| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Find minimum/maximum n for probability condition |
| Difficulty | Standard +0.3 Part (i) requires solving 0.8^n < 0.001 using logarithms, which is straightforward algebra. Part (ii) is a standard normal approximation to binomial with continuity correction (np=24, npq=19.2), requiring routine application of the formula and z-tables. Both parts are slightly easier than average A-level questions as they follow textbook procedures without requiring problem-solving insight. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((0.8)^n < 0.001\) | M1 | Eqn or inequ involving 0.8ⁿ or 0.2ⁿ and 0.001 or 0.999 |
| \(n > 30.9\) | M1 | Trial and error or logs (can be implied) |
| \(n = 31\) | A1 [3] | Correct answer; MR 0.01, max available M1M1A0 |
| (ii) \(\mu = 120 \times 0.2 = 24\) | B1 | 24 and 19.2 or \(\sqrt{19.2}\) seen |
| \(\sigma^2 = 120 \times 0.2 \times 0.8 = 19.2\) | M1 | Standardising with or without cc, must have sq rt in denom |
| \(P(x < 33) = P\left(z < \frac{32.5-24}{\sqrt{19.2}}\right)\) | M1 | Continuity correction 32.5 or 33.5 |
| \(= P(z < 1.9398) = 0.974\) | A1 [4] | Correct answer |
**(i)** $(0.8)^n < 0.001$ | M1 | Eqn or inequ involving 0.8ⁿ or 0.2ⁿ and 0.001 or 0.999
$n > 30.9$ | M1 | Trial and error or logs (can be implied)
$n = 31$ | A1 [3] | Correct answer; MR 0.01, max available M1M1A0
**(ii)** $\mu = 120 \times 0.2 = 24$ | B1 | 24 and 19.2 or $\sqrt{19.2}$ seen
$\sigma^2 = 120 \times 0.2 \times 0.8 = 19.2$ | M1 | Standardising with or without cc, must have sq rt in denom
$P(x < 33) = P\left(z < \frac{32.5-24}{\sqrt{19.2}}\right)$ | M1 | Continuity correction 32.5 or 33.5
$= P(z < 1.9398) = 0.974$ | A1 [4] | Correct answer4 In a certain country, on average one student in five has blue eyes.\\
(i) For a random selection of $n$ students, the probability that none of the students has blue eyes is less than 0.001 . Find the least possible value of $n$.\\
(ii) For a random selection of 120 students, find the probability that fewer than 33 have blue eyes.
\hfill \mbox{\textit{CAIE S1 2013 Q4 [7]}}