| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | November |
| Marks | 9 |
| 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 with a normal approximation in part (iii). Part (i) requires identifying p=0.25 and calculating P(X=5) for n=10, part (ii) is a simple binomial calculation with the probability from (i), and part (iii) is a standard normal approximation with continuity correction. All steps are routine S1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i): \(P(\text{equal}) = (0.25)^x (0.75)^x \times nC_5 = 0.0584\) | M1 | For \((0.25)^x \cdot (0.75)^x\) must be \(0.25, 0.75\) |
| A1 | 2 | For correct answer. A0 if subsequently doubled |
| Part (ii): \((0.0584)^1 \times (0.9416)^x \times nC_1 = 0.307\) | M1 | For \((their (i))^1 \times (1 - their (a))^x \times nC_1\) |
| A1 ft | 2 | For correct answer from their ans to (i). Accept anything from 0.304 to 0.307 for the ft if they have lost the A1 in (i) from PA |
| Part (iii): \(\mu = 120 \times 0.25 = 30, \quad \sigma^2 = 30 \times 0.75 = 22.5\) | M1 | For both mean and variance correct from any sensible \(p\) |
| M1 | For correct standardisation with or without cc | |
| \(P(X < 35) = \Phi\left(\frac{34.5 - 30}{\sqrt{22.5}}\right) = \Phi(0.949)\) | B1 | For correct use of continuity correction 34.5 |
| M1 | For use of tables based on their \(z\) value either end NB can't get if \(z\) is too large or too small | |
| \(= 0.829\) | A1 | 5 |
**Part (i):** $P(\text{equal}) = (0.25)^x (0.75)^x \times nC_5 = 0.0584$ | M1 | For $(0.25)^x \cdot (0.75)^x$ must be $0.25, 0.75$
| A1 | 2 | For correct answer. A0 if subsequently doubled
**Part (ii):** $(0.0584)^1 \times (0.9416)^x \times nC_1 = 0.307$ | M1 | For $(their (i))^1 \times (1 - their (a))^x \times nC_1$
| A1 ft | 2 | For correct answer from their ans to (i). Accept anything from 0.304 to 0.307 for the ft if they have lost the A1 in (i) from PA
**Part (iii):** $\mu = 120 \times 0.25 = 30, \quad \sigma^2 = 30 \times 0.75 = 22.5$ | M1 | For both mean and variance correct from any sensible $p$
| M1 | For correct standardisation with or without cc
$P(X < 35) = \Phi\left(\frac{34.5 - 30}{\sqrt{22.5}}\right) = \Phi(0.949)$ | B1 | For correct use of continuity correction 34.5
| M1 | For use of tables based on their $z$ value either end NB can't get if $z$ is too large or too small
$= 0.829$ | A1 | 5 | For correct answer
---6 (i) A manufacturer of biscuits produces 3 times as many cream ones as chocolate ones. Biscuits are chosen randomly and packed into boxes of 10 . Find the probability that a box contains equal numbers of cream biscuits and chocolate biscuits.\\
(ii) A random sample of 8 boxes is taken. Find the probability that exactly 1 of them contains equal numbers of cream biscuits and chocolate biscuits.\\
(iii) A large box of randomly chosen biscuits contains 120 biscuits. Using a suitable approximation, find the probability that it contains fewer than 35 chocolate biscuits.
\hfill \mbox{\textit{CAIE S1 2002 Q6 [9]}}