| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single scaled time period |
| Difficulty | Easy -1.2 Part (i) is pure recall of Poisson conditions (1 mark). Parts (ii) and (iii) are routine Poisson probability calculations using standard formulas and tables, with (iii) requiring the simple scaling of λ. No problem-solving or novel insight required—entirely procedural S2 content. |
| Spec | 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Snakes must occur independently of one another | B1 | Contextualised ("snakes" must be mentioned); not *just* "singly" but allow both independent and singly. Allow explanation e.g. "Occurrence of one snake doesn't affect occurrences of others". Allow "snakes must occur randomly". Otherwise, more than one condition, e.g. "randomly, independently, singly and at constant rate": 0. |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 - P(\leq 5)\) | M1 | Give M1 for 0.3712, 0.1107 or 0.2307. Answer 0.7851 is M0. |
| \(= 1 - 0.7851 = \mathbf{0.2149}\) | A1 [2] | Answer a.r.t. 0.215 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Po(3.08)\) | M1 | \(Po(3.08)\) stated or implied. [Just \(\lambda = 3.08\) is M0 unless Poisson later.] |
| \(e^{-3.08}\!\left(\dfrac{3.08^2}{2!} + \dfrac{3.08^3}{3!}\right)\) \([= 0.2180 + 0.2238]\) | M1 | Correct formula for \(Po\) (\(r > 0\)) used at least once, can be implied |
| A1ft | Completely correct formula for their \(\lambda\) (not 4), can be implied | |
| \(= \mathbf{0.4418}\) | A1 | Final answer a.r.t. 0.442 |
| [4] | No working: last 3 marks either 0 or 3, no "nearly right". |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Snakes must occur independently of one another | B1 | Contextualised ("snakes" must be mentioned); not *just* "singly" but allow both independent and singly. Allow explanation e.g. "Occurrence of one snake doesn't affect occurrences of others". Allow "snakes must occur randomly". Otherwise, more than one condition, e.g. "randomly, independently, singly and at constant rate": 0. |
| | **[1]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - P(\leq 5)$ | M1 | Give M1 for 0.3712, 0.1107 or 0.2307. Answer 0.7851 is M0. |
| $= 1 - 0.7851 = \mathbf{0.2149}$ | A1 **[2]** | Answer a.r.t. 0.215 |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Po(3.08)$ | M1 | $Po(3.08)$ stated or implied. [Just $\lambda = 3.08$ is M0 unless Poisson later.] |
| $e^{-3.08}\!\left(\dfrac{3.08^2}{2!} + \dfrac{3.08^3}{3!}\right)$ $[= 0.2180 + 0.2238]$ | M1 | Correct formula for $Po$ ($r > 0$) used at least once, can be implied |
| | A1ft | Completely correct formula for their $\lambda$ (not 4), can be implied |
| $= \mathbf{0.4418}$ | A1 | Final answer a.r.t. 0.442 |
| | **[4]** | No working: last 3 marks either 0 or 3, no "nearly right". |
---4 A zoologist investigates the number of snakes found in a given region of land. The zoologist intends to use a Poisson distribution to model the number of snakes.\\[0pt]
(i) One condition for a Poisson distribution to be valid is that snakes must occur at constant average rate. State another condition needed for a Poisson distribution to be valid. [1]
Assume now that the number of snakes found in 1 acre of a region can be modelled by the distribution Po(4).\\[0pt]
(ii) Find the probability that, in 1 acre of the region, at least 6 snakes are found. [2]\\[0pt]
(iii) Find the probability that, in 0.77 acres of the region, the number of snakes found is either 2 or 3. [4]
\hfill \mbox{\textit{OCR S2 2014 Q4 [7]}}