| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Poisson to the Normal distribution |
| Type | Scaled Poisson over time period |
| Difficulty | Standard +0.3 This is a straightforward two-stage approximation problem: Poisson→Normal for part (i), then Binomial→Normal for part (ii). Both approximations are standard S2 techniques with clear conditions met (large λ, large n). The calculations require continuity correction and z-table lookup but no conceptual insight beyond recognizing which approximations to apply. |
| Spec | 2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Po(4200) \approx N(4200,\ 4200)\) | M1, M1 | \(Po(60\lambda)\) stated or implied. \(N(60\lambda,\ 60\lambda)\) |
| \(1 - \Phi\!\left(\dfrac{4350.5 - 4200}{\sqrt{4200}}\right)\) | M1, A1 | Standardise with their \(60\lambda\) and \(\sqrt{60\lambda}\) or \(60\lambda\). 4350.5 explicitly seen and \(\sqrt{60\lambda}\) not wrong. Allow wrong or no cc, or no \(\sqrt{}\); \(\sqrt{60\lambda}\) needn't be explicit |
| \(= 1 - \Phi(2.322) = \mathbf{0.010}(1)\) | A1 | Answer, allow a.r.t. 0.010. Allow [0.0103, 0.0106] from no CC, but *not* 0.0105 from wrong CC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(B(30,\ 0.010(1))\) | M1 | \(B(30,\) their \((i))\) stated or implied. Exact binomial [0.00022]: M1A0A0 |
| \(\approx Po(0.30(3))\) | A1 | \(Po(0.3)\) or 0.303 etc. \([0.30\to0.000266.\ 0.303\to0.000276.\ 0.309\to0.000297]\) |
| \(1 - 0.9997 = \mathbf{0.0003}\) | A1 | Final answer a.r.t. 0.0003 |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Po(4200) \approx N(4200,\ 4200)$ | M1, M1 | $Po(60\lambda)$ stated or implied. $N(60\lambda,\ 60\lambda)$ |
| $1 - \Phi\!\left(\dfrac{4350.5 - 4200}{\sqrt{4200}}\right)$ | M1, A1 | Standardise with their $60\lambda$ and $\sqrt{60\lambda}$ or $60\lambda$. 4350.5 explicitly seen and $\sqrt{60\lambda}$ not wrong. Allow wrong or no cc, or no $\sqrt{}$; $\sqrt{60\lambda}$ needn't be explicit |
| $= 1 - \Phi(2.322) = \mathbf{0.010}(1)$ | A1 | Answer, allow a.r.t. 0.010. Allow [0.0103, 0.0106] from no CC, but *not* 0.0105 from wrong CC |
**[5 marks]**
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $B(30,\ 0.010(1))$ | M1 | $B(30,$ their $(i))$ stated or implied. Exact binomial [0.00022]: M1A0A0 |
| $\approx Po(0.30(3))$ | A1 | $Po(0.3)$ or 0.303 etc. $[0.30\to0.000266.\ 0.303\to0.000276.\ 0.309\to0.000297]$ |
| $1 - 0.9997 = \mathbf{0.0003}$ | A1 | Final answer a.r.t. 0.0003 |
**[3 marks]**
---5 In a mine, a deposit of the substance pitchblende emits radioactive particles. The number of particles emitted has a Poisson distribution with mean 70 particles per second. The warning level is reached if the total number of particles emitted in one minute is more than 4350.\\
(i) A one-minute period is chosen at random. Use a suitable approximation to show that the probability that the warning level is reached during this period is 0.01 , correct to 2 decimal places. You should calculate the answer correct to 4 decimal places.\\
(ii) Use a suitable approximation to find the probability that in 30 one-minute periods the warning level is reached on at least 4 occasions. (You should use the given rounded value of 0.01 from part (i) in your calculation.)
\hfill \mbox{\textit{OCR S2 2013 Q5 [8]}}