| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Explain or apply conditions in context |
| Difficulty | Standard +0.3 This question tests understanding of Poisson conditions and basic probability calculations. Parts (i)-(ii) require conceptual understanding of the constant rate assumption, which is accessible A-level reasoning. Part (iii) is straightforward table lookup. Part (iv) involves algebraic manipulation to derive the equation and simple numerical evaluation—all standard techniques with clear guidance. The multi-part structure and marks suggest moderate length, but no step requires novel insight or complex problem-solving beyond typical S2 level. |
| Spec | 1.09b Sign change methods: understand failure cases5.02i 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 |
|---|---|---|
| Answer | Mark | Guidance |
| No because breakdowns more likely in rush hours, etc | B1 | Any plausible reason for either "yes" or "no" that shows understanding of what the *statistical* concept means. Not "equally likely". *Not* reason for (in)dependence, unless [\*], which needs *both* conditions if affirmed |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Constant *average* rate; *or* [\*] same statement *plus* "breakdowns independent" | B1 | State "average" or equiv, "random" or "uniform". No extras apart from independence (ignore "singly") |
| Otherwise it means that they occur at exactly regular intervals | B1 | Correct explanation. Can't get from [\*] |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 13\) | B1 | |
| \(P = 0.0739\) | B1 | 0.074 or a.r.t. 0.0739. Marks independent |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-\lambda}\dfrac{\lambda^2}{2!} = 0.0072\) | M1* | Correct formula = their 0.0072 seen |
| \(\lambda = \sqrt{0.0144e^{\lambda}}\) | M1dep | Rearrange \(e^{-\lambda}\) and square root, to get \(\lambda = f(\lambda)\). Allow even if left with \(e^{\lambda}\) or \(e^{-\lambda}\) or exact equivalent |
| \(= 0.12e^{\lambda/2}\) | A1 | Correctly obtain AG, with \(k = 0.5\) |
| \(8.5 \to 8.4126\); \(\quad 8.6 \to 8.8440\) | A1 | Two correct evaluations to 4 dp at least. 4 dp explicitly required |
| Therefore solution between 8.5 and 8.6 | A1 | All completely correct and deduction stated. CWO, except allow if only 3 SF |
| Total: 5 |
## Question 9(ii) [previous page - partial]:
| Answer | Mark | Guidance |
|--------|------|----------|
| No because breakdowns more likely in rush hours, etc | B1 | Any plausible reason for either "yes" or "no" that shows understanding of what the *statistical* concept means. Not "equally likely". *Not* reason for (in)dependence, unless [\*], which needs *both* conditions if affirmed |
| **Total: 1** | | |
---
## Question 9(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Constant *average* rate; *or* [\*] same statement *plus* "breakdowns independent" | B1 | State "average" or equiv, "random" or "uniform". No extras apart from independence (ignore "singly") |
| Otherwise it means that they occur at exactly regular intervals | B1 | Correct explanation. Can't get from [\*] |
| **Total: 2** | | |
---
## Question 9(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 13$ | B1 | |
| $P = 0.0739$ | B1 | 0.074 or a.r.t. 0.0739. Marks independent |
| **Total: 2** | | |
---
## Question 9(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-\lambda}\dfrac{\lambda^2}{2!} = 0.0072$ | M1* | Correct formula = their 0.0072 seen |
| $\lambda = \sqrt{0.0144e^{\lambda}}$ | M1dep | Rearrange $e^{-\lambda}$ and square root, to get $\lambda = f(\lambda)$. Allow even if left with $e^{\lambda}$ or $e^{-\lambda}$ or exact equivalent |
| $= 0.12e^{\lambda/2}$ | A1 | Correctly obtain AG, with $k = 0.5$ |
| $8.5 \to 8.4126$; $\quad 8.6 \to 8.8440$ | A1 | Two correct evaluations to 4 dp at least. 4 dp explicitly required |
| Therefore solution between 8.5 and 8.6 | A1 | All completely correct and deduction stated. CWO, except allow if only 3 SF |
| **Total: 5** | | |9 The managers of a car breakdown recovery service are discussing whether the number of breakdowns per day can be modelled by a Poisson distribution. They agree that breakdowns occur randomly. Manager $A$ says, "it must be assumed that breakdowns occur at a constant rate throughout the day".\\
(i) Give an improved version of Manager $A$ 's statement, and explain why the improvement is necessary.\\
(ii) Explain whether you think your improved statement is likely to hold in this context.
Assume now that the number $B$ of breakdowns per day can be modelled by the distribution $\operatorname { Po } ( \lambda )$.\\
(iii) Given that $\lambda = 9.0$ and $\mathrm { P } \left( B > B _ { 0 } \right) < 0.1$, use tables to find the smallest possible value of $B _ { 0 }$, and state the corresponding value of $\mathrm { P } \left( B > B _ { 0 } \right)$.\\
(iv) Given that $\mathrm { P } ( B = 2 ) = 0.0072$, show that $\lambda$ satisfies an equation of the form $\lambda = 0.12 \mathrm { e } ^ { k \lambda }$, for a value of $k$ to be stated. Evaluate the expression $0.12 \mathrm { e } ^ { k \lambda }$ for $\lambda = 8.5$ and $\lambda = 8.6$, giving your answers correct to 4 decimal places. What can be deduced about a possible value of $\lambda$ ?
\hfill \mbox{\textit{OCR S2 2013 Q9 [10]}}