| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Topic | Central limit theorem |
| Type | Normal population, known parameters |
| Difficulty | Moderate -0.3 This is a straightforward application of sampling distributions with normal populations. Part (i) requires finding the distribution of the difference of two sample means (routine calculation with known formulas), and part (ii) tests basic theoretical knowledge. The populations are already normal, so no CLT invocation is needed, making this easier than typical sampling distribution problems. |
| Spec | 5.04b Linear combinations: of normal distributions5.05a Sample mean distribution: central limit theorem |
(i) $\bar{B} \sim \left(156, \frac{64}{9}\right)$, $\bar{G} \sim \left(160, \frac{49}{16}\right)$ (Can be implied by working.) **B1B1**
$\bar{B} - \bar{G} \sim N\left(-4, \frac{1465}{144}\right)$ **B1M1A1**
$z = \frac{0-(-4)}{\sqrt{\frac{1465}{144}}} = 1.254$ **M1A1**
$1 - \phi(1.254) = 0.1049 = 0.105$ (3sf) (AWRT 0.105) **A1** [8]
(ii) Samples are taken from underlying normal distributions $\Rightarrow$ distributions of sample means are normal. **B1** [1]2 (i) The heights of boys in Year 9 are normally distributed with mean 156 cm and standard deviation 8 cm . The heights of girls in Year 10 are, independently, normally distributed with mean 160 cm and standard deviation 7 cm . Find the probability that the mean height of a random sample of 9 boys in Year 9 exceeds the mean height of a random sample of 16 girls in Year 10.\\
(ii) State why the distributions of the sample means are normally distributed.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q2 [9]}}