| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Single tail probability P(X < a) or P(X > a) |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question with known standard deviation. Part (a) requires standard formula application (z-value lookup and calculation), part (b) is basic interpretation, and part (c) tests understanding that normality is already assumed. Slightly easier than average due to being routine bookwork with no problem-solving required, though the multi-part structure and interpretation elements keep it close to typical difficulty. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution |
| 155 | 148 | 156 | 149 | 147 | 156 |
| 157 | 156 | 150 | 154 | 148 | 154 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Volume \(\sim N(\mu, 3.5^2)\) | — | — |
| \(\bar{x} = \frac{1830}{12} = 152.5\) | B1 | CAO; \((s_{n-1} = 3.778,\ s_n = 3.617)\) |
| \(98\% \Rightarrow z = 2.3263\) | B1 | AWFW 2.32 to 2.33 |
| CI for \(\mu\) is \(\bar{x} \pm z \times \frac{(\sigma \text{ or } s)}{\sqrt{n}}\) | M1 | Must have \((\div\sqrt{n})\) with \(n > 1\) |
| \(152.5 \pm 2.3263 \times \frac{3.5}{\sqrt{12}}\) | A1\(\checkmark\) | ft on \(\bar{x}\) and \(z\) only |
| \((150.1 \text{ to } 150.2,\ 154.8 \text{ to } 154.9)\) | A1 | AWFW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Evidence from CI that mean volume is (above) 150 ml | B1\(\checkmark\) | ft on CI in part (a); must be clear comparison of mean of 150 with CI; or reference to range of can volumes in sample |
| In sample, some cans have volumes less than 150 ml | B1 | Dependent upon making some comment about mean volume and some comment about individual can volume or range of can volumes |
| Thus claim of 150 ml is not justified | B1dep | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Volume is normally distributed | E1 | Accept 'population' or '\(X\)'; but not 'it' or '\(\bar{X}\)' etc; must be clear statement; sample too small \(\Rightarrow\) E0 |
## Question 2:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Volume $\sim N(\mu, 3.5^2)$ | — | — |
| $\bar{x} = \frac{1830}{12} = 152.5$ | B1 | CAO; $(s_{n-1} = 3.778,\ s_n = 3.617)$ |
| $98\% \Rightarrow z = 2.3263$ | B1 | AWFW 2.32 to 2.33 |
| CI for $\mu$ is $\bar{x} \pm z \times \frac{(\sigma \text{ or } s)}{\sqrt{n}}$ | M1 | Must have $(\div\sqrt{n})$ with $n > 1$ |
| $152.5 \pm 2.3263 \times \frac{3.5}{\sqrt{12}}$ | A1$\checkmark$ | ft on $\bar{x}$ and $z$ only |
| $(150.1 \text{ to } 150.2,\ 154.8 \text{ to } 154.9)$ | A1 | AWFW |
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Evidence from CI that mean volume is (above) 150 ml | B1$\checkmark$ | ft on CI in part (a); must be clear comparison of mean of 150 with CI; or reference to range of can volumes in sample |
| In sample, some cans have volumes less than 150 ml | B1 | Dependent upon making some comment about mean volume and some comment about individual can volume or range of can volumes |
| Thus claim of 150 ml is not justified | B1dep | — |
### Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Volume is normally distributed | E1 | Accept 'population' or '$X$'; but not 'it' or '$\bar{X}$' etc; must be clear statement; sample too small $\Rightarrow$ E0 |
---2 The volume, in millilitres, of lemonade in mini-cans may be assumed to be normally distributed with a standard deviation of 3.5.
The volumes, in millilitres, of lemonade in a random sample of 12 mini-cans were as follows.
\begin{center}
\begin{tabular}{ l l l l l l }
155 & 148 & 156 & 149 & 147 & 156 \\
157 & 156 & 150 & 154 & 148 & 154 \\
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Construct a $98 \%$ confidence interval for the mean volume of lemonade in a mini-can, giving the limits to one decimal place.
\item On each mini-can is printed " 150 ml ". Comment on this, using the given sample and your confidence interval in part (a).
\item State why, in part (a), use of the Central Limit Theorem was not necessary.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2005 Q2 [9]}}