| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Known variance confidence intervals |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question with known variance. Part (a) is trivial arithmetic, parts (b) and (c) are standard bookwork applications of normal distribution and confidence intervals, and part (d) tests understanding rather than calculation. The multi-part structure adds marks but not conceptual difficulty—all parts follow directly from formulas with no problem-solving required. |
| Spec | 2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Mean = \(\frac{1620}{30} = 54\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(98\% (0.98) ⟹ z = 2.32\) to \(2.33\); CI for μ is \(\bar{x} \pm z×\frac{\sigma}{\sqrt{n}}\); Thus \(54 \pm 2.3263 \times \frac{8}{\sqrt{30}}\); Hence \(54 \pm (3.38\) to \(3.42)\) or \((50.58\) to \(50.62, 57.38\) to \(57.42)\) | B1 (M1) (A1F) (A1) | 4 |
| Notes: Use of \(n = 1\) in (b) must not be deemed as answer to (c); Use of \(n = 1\) in (b) followed by use of \(n = 1\) in (c) ⟹ (b) B1, (c) M1 A1 max; Use of \(n = 1\) with (b) or (c) not identified ⟹ (b) B1, (c) 0 max |
| Answer | Marks | Guidance |
|---|---|---|
| Repeat of structure in (b) but with \(n = 1\) and \(1.96 \leq z \leq 3.03\); Thus \(54 \pm (18.56\) to \(18.64)\) or \((35.36\) to \(35.44, 72.56\) to \(72.64)\) | M1 (A1F) | 2 |
| Note: Accept sensible non-symmetric intervals such as: \((0, 54 + 2.0537 \times 8) = (0, 70.4\) to \(70.5)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Nowhere or No | B1 | 1 |
**5(a)**
| Mean = $\frac{1620}{30} = 54$ | B1 | 1 | CAO; cannot be gained in (b) |
**5(b)**
| $98\% (0.98) ⟹ z = 2.32$ to $2.33$; CI for μ is $\bar{x} \pm z×\frac{\sigma}{\sqrt{n}}$; Thus $54 \pm 2.3263 \times \frac{8}{\sqrt{30}}$; Hence $54 \pm (3.38$ to $3.42)$ or $(50.58$ to $50.62, 57.38$ to $57.42)$ | B1 (M1) (A1F) (A1) | 4 | AWFW (2.3263); Used; Must have $\sqrt{n}$ with $n > 1$; F on $\bar{x}$ (but not 1620) and z only; Allow $\bar{x} = 54$ even if B0 in (a); CAO & AWFW (50.6, 57.4) |
| Notes: Use of $n = 1$ in (b) must not be deemed as answer to (c); Use of $n = 1$ in (b) followed by use of $n = 1$ in (c) ⟹ (b) B1, (c) M1 A1 max; Use of $n = 1$ with (b) or (c) not identified ⟹ (b) B1, (c) 0 max | | | |
**5(c)**
| Repeat of structure in (b) but with $n = 1$ and $1.96 \leq z \leq 3.03$; Thus $54 \pm (18.56$ to $18.64)$ or $(35.36$ to $35.44, 72.56$ to $72.64)$ | M1 (A1F) | 2 | Or equivalent; CAO & AWFW (54 & 18.6); If z-value incorrect, then must use $54 \pm 8×[\text{z from (b)}]$; AWFW (35.4, 72.6) |
| Note: Accept sensible non-symmetric intervals such as: $(0, 54 + 2.0537 \times 8) = (0, 70.4$ to $70.5)$ | | | |
**5(d)**
| Nowhere or No | B1 | 1 | CAO; or equivalent (ignore any reasoning) |
**Total for Q5: 8 marks**
---5 The times taken by new recruits to complete an assault course may be modelled by a normal distribution with a standard deviation of 8 minutes.
A group of 30 new recruits takes a total time of 1620 minutes to complete the course.
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean time taken by these 30 new recruits.
\item Assuming that the 30 recruits may be considered to be a random sample, construct a $98 \%$ confidence interval for the mean time taken by new recruits to complete the course.
\item Construct an interval within which approximately $98 \%$ of the times taken by individual new recruits to complete the course will lie.
\item State where, if at all, in this question you made use of the Central Limit Theorem.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2009 Q5 [8]}}