| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Known variance confidence intervals |
| Difficulty | Moderate -0.3 This is a straightforward S1 confidence interval question with standard bookwork applications. Part (a) involves routine calculation of a confidence interval with known variance (given sample mean, n=25, σ=0.4), a simple interpretation, and recall of when CLT is needed. Part (b) applies standard normal distribution to sample means and individual observations. All steps are direct applications of formulas with no problem-solving insight required, though the multi-part structure and need for careful reading places it slightly below average difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = \frac{497.5}{25} = 19.9\) | B1 | Correct sample mean |
| \(z = 2.326\) (or \(2.33\)) for 98% CI | B1 | Correct \(z\) value |
| \(19.9 \pm 2.326 \times \frac{0.4}{\sqrt{25}}\) | M1 | Correct structure of CI |
| \(= 19.9 \pm 2.326 \times 0.08\) | A1 | Correct SE = \(\frac{0.4}{5} = 0.08\) |
| \(= (19.714, 20.086)\) or \((19.71, 20.09)\) | A1 | Correct interval (allow awrt) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(20\) is within/inside the confidence interval | M1 | Correct comparison stated |
| Therefore the claim is supported / plausible | A1 | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Because \(X\) is already normally distributed (the population is normal) | B1 | Must reference normality of population/distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{Y} \sim N\!\left(25.25, \frac{0.35^2}{10}\right)\) | M1 | Correct distribution for sample mean |
| \(= N(25.25, 0.01225)\) | A1 | Correct variance |
| \(P(\bar{Y} < 25) = P\!\left(Z < \frac{25 - 25.25}{0.35/\sqrt{10}}\right)\) | M1 | Standardising correctly |
| \(= P(Z < -2.2587...)= P(Z < -2.26)\) | ||
| \(= 1 - \Phi(2.26) = 1 - 0.9881 = 0.0119\) | A1 | Correct probability (awrt \(0.0119\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(Y > 25) = P\!\left(Z > \frac{25 - 25.25}{0.35}\right)\) | M1 | Using individual bag distribution |
| \(= P(Z > -0.7143)\) | A1 | Correct standardisation |
| \(= \Phi(0.7143) = 0.7625\) (awrt \(0.7625\)) | A1 | Correct single bag probability |
| \([P(Y>25)]^{10} = (0.7625)^{10}\) | M1 | Raising to power 10 |
| \(= 0.0特\) awrt \(0.0611\) | A1 | Correct final answer |
# Question 6:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = \frac{497.5}{25} = 19.9$ | B1 | Correct sample mean |
| $z = 2.326$ (or $2.33$) for 98% CI | B1 | Correct $z$ value |
| $19.9 \pm 2.326 \times \frac{0.4}{\sqrt{25}}$ | M1 | Correct structure of CI |
| $= 19.9 \pm 2.326 \times 0.08$ | A1 | Correct SE = $\frac{0.4}{5} = 0.08$ |
| $= (19.714, 20.086)$ or $(19.71, 20.09)$ | A1 | Correct interval (allow awrt) |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $20$ is within/inside the confidence interval | M1 | Correct comparison stated |
| Therefore the claim is supported / plausible | A1 | Correct conclusion in context |
## Part (a)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Because $X$ is already normally distributed (the population is normal) | B1 | Must reference normality of population/distribution |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{Y} \sim N\!\left(25.25, \frac{0.35^2}{10}\right)$ | M1 | Correct distribution for sample mean |
| $= N(25.25, 0.01225)$ | A1 | Correct variance |
| $P(\bar{Y} < 25) = P\!\left(Z < \frac{25 - 25.25}{0.35/\sqrt{10}}\right)$ | M1 | Standardising correctly |
| $= P(Z < -2.2587...)= P(Z < -2.26)$ | | |
| $= 1 - \Phi(2.26) = 1 - 0.9881 = 0.0119$ | A1 | Correct probability (awrt $0.0119$) |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(Y > 25) = P\!\left(Z > \frac{25 - 25.25}{0.35}\right)$ | M1 | Using individual bag distribution |
| $= P(Z > -0.7143)$ | A1 | Correct standardisation |
| $= \Phi(0.7143) = 0.7625$ (awrt $0.7625$) | A1 | Correct single bag probability |
| $[P(Y>25)]^{10} = (0.7625)^{10}$ | M1 | Raising to power 10 |
| $= 0.0特$ awrt $0.0611$ | A1 | Correct final answer |
These pages (22, 23, and 24) are **answer space pages** from an AQA examination paper (P61084/Jun13/MS/SS1B). They contain:
- Blank lined answer spaces for **Question 6**
- Page 24 is a blank page with "DO NOT WRITE ON THIS PAGE" instruction
**There is no mark scheme content on these pages.** These are student answer booklet pages, not mark scheme pages. The mark scheme would be a separate document published by AQA.
To find the actual mark scheme for this paper (AQA SS1B June 2013), you would need to access the official AQA mark scheme document directly from the AQA website or a licensed resource.6 The weight, $X$ kilograms, of sand in a bag can be modelled by a normal random variable with unknown mean $\mu$ and known standard deviation 0.4 .
\begin{enumerate}[label=(\alph*)]
\item The sand in each of a random sample of 25 bags from a large batch is weighed. The total weight of sand in these 25 bags is found to be 497.5 kg .
\begin{enumerate}[label=(\roman*)]
\item Construct a 98\% confidence interval for the mean weight of sand in bags in the batch.
\item Hence comment on the claim that bags in the batch contain an average of 20 kg of sand.
\item State why use of the Central Limit Theorem is not required in answering part (a)(i).
\end{enumerate}\item The weight, $Y$ kilograms, of cement in a bag can be modelled by a normal random variable with mean 25.25 and standard deviation 0.35.
A firm purchases 10 such bags. These bags may be considered to be a random sample.
\begin{enumerate}[label=(\roman*)]
\item Determine the probability that the mean weight of cement in the 10 bags is less than 25 kg .
\item Calculate the probability that the weight of cement in each of the 10 bags is more than 25 kg .
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{fbee7665-54e4-4805-9ce0-6244a4ba043c-20_1111_1707_1592_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{fbee7665-54e4-4805-9ce0-6244a4ba043c-23_2351_1707_219_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2013 Q6 [16]}}