| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Topic | Confidence intervals |
| Type | Validity or suitability of sample |
| Difficulty | Moderate -0.5 Part (i) is a straightforward application of the large-sample confidence interval formula (mean ± 1.96×SE), requiring only basic substitution. Part (ii) tests understanding of the Central Limit Theorem (n=500 is large enough that normality assumption is unnecessary), which is conceptual but standard knowledge. Overall easier than average due to routine calculation and well-known theoretical point. |
| Spec | 5.05d Confidence intervals: using normal distribution |
(i) $75 \pm 1.96\sqrt{\dfrac{40^2}{500} \times \dfrac{500}{499}}$
$= (71.5, 78.5)$
- M1: $75 \pm zs$, $s$ involving 500
- B1: $z = 1.96$, allow from no 500
- A1: Variance correct
- A1: Both limits correct to 3sf (Condone omission of $\dfrac{500}{499}$)
**[4]**
(ii) No, as the Central Limit Theorem applies *OR* as $n$ is large
- B1: "No" and mention CLT or large sample size; focus on different distributions; no irrelevancies
**[1]**1 An investigation was carried out of the lengths of commuters' journeys. For a random sample of 500 commuters, the mean journey time was 75 minutes, and the standard deviation was 40 minutes.\\
(i) Calculate a 95\% confidence interval for the mean journey time.\\
(ii) Explain whether you need to assume that journey times are normally distributed.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q1 [5]}}