|---|
| …at a uniform rate S | e |
| Answer | Marks |
|---|---|
| [1] | 1.2 |
| 1.2 | Not answers equivalent to |
| “random” oe | Allow “constant average rate” but |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (ii) | E.g. Traffic lights or roundabouts might make cars |
| Answer | Marks |
|---|---|
| commuting times, than in the middle of the day. | E1 |
| Answer | Marks |
|---|---|
| [1] | 2.4 |
| 2.4 | One reason why cars might not |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (iv) | (a) |
| Answer | Marks | Guidance |
|---|---|---|
| r! | B1 | |
| [1] | 3.4 | n |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (iv) | (b) |
| Answer | Marks | Guidance |
|---|---|---|
| 2! | B1 | |
| [1] | 1.1 | m |
| 6 | (v) | X~Po(cid:11)24(cid:12) |
| Answer | Marks |
|---|---|
| (cid:32)0.232to 3sf | c |
| Answer | Marks |
|---|---|
| [3] | i |
| Answer | Marks |
|---|---|
| 1.1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (vi) | p |
| Answer | Marks |
|---|---|
| (cid:32)0.328(cid:11)2(cid:12) | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.2 |
| Answer | Marks |
|---|---|
| 1.1 | Attempt at Po(cid:11)10(cid:117)0.8(cid:14)2(cid:117)1.5(cid:12) |
| Answer | Marks |
|---|---|
| BC | Allow M1M1A0 for 0.2187 |
| Answer | Marks |
|---|---|
| 6(iii)(a) | e |
| Answer | Marks |
|---|---|
| 6(v) | i |
Question 6:
6 | (i) | p
Cars must pass independently of one another…
…at a uniform rate S | e
E1
E1
[1] | 1.2
1.2 | Not answers equivalent to
“random” oe | Allow “constant average rate” but
not “constant rate”
E0 for any answer that implies
fixed numbers in given time
E0 for “events must occur
randomly”, “independently”,
“singly” or “at constant rate” oe
6 | (ii) | E.g. Traffic lights or roundabouts might make cars
go past in groups, not independently
E.g. The rate at which cars pass will be higher at
commuting times, than in the middle of the day. | E1
E1
[1] | 2.4
2.4 | One reason why cars might not
act independently, in context
One reason why the rate may not
be uniform, in context
6 | (iv) | (a) | 0.8r
e(cid:16)0.8
r! | B1
[1] | 3.4 | n
e
6 | (iv) | (b) | 0.82
e(cid:16)0.8 (cid:32)0.144
2! | B1
[1] | 1.1 | m | (cid:62)0.143785(cid:64)
6 | (v) | X~Po(cid:11)24(cid:12)
P(cid:11)X (cid:116)28(cid:12)(cid:32)1(cid:16)P(cid:11)X (cid:100)27(cid:12)
(cid:32)0.232to 3sf | c
M1
eM1
A1
[3] | i
1.1a
1.1
1.1 | BC
awrt 0.232
6 | (vi) | p
Distributions must be independent
S
Y~Po(cid:11)11(cid:12)
P(cid:11)Y (cid:100)15(cid:12)(cid:16)P(cid:11)Y (cid:100)11(cid:12)(cid:32)0.9074(cid:16)0.5792
(cid:32)0.328(cid:11)2(cid:12) | B1
M1
M1
A1
[4] | 1.2
1.1
3.4
1.1 | Attempt at Po(cid:11)10(cid:117)0.8(cid:14)2(cid:117)1.5(cid:12)
soi
BC | Allow M1M1A0 for 0.2187
--- 6(i) ---
6(i)
--- 6(ii) ---
6(ii)
n
e
m
i
c
6(iii)(a) | e
p
S
6(iii)(b)
--- 6(iv) ---
6(iv)
n
e
m
--- 6(v) ---
6(v) | i
c
e
p
SSabrina counts the number of cars passing her house during randomly chosen one minute intervals. Two assumptions are needed for the number of cars passing her house in a fixed time interval to be well modelled by a Poisson distribution.
\begin{enumerate}[label=(\roman*)]
\item State these two assumptions. [2]
\item For each assumption in part (i) give a reason why it might not be a reasonable assumption for this context. [2]
\end{enumerate}
Assume now that the number of cars that pass Sabrina's house in one minute can be well modelled by the distribution $\mathrm{Po}(0.8)$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item
\begin{enumerate}[label=(\alph*)]
\item Write down an expression for the probability that, in a given one minute period, exactly $r$ cars pass Sabrina's house. [1]
\item Hence find the probability that, in a given one minute period, exactly 2 cars pass Sabrina's house. [1]
\end{enumerate}
\item Find the probability that, in a given 30 minute period, at least 28 cars pass Sabrina's house. [3]
\item The number of bicycles that pass Sabrina's house in a 5 minute period is a random variable with the distribution $\mathrm{Po}(1.5)$. Find the probability that, in a given 10 minute period, the total number of cars and bicycles that pass Sabrina's house is between 12 and 15 inclusive. State a necessary condition. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS Q6 [13]}}