| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Distribution of sample mean |
| Difficulty | Moderate -0.8 This is a straightforward application of the sampling distribution of the mean with direct parameter substitution. Part (i) requires recalling that X̄ ~ N(μ, σ²/n), and part (ii) is a routine normal probability calculation with standardization. No problem-solving or conceptual insight needed beyond textbook formulas. |
| Spec | 5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(\bar{X} \sim N(48.8, 15.6^2/5)\) | B1 B1 [2] |
| (ii) | \(P(\bar{X} < 50) = \Phi\left(\frac{50-48.8}{\sqrt{15.6/\sqrt{5}}}\right)\) o.e. | M1 |
| \(= \Phi(0.1720) = 0.568\) | M1 A1 [3] | Correct area > 0.5. Correct answer |
(i) | $\bar{X} \sim N(48.8, 15.6^2/5)$ | B1 B1 [2] | For normal. Correct mean and variance/s.d.
(ii) | $P(\bar{X} < 50) = \Phi\left(\frac{50-48.8}{\sqrt{15.6/\sqrt{5}}}\right)$ o.e. | M1 | Standardising with sq root
$= \Phi(0.1720) = 0.568$ | M1 A1 [3] | Correct area > 0.5. Correct answer
---2 The lengths of time people take to complete a certain type of puzzle are normally distributed with mean 48.8 minutes and standard deviation 15.6 minutes. The random variable $X$ represents the time taken in minutes by a randomly chosen person to solve this type of puzzle. The times taken by random samples of 5 people are noted. The mean time $\bar { X }$ is calculated for each sample.\\
(i) State the distribution of $\bar { X }$, giving the values of any parameters.\\
(ii) Find $\mathrm { P } ( \bar { X } < 50 )$.
\hfill \mbox{\textit{CAIE S2 2008 Q2 [5]}}