| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct comparison of probabilities |
| Difficulty | Moderate -0.3 This is a standard S1 question testing routine application of sampling distributions and confidence intervals. Part (a) requires straightforward use of the Central Limit Theorem with known σ. Part (b) involves textbook confidence interval construction and interpretation of claims using the interval. All techniques are direct applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.04b Linear combinations: of normal distributions5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{X} \sim N\left(10.15, \frac{0.3^2}{12}\right)\) | M1 | Correct distribution for sample mean, may be implied |
| \(P(\bar{X} < 10) = P\left(Z < \frac{10 - 10.15}{0.3/\sqrt{12}}\right)\) | M1 | Standardising with \(\sqrt{12}\) in denominator |
| \(= P(Z < -1.732...)\) | A1 | Correct z-value |
| \(= 1 - \Phi(1.732) = 1 - 0.9584 = 0.0416\) | A1 | Correct probability (accept 0.0418) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} \pm z \times \frac{s}{\sqrt{n}}\) | M1 | Correct structure of confidence interval |
| \(z = 2.576\) for 99% | B1 | Correct z-value |
| \(304.6 \pm 2.576 \times \frac{5.37}{\sqrt{40}}\) | A1 | Correct values substituted |
| \((302.4, 306.8)\) | A1 | Both limits correct to 1 d.p. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Claim 1: 300 is below the lower limit of 302.4, so there is evidence to support the claim that the mean weight is more than 300g | B1 | Must reference the confidence interval and 300 being outside/below it |
| Claim 1 is supported / accepted | B1 | Correct conclusion for Claim 1 with justification |
| Claim 2: The confidence interval is about the mean, not individual values; the distribution of individual pouches has variation, so we cannot conclude all pouches contain more than 300g | B1 | Must distinguish between mean and individual values |
| Claim 2 is not justified / cannot be supported | B1 | Correct conclusion for Claim 2 with justification |
# Question 7:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{X} \sim N\left(10.15, \frac{0.3^2}{12}\right)$ | M1 | Correct distribution for sample mean, may be implied |
| $P(\bar{X} < 10) = P\left(Z < \frac{10 - 10.15}{0.3/\sqrt{12}}\right)$ | M1 | Standardising with $\sqrt{12}$ in denominator |
| $= P(Z < -1.732...)$ | A1 | Correct z-value |
| $= 1 - \Phi(1.732) = 1 - 0.9584 = 0.0416$ | A1 | Correct probability (accept 0.0418) |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} \pm z \times \frac{s}{\sqrt{n}}$ | M1 | Correct structure of confidence interval |
| $z = 2.576$ for 99% | B1 | Correct z-value |
| $304.6 \pm 2.576 \times \frac{5.37}{\sqrt{40}}$ | A1 | Correct values substituted |
| $(302.4, 306.8)$ | A1 | Both limits correct to 1 d.p. |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Claim 1:** 300 is below the lower limit of 302.4, so there is evidence to support the claim that the mean weight is more than 300g | B1 | Must reference the confidence interval and 300 being outside/below it |
| Claim 1 is supported / accepted | B1 | Correct conclusion for Claim 1 with justification |
| **Claim 2:** The confidence interval is about the **mean**, not individual values; the distribution of individual pouches has variation, so we cannot conclude all pouches contain more than 300g | B1 | Must distinguish between mean and individual values |
| Claim 2 is not justified / cannot be supported | B1 | Correct conclusion for Claim 2 with justification |7
\begin{enumerate}[label=(\alph*)]
\item The weight of a sack of mixed dog biscuits can be modelled by a normal distribution with a mean of 10.15 kg and a standard deviation of 0.3 kg .
A pet shop purchases 12 such sacks that can be considered to be a random sample.\\
Calculate the probability that the mean weight of the 12 sacks is less than 10 kg .
\item The weight of dry cat food in a pouch can also be modelled by a normal distribution.
The contents, $x$ grams, of each of a random sample of 40 pouches were weighed. Subsequent analysis of these weights gave
$$\bar { x } = 304.6 \quad \text { and } \quad s = 5.37$$
\begin{enumerate}[label=(\roman*)]
\item Construct a $99 \%$ confidence interval for the mean weight of dry cat food in a pouch. Give the limits to one decimal place.
\item Comment, with justification, on each of the following two claims.
Claim 1: The mean weight of dry cat food in a pouch is more than 300 grams.\\
Claim 2: All pouches contain more than 300 grams of dry cat food.\\[0pt]
[4 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{6fbb8891-e6de-42fe-a195-ea643552fdcf-24_2288_1705_221_155}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2015 Q7 [12]}}