| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Non-normal population sample mean (CLT) |
| Difficulty | Standard +0.8 This question requires understanding of the Central Limit Theorem applied to a binomial distribution, including calculating the mean and variance of the sample mean distribution, then using normal approximation. While the calculations are straightforward once set up, recognizing that CLT applies to binomial samples and correctly deriving the parameters (μ = 6, σ²/n = 15×0.4×0.6/120) requires solid conceptual understanding beyond routine application, placing it moderately above average difficulty. |
| Spec | 2.04d Normal approximation to binomial5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Normal, mean \(6\), variance \(3.6/120\) \((0.03)\) | B1, B1, B1 (3 marks) | Correct mean. Correct variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\bar{X} > 6.1) = 1 - \Phi\left(\frac{6.1-6}{\sqrt{3.6/120}}\right) = 1 - \Phi(0.5773) = 1 - 0.7181 = 0.282\) | M1, M1, A1 (3 marks) | Standardising must have \(\sqrt{120}\). Correct area is \(< 0.5\). Correct answer |
**Part (i):**
Normal, mean $6$, variance $3.6/120$ $(0.03)$ | B1, B1, B1 (3 marks) | Correct mean. Correct variance
**Part (ii):**
$P(\bar{X} > 6.1) = 1 - \Phi\left(\frac{6.1-6}{\sqrt{3.6/120}}\right) = 1 - \Phi(0.5773) = 1 - 0.7181 = 0.282$ | M1, M1, A1 (3 marks) | Standardising must have $\sqrt{120}$. Correct area is $< 0.5$. Correct answer3 Random samples of size 120 are taken from the distribution $\mathrm { B } ( 15,0.4 )$.\\
(i) Describe fully the distribution of the sample mean.\\
(ii) Find the probability that the mean of a random sample of size 120 is greater than 6.1.
\hfill \mbox{\textit{CAIE S2 2006 Q3 [6]}}