| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding maximum n for P(X=0) threshold |
| Difficulty | Moderate -0.8 Part (i) is pure recall of standard Poisson conditions requiring no calculation or problem-solving. Parts (ii) and (iii) involve routine application of Poisson distribution properties (scaling parameter and solving inequalities), but are straightforward textbook exercises with no conceptual challenges beyond knowing the basic formulas. |
| Spec | 5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Bricks scattered at constant average rate | B1 | B1 for each of 2 different reasons, in context |
| Independently of one another | B1 2 | Treat "randomly" ≡ "singly" ≡ "independently" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(Po(12)\) | B1 | \(Po(12)\) stated or implied |
| \(P(\leq 14) - P(\leq 7) = .7720 - .0895\) | M1 | Allow one out at either end or both, eg 0.617, or wrong column, but *not* from \(Po(3)\) nor, eg, \(.9105 - .7720\) |
| \([= P(8) + P(9) + \ldots + P(14)]\) | ||
| \(= 0.6825\) | A1 3 | Answer in range \([0.682, 0.683]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-\lambda} = 0.4\) | B1 | This equation, aef, can be implied by, eg 0.9 |
| \(\lambda = -\ln(0.4)\) | M1 | Take ln, or 0.91 by T & I |
| \(= 0.9163\) | A1 | \(\lambda\) art 0.916 or 0.92, can be implied |
| \(\text{Volume} = 0.9163 \div 3 = \mathbf{0.305}\) | M1 4 | Divide their \(\lambda\) value by 3. [SR: Tables, eg \(0.9 \div 3\): B1 M0 A0 M1] |
# Question 5:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Bricks scattered at constant average rate | B1 | B1 for each of 2 different reasons, in context |
| Independently of one another | B1 **2** | Treat "randomly" ≡ "singly" ≡ "independently" |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $Po(12)$ | B1 | $Po(12)$ stated or implied |
| $P(\leq 14) - P(\leq 7) = .7720 - .0895$ | M1 | Allow one out at either end or both, eg 0.617, or wrong column, but *not* from $Po(3)$ nor, eg, $.9105 - .7720$ |
| $[= P(8) + P(9) + \ldots + P(14)]$ | | |
| $= 0.6825$ | A1 **3** | Answer in range $[0.682, 0.683]$ |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-\lambda} = 0.4$ | B1 | This equation, aef, can be implied by, eg 0.9 |
| $\lambda = -\ln(0.4)$ | M1 | Take ln, or 0.91 by T & I |
| $= 0.9163$ | A1 | $\lambda$ art 0.916 or 0.92, can be implied |
| $\text{Volume} = 0.9163 \div 3 = \mathbf{0.305}$ | M1 **4** | Divide their $\lambda$ value by 3. [SR: Tables, eg $0.9 \div 3$: B1 M0 A0 M1] |
---5 In a large region of derelict land, bricks are found scattered in the earth.\\
(i) State two conditions needed for the number of bricks per cubic metre to be modelled by a Poisson distribution.
Assume now that the number of bricks in 1 cubic metre of earth can be modelled by the distribution Po(3).\\
(ii) Find the probability that the number of bricks in 4 cubic metres of earth is between 8 and 14 inclusive.\\
(iii) Find the size of the largest volume of earth for which the probability that no bricks are found is at least 0.4.
\hfill \mbox{\textit{OCR S2 2009 Q5 [9]}}