| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Sample size determination |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question requiring standard formula manipulation. Part (a) involves using the CI width formula to solve for n (routine algebra with known variance), and part (b) is a simple interpretation check. While it's Further Maths content, the mechanics are direct application of formulas without conceptual difficulty or multi-step problem-solving. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = 2.5758\) | B1 | AWRT 2.58; PI |
| \(\dfrac{5.429 - 5.239}{2} = 2.5758 \times \sqrt{\dfrac{0.7}{n}}\) | M1 | Forms an equation containing their \(2.5758 \times \sqrt{\dfrac{0.7}{n}}\); PI |
| \(n = 515\) | A1 | Finds correct value of \(n\); whole number from 510 to 520 inclusive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The confidence interval supports the claim as 5.3 is within the interval | E1 | Infers that the confidence interval supports Joey's claim as 5.3 lies within the interval; condone use of "it" for 5.3; condone "between the values" for being within the interval |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = 2.5758$ | B1 | AWRT 2.58; PI |
| $\dfrac{5.429 - 5.239}{2} = 2.5758 \times \sqrt{\dfrac{0.7}{n}}$ | M1 | Forms an equation containing their $2.5758 \times \sqrt{\dfrac{0.7}{n}}$; PI |
| $n = 515$ | A1 | Finds correct value of $n$; whole number from 510 to 520 inclusive |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The confidence interval supports the claim as 5.3 is within the interval | E1 | Infers that the confidence interval supports Joey's claim as 5.3 lies within the interval; condone use of "it" for 5.3; condone "between the values" for being within the interval |
---4 The height of lilac trees, in metres, can be modelled by a normal distribution with variance 0.7
A random sample of $n$ lilac trees is taken and used to construct a 99\% confidence interval for the population mean.
This confidence interval is $( 5.239,5.429 )$\\
4
\begin{enumerate}[label=(\alph*)]
\item Find the value of $n$\\
4
\item Joey claims that the mean height of lilac trees is 5.3 metres.\\
State, with a reason, whether the confidence interval supports Joey's claim.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2022 Q4 [4]}}