| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct comparison of probabilities |
| Difficulty | Moderate -0.3 This is a straightforward application of sampling distribution of the mean (part a) and confidence interval construction (part b). Both require standard procedures: calculating standard error, using z-tables, and interpreting results. The comparison in (b)(ii) is routine. Slightly easier than average due to being direct application of formulas with no conceptual complications. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.01a Permutations and combinations: evaluate probabilities5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\bar{X} \sim N\left(1.16, \frac{0.43^2}{10}\right)\) | M1 | Using sampling distribution |
| \(P(\bar{X} > 1.25) = P\left(Z > \frac{1.25 - 1.16}{0.43/\sqrt{10}}\right)\) | M1 | Standardising |
| \(= P(Z > 0.6621...)\) | A1 | Correct z-value |
| \(= 1 - \Phi(0.6621) \approx 0.2540\) | A1 | Correct probability |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0.86 \pm 2.0537 \times \frac{0.65}{\sqrt{40}}\) | M1+B1+M1 | 96% CI; z = 2.0537 |
| \((0.649, 1.071)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Mean apples weight 1.16 kg lies outside/above the CI for pears | B1 | |
| Claim is supported — apples heavier on average | B1 |
## Question 7(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\bar{X} \sim N\left(1.16, \frac{0.43^2}{10}\right)$ | M1 | Using sampling distribution |
| $P(\bar{X} > 1.25) = P\left(Z > \frac{1.25 - 1.16}{0.43/\sqrt{10}}\right)$ | M1 | Standardising |
| $= P(Z > 0.6621...)$ | A1 | Correct z-value |
| $= 1 - \Phi(0.6621) \approx 0.2540$ | A1 | Correct probability |
## Question 7(b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0.86 \pm 2.0537 \times \frac{0.65}{\sqrt{40}}$ | M1+B1+M1 | 96% CI; z = 2.0537 |
| $(0.649, 1.071)$ | A1 | |
## Question 7(b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Mean apples weight 1.16 kg lies **outside/above** the CI for pears | B1 | |
| Claim is **supported** — apples heavier on average | B1 | |7
\begin{enumerate}[label=(\alph*)]
\item A greengrocer displays apples in trays. Each customer selects the apples he or she wishes to buy and puts them into a bag.
Records show that the weight of such bags of apples may be modelled by a normal distribution with mean 1.16 kg and standard deviation 0.43 kg .
Determine the probability that the mean weight of a random sample of 10 such bags of apples exceeds 1.25 kg .
\item The greengrocer also displays pears in trays. Each customer selects the pears he or she wishes to buy and puts them into a bag.
A random sample of 40 such bags of pears had a mean weight of 0.86 kg and a standard deviation of 0.65 kg .
\begin{enumerate}[label=(\roman*)]
\item Construct a $\mathbf { 9 6 \% }$ confidence interval for the mean weight of a bag of pears.
\item Hence comment on a claim that customers wish to buy, on average, a greater weight of apples than of pears.\\[0pt]
[2 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2015 Q7 [10]}}