| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Standard +0.3 This is a straightforward application of binomial distribution for parts (i)-(ii) with simple probability calculations, followed by a routine normal approximation to binomial in part (iii). The question explicitly tells students to use an approximation, requires only standard continuity correction, and involves no conceptual challenges beyond textbook exercises. Slightly easier than average due to clear signposting and mechanical application. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 3 term binomial expression involving \({}_{20}C_{\text{something}}\) and powers summing to 20 | |
| A1 | 2 marks | Correct final answer |
| Answer | Marks |
|---|---|
| M1 | Summing three or 4 binomial expressions |
| A1 | One correct unsimplified expression allow \(0.4 - 0.6\) muddle |
| A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Standardising, cc \(16.5\) or \(17.5\), their mean, \(\sqrt{\text{(their var)}}\) | |
| A1 | 2.51 seen | |
| A1 | 3 marks | 0.0060 seen must be 0.0060 |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | For seeing 90 and 36 | |
| M1 | For standardising, with or without cc, must have sq rt on denom | |
| M1 | One continuity correction \(97.5\) or \(96.5\) or \(87.5\) or \(88.5\) | |
| A1 | \(0.8944\) or \(0.6616\) or \(0.3384\) or \(0.3944\) or \(0.1616\) seen | |
| M1 | Subtracting a probability from their standardised 97 prob | |
| A1 | 6 marks | Correct answer |
**(i)** $(0.6)^{10} \times (0.4)^{10} \times {}_{20}C_{10} = 0.117$
| M1 | 3 term binomial expression involving ${}_{20}C_{\text{something}}$ and powers summing to 20 |
| A1 | 2 marks | Correct final answer |
**(ii)** $P(18, 19, 20) = (0.6)^{18}(0.4)^2 {}_{20}C_2 + (0.6)^{19}(0.4)^1 {}_{21}C_1 + (0.6)^{20}$
$= 0.003087 + 0.000487 + 0.00003635 = 0.00361$
| M1 | Summing three or 4 binomial expressions |
| A1 | One correct unsimplified expression allow $0.4 - 0.6$ muddle |
| A1 | Correct answer |
OR using normal approx $N(12, 4.8)$
$z = \frac{17.5 - 12}{\sqrt{4.8}} = 2.51$
Prob $= 1 - 0.9940 = 0.0060$
| M1 | Standardising, cc $16.5$ or $17.5$, their mean, $\sqrt{\text{(their var)}}$ |
| A1 | 2.51 seen |
| A1 | 3 marks | 0.0060 seen must be 0.0060 |
**(iii)** $\mu = 150 \times 0.60 = 90$
$\sigma^2 = 150 \times 0.60 \times 0.40 = 36$
$P(88 < X < 97) = \Phi\left(\frac{97.5 - 90}{6}\right) - \Phi\left(\frac{87.5 - 90}{6}\right)$
$= \Phi(1.25) - \Phi(-0.4166)$
$= 0.8944 - (1 - 0.6616) = 0.556$
| B1 | For seeing 90 and 36 |
| M1 | For standardising, with or without cc, must have sq rt on denom |
| M1 | One continuity correction $97.5$ or $96.5$ or $87.5$ or $88.5$ |
| A1 | $0.8944$ or $0.6616$ or $0.3384$ or $0.3944$ or $0.1616$ seen |
| M1 | Subtracting a probability from their standardised 97 prob |
| A1 | 6 marks | Correct answer |7 A manufacturer makes two sizes of elastic bands: large and small. $40 \%$ of the bands produced are large bands and $60 \%$ are small bands. Assuming that each pack of these elastic bands contains a random selection, calculate the probability that, in a pack containing 20 bands, there are\\
(i) equal numbers of large and small bands,\\
(ii) more than 17 small bands.
An office pack contains 150 elastic bands.\\
(iii) Using a suitable approximation, calculate the probability that the number of small bands in the office pack is between 88 and 97 inclusive.
\hfill \mbox{\textit{CAIE S1 2006 Q7 [11]}}