| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for sampling distribution |
| Difficulty | Moderate -0.3 This is a straightforward application of the Central Limit Theorem with standard bookwork justifications. Part (a)(i) requires recognizing that volume cannot be negative (using the empirical rule), part (a)(ii) is direct recall of CLT conditions (n=80 is large), and part (b) involves routine confidence interval construction and interpretation. While it requires understanding of several concepts, each step follows a standard template with no novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V\) cannot be negative / volume must be positive, but normal distribution extends to \(-\infty\) | B1 | Accept: the standard deviation (65) is large relative to mean, so normal would give negative values with non-negligible probability |
| \(V\) is likely to be skewed (e.g. most people use a small amount but some use very large amounts) | B1 | Need numerical justification e.g. \(\mu - 2\sigma\) could be negative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| By the Central Limit Theorem | B1 | Must mention CLT |
| The sample size \(n = 80\) is large (sufficiently large) | B1 | Must reference large sample size |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(98\%\) CI so \(z = 2.3263\) | B1 | Accept \(z = 2.326\) or \(2.33\) |
| \(\bar{v} \pm z \times \dfrac{\sigma}{\sqrt{n}} = 118 \pm 2.3263 \times \dfrac{65}{\sqrt{80}}\) | M1 | Correct structure of CI |
| \(= 118 \pm 16.9...\) | A1 | Correct calculation |
| \((101, 135)\) or \((101.1, 134.9)\) | A1 | Accept equivalent correct interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(140\) lies outside the confidence interval \((101, 135)\) | M1 | Must reference their CI from (b)(i) |
| There is evidence to doubt/reject the claim that \(\mu = 140\) | A1 | Must be a contextualised conclusion; follow through from their CI |
# Question 7:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $V$ cannot be negative / volume must be positive, but normal distribution extends to $-\infty$ | B1 | Accept: the standard deviation (65) is large relative to mean, so normal would give negative values with non-negligible probability |
| $V$ is likely to be skewed (e.g. most people use a small amount but some use very large amounts) | B1 | Need numerical justification e.g. $\mu - 2\sigma$ could be negative |
**[2 marks]**
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| By the Central Limit Theorem | B1 | Must mention CLT |
| The sample size $n = 80$ is large (sufficiently large) | B1 | Must reference large sample size |
**[2 marks]**
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $98\%$ CI so $z = 2.3263$ | B1 | Accept $z = 2.326$ or $2.33$ |
| $\bar{v} \pm z \times \dfrac{\sigma}{\sqrt{n}} = 118 \pm 2.3263 \times \dfrac{65}{\sqrt{80}}$ | M1 | Correct structure of CI |
| $= 118 \pm 16.9...$ | A1 | Correct calculation |
| $(101, 135)$ or $(101.1, 134.9)$ | A1 | Accept equivalent correct interval |
**[4 marks]**
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $140$ lies outside the confidence interval $(101, 135)$ | M1 | Must reference their CI from (b)(i) |
| There is evidence to doubt/reject the claim that $\mu = 140$ | A1 | Must be a contextualised conclusion; follow through from their CI |
**[2 marks]**
These pages (27 and 28) of the AQA June 2014 MS1B mark scheme are blank continuation/overflow pages containing no questions or marking content. They simply state:
- "There are no questions printed on this page"
- "DO NOT WRITE ON THIS PAGE / ANSWER IN THE SPACES PROVIDED"
There is no mark scheme content to extract from these pages.7 The volume of water, $V$, used by a guest in an en suite shower room at a small guest house may be modelled by a random variable with mean $\mu$ litres and standard deviation 65 litres.
A random sample of 80 guests using this shower room showed a mean usage of 118 litres of water.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Give a numerical justification as to why $V$ is unlikely to be normally distributed.
\item Explain why $\bar { V }$, the mean of a random sample of 80 observations of $V$, may be assumed to be approximately normally distributed.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Construct a $98 \%$ confidence interval for $\mu$.
\item Hence comment on a claim that $\mu$ is 140 .
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-24_1526_1709_1181_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-25_2484_1707_221_153}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-27_2490_1719_217_150}\\
\includegraphics[max width=\textwidth, alt={}, center]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-28_2486_1728_221_141}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2014 Q7 [10]}}