| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Justify normal approximation |
| Difficulty | Moderate -0.3 This is a straightforward application of normal approximation to binomial with standard conditions (np and nq both >5) and continuity correction. Part (i) is routine single probability calculation; part (ii) adds CLT for sample means but follows the same template. Both require justification but this is formulaic for S2 level, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(N(14.7, 4.41)\) | M1, A1 | Normal, attempt at \(np\); Both parameters correct |
| Valid because \(np = 14.7 > 5;\ nq = 6.3 > 5\) | B1, B1 | Check \(np > 5\); if both asserted but not both \(nq\) or \(npq > 5\): B1 only; [Allow "\(n\) large, \(p\) close to \(\frac{1}{2}\)"] |
| \(1 - \Phi\!\left(\frac{15.5-14.7}{\sqrt{4.41}}\right) = 1 - \Phi(0.381)\) | M1, A1 | Standardise, answer \(< 0.5\), no \(\sqrt{n}\); \(z\), a.r.t. 0.381 |
| \(= 1 - 0.6484 = \mathbf{0.3516}\) | A1 7 | Answer in range [0.351, 0.352] [Exact: M0] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{K} \sim N(14.7,\ 4.41/36)\ [= N(14.7,\ 0.35^2)]\) | M1, A1\(\sqrt{}\) | Normal, their \(np\) from (i); Their variance/36 |
| Valid by Central Limit Theorem as 36 is large | B1 | Refer to CLT or large \(n\) (\(= 36\), *not* 21), or "\(K \sim N\) so \(\bar{K} \sim N\)", *not* same as (i), *not* \(np > 5\), \(nq > 5\) for \(\bar{K}\) |
| \(\Phi\!\left(\frac{14.0 + \frac{1}{72} - 14.7}{\sqrt{4.41/36}}\right) = \Phi(-1.96)\) | M1, A1, A1 | Standardise 14.0 with 36 or \(\sqrt{36}\); cc included, allow 0.5 here, e.g. \(14.5 - 14.7\); \(z = -1.96\) or \(-2.00\) or \(-2.04\), allow \(+\) if answer \(< 0.5\) |
| \(= \mathbf{0.025}\) | A1 7 | 0.025 or 0.0228 |
| *OR:* \(B(756, 0.7) \approx N(529.2, 158.76)\) | M1M1A1 | \(\times 36\); \(N(529.6,\ldots)\); 158.76 |
| \(\Phi\!\left(\frac{504.5 - 529.2}{\sqrt{158.76}}\right) = \Phi(-1.96)\) | M1, A1, A1 | CLT as above, or \(np > 5\), \(nq > 5\); Standardise \(14\times36\); cc correct and \(\sqrt{npq}\) |
| \(= \mathbf{0.025}\) | A1 | 0.025 or 0.0228 |
# Question 8:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $N(14.7, 4.41)$ | M1, A1 | Normal, attempt at $np$; Both parameters correct |
| Valid because $np = 14.7 > 5;\ nq = 6.3 > 5$ | B1, B1 | Check $np > 5$; if both asserted but not both $nq$ or $npq > 5$: B1 only; [Allow "$n$ large, $p$ close to $\frac{1}{2}$"] |
| $1 - \Phi\!\left(\frac{15.5-14.7}{\sqrt{4.41}}\right) = 1 - \Phi(0.381)$ | M1, A1 | Standardise, answer $< 0.5$, no $\sqrt{n}$; $z$, a.r.t. 0.381 |
| $= 1 - 0.6484 = \mathbf{0.3516}$ | A1 **7** | Answer in range [0.351, 0.352] [Exact: M0] |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{K} \sim N(14.7,\ 4.41/36)\ [= N(14.7,\ 0.35^2)]$ | M1, A1$\sqrt{}$ | Normal, their $np$ from (i); Their variance/36 |
| Valid by Central Limit Theorem as 36 is large | B1 | Refer to CLT or large $n$ ($= 36$, *not* 21), or "$K \sim N$ so $\bar{K} \sim N$", *not* same as (i), *not* $np > 5$, $nq > 5$ for $\bar{K}$ |
| $\Phi\!\left(\frac{14.0 + \frac{1}{72} - 14.7}{\sqrt{4.41/36}}\right) = \Phi(-1.96)$ | M1, A1, A1 | Standardise 14.0 with 36 or $\sqrt{36}$; cc included, allow 0.5 here, e.g. $14.5 - 14.7$; $z = -1.96$ or $-2.00$ or $-2.04$, allow $+$ if answer $< 0.5$ |
| $= \mathbf{0.025}$ | A1 **7** | 0.025 or 0.0228 |
| *OR:* $B(756, 0.7) \approx N(529.2, 158.76)$ | M1M1A1 | $\times 36$; $N(529.6,\ldots)$; 158.76 |
| $\Phi\!\left(\frac{504.5 - 529.2}{\sqrt{158.76}}\right) = \Phi(-1.96)$ | M1, A1, A1 | CLT as above, or $np > 5$, $nq > 5$; Standardise $14\times36$; cc correct and $\sqrt{npq}$ |
| $= \mathbf{0.025}$ | A1 | 0.025 or 0.0228 |8 A sales office employs 21 representatives. Each day, for each representative, the probability that he or she achieves a sale is 0.7 , independently of other representatives. The total number of representatives who achieve a sale on any one day is denoted by $K$.\\
(i) Using a suitable approximation (which should be justified), find $\mathrm { P } ( K \geqslant 16 )$.\\
(ii) Using a suitable approximation (which should be justified), find the probability that the mean of 36 observations of $K$ is less than or equal to 14.0 .
4
\hfill \mbox{\textit{OCR S2 2009 Q8 [14]}}