| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Two independent Poisson sums |
| Difficulty | Moderate -0.8 This question tests basic Poisson distribution properties and standard applications. Parts (a)-(b) require simple recall (constant rate assumption, SD = √λ), part (c) is routine scaling and calculation, part (d) uses standard addition of independent Poisson variables, and part (e) requires a straightforward comparison. All techniques are textbook exercises with no novel problem-solving required, though the multi-part structure and context provide some substance. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | Punctures must occur at constant average |
| rate throughout the period of 24 hours | B1 | |
| [1] | 3.3 | Constant average rate (or uniform rate), with |
| context (not events must occur …), no extras | Not constant rate, not constant probability, |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | 2.7 = 1.64(3…) | B1 |
| 4 | (c) | Po(18.9) |
| Answer | Marks |
|---|---|
| = 0.267 | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Po(72.7) stated or implied |
| Answer | Marks |
|---|---|
| Awrt 0.267 | E.g. 1 – P( 22) = 0.2(0015) M1M1A0 |
| (d) | Po(3.5) |
| P(> 6) = 0.0653 | M1 |
| A1 | 1.1 |
| 3.4 | Stated or implied, e.g. by 1 – 0.858 = 0.142 |
| Answer | Marks |
|---|---|
| (e) | 0.0653 is some way from 0.12 |
| Answer | Marks |
|---|---|
| as 100 days may not be enough evidence) | B1ft |
| Answer | Marks |
|---|---|
| [2] | 3.5b |
| 3.5a | Compare 0.12 with answer to (d), or other |
| Answer | Marks |
|---|---|
| calculation, if used). FT on their (d) | Or: B(100, 0.0653), P( 12) = 0.03 or |
Question 4:
4 | (a) | Punctures must occur at constant average
rate throughout the period of 24 hours | B1
[1] | 3.3 | Constant average rate (or uniform rate), with
context (not events must occur …), no extras | Not constant rate, not constant probability,
not average constant rate, not singly
(b) | 2.7 = 1.64(3…) | B1 | 1.1 | Any equivalent answer, e.g. 330/10
4 | (c) | Po(18.9)
1 – P( 21)
= 0.267 | M1
M1
A1 | 1.1
3.4
1.1 | Po(72.7) stated or implied
Probability from their Po(72.7)
Awrt 0.267 | E.g. 1 – P( 22) = 0.2(0015) M1M1A0
(d) | Po(3.5)
P(> 6) = 0.0653 | M1
A1 | 1.1
3.4 | Stated or implied, e.g. by 1 – 0.858 = 0.142
Correct calculation, awrt 0.0653
(e) | 0.0653 is some way from 0.12
so assumptions look doubtful (but possible
as 100 days may not be enough evidence) | B1ft
B1ft
[2] | 3.5b
3.5a | Compare 0.12 with answer to (d), or other
appropriate calculation and comparison
Appropriate assessment, not definite (and correct
calculation, if used). FT on their (d) | Or: B(100, 0.0653), P( 12) = 0.03 or
E(R) = 6.5, etc
See Appendix for exemplars4 The manager of a car breakdown service uses the distribution $\operatorname { Po } ( 2.7 )$ to model the number of punctures, $R$, in a 24-hour period in a given rural area. The manager knows that, for this model to be valid, punctures must occur randomly and independently of one another.
\begin{enumerate}[label=(\alph*)]
\item State a further assumption needed for the Poisson model to be valid.
\item State the value of the standard deviation of $R$.
\item Use the model to calculate the probability that, in a randomly chosen period of 168 hours, at least 22 punctures occur.
The manager uses the distribution $\operatorname { Po } ( 0.8 )$ to model the number of flat batteries in a 24 -hour period in the same rural area, and he assumes that instances of flat batteries are independent of punctures. A day begins and ends at midnight, and a "bad" day is a day on which there are more than 6 instances, in total, of punctures and flat batteries.
\item Assume first that both the manager's models are correct.
Calculate the probability that a randomly chosen day is a "bad" day.
\item It is found that 12 of the next 100 days are "bad" days.
Comment on whether this casts doubt on the validity of the manager's models.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2022 Q4 [9]}}