| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| 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 straightforward S2 question testing standard binomial-to-normal approximation procedures: direct binomial calculation, applying the np>5 and nq>5 rule, and using continuity correction. All steps are routine textbook exercises requiring only recall of standard methods with no problem-solving insight needed. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(^{30}C_{10}(0.4)^{10}(0.6)^{20} + 0.2915 - 0.1763 = 0.1152\) | M1 | Correct formula or use of tables |
| A1 2 | Answer, a.r.t. 0.115 | |
| (ii) \(30p > 5\) so \(p > \frac{1}{6}\) | M1 | 30p or 30pq used |
| \(30q > 5\) so \(q > \frac{1}{6}\) | M1 | 30q or both solutions from 30pq used |
| \(\frac{1}{6} < p < \frac{5}{6}\) | A1 3 | Either \(\frac{1}{6} < p < \frac{5}{6}\) or \([\frac{1}{4} - \frac{\sqrt{3}}{6} < p < \frac{1}{4} + \frac{\sqrt{3}}{6}]\) [0.211 \(\leq p \leq\) 0.789], allow \(\leq\) |
| (iii) \(N(12, 7.2)\) | B1 | 12 seen |
| \(10.5 - np\) and \(\frac{9.5 - np}{\sqrt{npq}}\) | B1 | 7.2 or 2.683 seen, allow 7.2² |
| \(\sqrt{npq}\) | M1 | Both standardised, allow wrong/no cc, npq |
| A1V | √npq, 10.5 and 9.5 correct, √ on their np, npq | |
| \(\Phi(-0.559) - \Phi(-0.9317) = 0.8243 - 0.7119 = 0.1124\) | M1 | Correct use of tails |
| A1 6 | Answer, in range [0.112, 0.113] | |
| [SR: \(\frac{1}{\sqrt{2\pi \times 7.2}} e^{-\frac{(10-12)^2}{7.2}} = 0.1124\), M1A1, answer A2] |
**(i)** $^{30}C_{10}(0.4)^{10}(0.6)^{20} + 0.2915 - 0.1763 = 0.1152$ | M1 | Correct formula or use of tables
| A1 2 | Answer, a.r.t. 0.115
**(ii)** $30p > 5$ so $p > \frac{1}{6}$ | M1 | 30p or 30pq used
$30q > 5$ so $q > \frac{1}{6}$ | M1 | 30q or both solutions from 30pq used
$\frac{1}{6} < p < \frac{5}{6}$ | A1 3 | Either $\frac{1}{6} < p < \frac{5}{6}$ or $[\frac{1}{4} - \frac{\sqrt{3}}{6} < p < \frac{1}{4} + \frac{\sqrt{3}}{6}]$ [0.211 $\leq p \leq$ 0.789], allow $\leq$
**(iii)** $N(12, 7.2)$ | B1 | 12 seen
$10.5 - np$ and $\frac{9.5 - np}{\sqrt{npq}}$ | B1 | 7.2 or 2.683 seen, allow 7.2²
$\sqrt{npq}$ | M1 | Both standardised, allow wrong/no cc, npq
| A1V | √npq, 10.5 and 9.5 correct, √ on their np, npq
$\Phi(-0.559) - \Phi(-0.9317) = 0.8243 - 0.7119 = 0.1124$ | M1 | Correct use of tails
| A1 6 | Answer, in range [0.112, 0.113]
| | [SR: $\frac{1}{\sqrt{2\pi \times 7.2}} e^{-\frac{(10-12)^2}{7.2}} = 0.1124$, M1A1, answer A2]5 The random variable $W$ has the distribution $\mathbf { B } ( 30 , p )$.\\
(i) Use the exact binomial distribution to calculate $\mathbf { P } ( W = 10 )$ when $p = 0.4$.\\
(ii) Find the range of values of $p$ for which you would expect that a normal distribution could be used as an approximation to the distribution of $W$.\\
(iii) Use a normal approximation to calculate $\mathrm { P } ( W = 10 )$ when $p = 0.4$.
\hfill \mbox{\textit{OCR S2 2005 Q5 [11]}}