| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown variance confidence intervals |
| Difficulty | Moderate -0.8 This is a straightforward application of a confidence interval formula with unknown variance. Students need to calculate the sample mean, use the given sum of squared deviations to find the standard error, look up the critical value for 96% confidence, and apply the formula. It's computational rather than conceptual, requiring only routine recall of the CLT-based confidence interval procedure with no problem-solving insight. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| % \(\begin{aligned | 4 \text { The continuous random variable } X \text { has probability density fu } | ||
| \qquad f ( x ) = \begin{cases} \frac { 4 } { 99 } \left( 12 x - x ^ { 2 } - x ^ { 3 } \right) | 0 \leq x \leq 3 | ||
| 0 | \text { otherwise } \end{cases} \end{aligned}\)} | ||
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = \frac{60255}{200} = 301.275\) | B1 | Calculates sample mean; OE accepted |
| \(s^2 = \frac{995}{199} = 5\), so \(s = \sqrt{5}\) (AWRT \(2.24\)) | B1 | Calculates sample variance \(\frac{995}{199}\) OE, or standard deviation \(\sqrt{\frac{995}{199}}\) OE |
| \(z = 2.05(374891\ldots)\) | B1 | \(z\) value to at least 3 significant figures; can be implied by a correct confidence interval |
| \(\bar{x} \pm z\sqrt{\frac{s^2}{n}} = 301.275 \pm 2.05\sqrt{\frac{5}{200}}\) | M1 | Uses formula for confidence interval (PI); values substituted must be clear |
| \(= (301.0,\ 301.6)\) | A1 | Correct CI, correct way round; AWRT values to 1 d.p.; allow \(301\) for \(301.0\); allow if population variance \(\frac{995}{200}\) used |
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = \frac{60255}{200} = 301.275$ | B1 | Calculates sample mean; OE accepted |
| $s^2 = \frac{995}{199} = 5$, so $s = \sqrt{5}$ (AWRT $2.24$) | B1 | Calculates sample variance $\frac{995}{199}$ OE, or standard deviation $\sqrt{\frac{995}{199}}$ OE |
| $z = 2.05(374891\ldots)$ | B1 | $z$ value to at least 3 significant figures; can be implied by a correct confidence interval |
| $\bar{x} \pm z\sqrt{\frac{s^2}{n}} = 301.275 \pm 2.05\sqrt{\frac{5}{200}}$ | M1 | Uses formula for confidence interval (PI); values substituted must be clear |
| $= (301.0,\ 301.6)$ | A1 | Correct CI, correct way round; AWRT values to 1 d.p.; allow $301$ for $301.0$; allow if population variance $\frac{995}{200}$ used |
**Total: 5 marks**3 Fiona is studying the heights of corn plants on a farm.
She measures the height, $x \mathrm {~cm}$, of a random sample of 200 corn plants on the farm.\\
The summarised results are as follows:
$$\sum x = 60255 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 995$$
Calculate a $96 \%$ confidence interval for the population mean of heights of corn plants on the farm, giving your values to one decimal place.\\
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{3}{|l|}{%
\(\begin{aligned} & 4 \text { The continuous random variable } X \text { has probability density fu } \\ & \qquad f ( x ) = \begin{cases} \frac { 4 } { 99 } \left( 12 x - x ^ { 2 } - x ^ { 3 } \right) & 0 \leq x \leq 3 \\ 0 & \text { otherwise } \end{cases} \end{aligned}\)} \\
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2019 Q3 [5]}}