| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Compare approximation methods |
| Difficulty | Moderate -0.3 This is a standard S2 question testing routine application of normal and Poisson approximations to binomial distributions. Part (i) requires normal approximation with continuity correction (n=32, p=0.4 satisfies np>5, nq>5), and part (ii) requires Poisson approximation (n=90, p=0.01 gives np=0.9<5). Both are textbook applications with clear conditions to check and straightforward calculations, making this slightly easier than average but still requiring proper statistical reasoning. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(B(32, 0.4)\) | B1 | \(B(32, 0.4)\) stated or implied, e.g. by \(Po(12.8)\). Poisson \([0.09888]\), or exact \([0.046269]\): B1max |
| \(\approx N(12.8, 7.68)\) | M1A1 | N(their attempt at \(np\), \(npq\)); \(N(12.8, 7.68)\). SC: \(B(12.8, 7.68/32)\): M1A0 |
| Valid as \(12.8\) and \(19.2 > 5\) | B1 | Or "\(n\) large and \(p\) close to 0.5". Not \(npq\) or \(7.68 > 5\). Allow \(np\) and \(nq\) both asserted \(> 5\). \(\div 32\): M0 |
| \(1 - \Phi\!\left(\dfrac{17.5 - 12.8}{\sqrt{7.68}}\right)\) | M1 | Standardise, their \(np, npq\), allow wrong/no cc or no \(\sqrt{}\) |
| A1 | 17.5 and \(\sqrt{npq}\) correct | |
| \([= 1 - \Phi(1.696)] = \mathbf{0.0449}\) | A1 | Answer, a.r.t. 0.045 |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(B(90, 0.01)\) | B1 | \(B(90, 0.01)\) stated or implied. Exact \([0.049003]\): B1 max. |
| \(\approx Po(0.9)\) | M1 | \(Po\)(their attempt at \(np\)). Don't treat \(p = 0.01\) as MR. If \(np > 5\), M0M0 |
| \(e^{-0.9}\dfrac{0.9^3}{3!} = \mathbf{0.0494}\) | M1 | Correct formula or use of tables, e.g. 0.1646 or 0.0112. No working, wrong answer \(\Rightarrow\) M0A0, but right answer \(\Rightarrow\) M1A1 provided clearly Po |
| A1 | Final answer in range \([0.049, 0.05]\) [i.e., *not* 0.05]. SC: \(B(90, 0.1)\), \(N(9, 8.1)\), \([0.015, 0.016]\) cwo B2 | |
| [4] |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $B(32, 0.4)$ | B1 | $B(32, 0.4)$ stated or implied, e.g. by $Po(12.8)$. Poisson $[0.09888]$, or exact $[0.046269]$: B1max |
| $\approx N(12.8, 7.68)$ | M1A1 | N(their attempt at $np$, $npq$); $N(12.8, 7.68)$. SC: $B(12.8, 7.68/32)$: M1A0 |
| Valid as $12.8$ and $19.2 > 5$ | B1 | Or "$n$ large and $p$ close to 0.5". Not $npq$ or $7.68 > 5$. Allow $np$ and $nq$ both asserted $> 5$. $\div 32$: M0 |
| $1 - \Phi\!\left(\dfrac{17.5 - 12.8}{\sqrt{7.68}}\right)$ | M1 | Standardise, their $np, npq$, allow wrong/no cc or no $\sqrt{}$ |
| | A1 | 17.5 and $\sqrt{npq}$ correct |
| $[= 1 - \Phi(1.696)] = \mathbf{0.0449}$ | A1 | Answer, a.r.t. 0.045 |
| | **[7]** | |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $B(90, 0.01)$ | B1 | $B(90, 0.01)$ stated or implied. Exact $[0.049003]$: B1 max. |
| $\approx Po(0.9)$ | M1 | $Po$(their attempt at $np$). Don't treat $p = 0.01$ as MR. If $np > 5$, M0M0 |
| $e^{-0.9}\dfrac{0.9^3}{3!} = \mathbf{0.0494}$ | M1 | Correct formula or use of tables, e.g. 0.1646 or 0.0112. No working, wrong answer $\Rightarrow$ M0A0, but right answer $\Rightarrow$ M1A1 provided clearly Po |
| | A1 | Final answer in range $[0.049, 0.05]$ [i.e., *not* 0.05]. SC: $B(90, 0.1)$, $N(9, 8.1)$, $[0.015, 0.016]$ cwo B2 |
| | **[4]** | |
---6 At a tourist car park, a survey is made of the regions from which cars come.\\
(i) It is given that $40 \%$ of cars come from the London region. Use a suitable approximation to find the probability that, in a random sample of 32 cars, more than 17 come from the London region. Justify your approximation.\\
(ii) It is given that $1 \%$ of cars come from France. Use a suitable approximation to find the probability that, in a random sample of 90 cars, exactly 3 come from France.
\hfill \mbox{\textit{OCR S2 2012 Q6 [11]}}