| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Estimate from grouped frequency data |
| Difficulty | Moderate -0.3 This is a standard S1 grouped data question requiring routine calculations (mean, standard deviation from grouped data) and straightforward confidence interval construction. Part (a) tests basic understanding of grouped data boundaries, (b) is mechanical calculation, (c) applies standard normal confidence interval formula, and (d) requires simple interpretation. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05c Hypothesis test: normal distribution for population mean |
| Volume ( \(\boldsymbol { x }\) litres) | Number of cartons (f) |
| 0.95-0.97 | 2 |
| 0.98-1.00 | 7 |
| 1.01-1.03 | 15 |
| 1.04-1.06 | 32 |
| 1.07-1.09 | 22 |
| 1.10-1.12 | 14 |
| 1.13-1.15 | 7 |
| 1.16-1.18 | 1 |
| Total | 100 |
| Answer | Marks | Guidance |
|---|---|---|
| Mid-point of 0.98–1.00 is \(\frac{0.98 + 1.00}{2} = 0.99\) | B1 | Show working clearly |
| Answer | Marks | Guidance |
|---|---|---|
| Minimum: \(x = 0.975\), Maximum: \(x = 1.005\) | B1B1 | B1 each value; accept 0.9750... and 1.0049... |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{\sum fx}{\sum f} = \frac{2(0.96)+7(0.99)+15(1.02)+32(1.05)+22(1.08)+14(1.11)+7(1.14)+1(1.17)}{100}\) | M1 | Using midpoints with frequencies |
| \(\bar{x} = \frac{105.39}{100} = 1.054\) (3 d.p.) | A1 | |
| \(s^2 = \frac{\sum fx^2}{\sum f} - \bar{x}^2\), \(s = \sqrt{\frac{\sum fx^2}{100} - \bar{x}^2}\) | M1 | Correct formula for standard deviation |
| \(s = 0.049\) (3 d.p.) | A1 | Accept \(\sigma = 0.049\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} \pm 2.576 \times \frac{s}{\sqrt{n}}\) | M1 | Correct structure of CI |
| \(1.054 \pm 2.576 \times \frac{0.049}{\sqrt{100}}\) | A1 | Correct values substituted |
| \(1.054 \pm 0.01262...\) | A1 | |
| \((1.041, 1.067)\) | A1 | Accept \((1.0414, 1.0666)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(X\) is already normally distributed (so \(\bar{X}\) is exactly normal for any \(n\)) | B1 | Must reference normality of \(X\) |
| Answer | Marks | Guidance |
|---|---|---|
| The standard deviation \(\sigma\) is unknown and estimated by \(s\) | B1 | Or: population variance unknown/estimated |
| Answer | Marks | Guidance |
|---|---|---|
| The confidence interval lies entirely above 1 (lower limit \(> 1\)), supporting \(\mu > 1\) | B1 | Or: sample mean \(= 1.054 > 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| The interval \((0.94, 1.16)\) covers approximately \(\mu \pm 2\sigma\) (since \(s \approx 0.049\), \(2s \approx 0.1\)), and for a normal distribution \(P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 0.954 \approx 1\) | B1 | Accept: range is approximately \(\mu \pm 2\sigma\), nearly all data lies within this interval |
# Question 3:
## Part (a)(i)
| Mid-point of 0.98–1.00 is $\frac{0.98 + 1.00}{2} = 0.99$ | B1 | Show working clearly |
## Part (a)(ii)
| Minimum: $x = 0.975$, Maximum: $x = 1.005$ | B1B1 | B1 each value; accept 0.9750... and 1.0049... |
## Part (b)
| $\bar{x} = \frac{\sum fx}{\sum f} = \frac{2(0.96)+7(0.99)+15(1.02)+32(1.05)+22(1.08)+14(1.11)+7(1.14)+1(1.17)}{100}$ | M1 | Using midpoints with frequencies |
| $\bar{x} = \frac{105.39}{100} = 1.054$ (3 d.p.) | A1 | |
| $s^2 = \frac{\sum fx^2}{\sum f} - \bar{x}^2$, $s = \sqrt{\frac{\sum fx^2}{100} - \bar{x}^2}$ | M1 | Correct formula for standard deviation |
| $s = 0.049$ (3 d.p.) | A1 | Accept $\sigma = 0.049$ |
## Part (c)(i)
| $\bar{x} \pm 2.576 \times \frac{s}{\sqrt{n}}$ | M1 | Correct structure of CI |
| $1.054 \pm 2.576 \times \frac{0.049}{\sqrt{100}}$ | A1 | Correct values substituted |
| $1.054 \pm 0.01262...$ | A1 | |
| $(1.041, 1.067)$ | A1 | Accept $(1.0414, 1.0666)$ |
## Part (c)(ii)
| $X$ is already normally distributed (so $\bar{X}$ is exactly normal for any $n$) | B1 | Must reference normality of $X$ |
## Part (c)(iii)
| The standard deviation $\sigma$ is unknown and estimated by $s$ | B1 | Or: population variance unknown/estimated |
## Part (d)(i)
| The confidence interval lies entirely above 1 (lower limit $> 1$), supporting $\mu > 1$ | B1 | Or: sample mean $= 1.054 > 1$ |
## Part (d)(ii)
| The interval $(0.94, 1.16)$ covers approximately $\mu \pm 2\sigma$ (since $s \approx 0.049$, $2s \approx 0.1$), and for a normal distribution $P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 0.954 \approx 1$ | B1 | Accept: range is approximately $\mu \pm 2\sigma$, nearly all data lies within this interval |3 The volume, $X$ litres, of orange juice in a 1-litre carton may be modelled by a normal distribution with unknown mean $\mu$.
The volumes, $x$ litres, recorded to the nearest 0.01 litre, in a random sample of 100 cartons are shown in the table.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Volume ( $\boldsymbol { x }$ litres) & Number of cartons (f) \\
\hline
0.95-0.97 & 2 \\
\hline
0.98-1.00 & 7 \\
\hline
1.01-1.03 & 15 \\
\hline
1.04-1.06 & 32 \\
\hline
1.07-1.09 & 22 \\
\hline
1.10-1.12 & 14 \\
\hline
1.13-1.15 & 7 \\
\hline
1.16-1.18 & 1 \\
\hline
Total & 100 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item For the group ' $0.98 - 1.00$ ':
\begin{enumerate}[label=(\roman*)]
\item show that it has a mid-point of 0.99 litres;
\item state the minimum and the maximum values of $x$ that could be included in this group.
\end{enumerate}\item Calculate, to three decimal places, estimates of the mean and the standard deviation of these 100 volumes.
\item \begin{enumerate}[label=(\roman*)]
\item Construct an approximate $99 \%$ confidence interval for $\mu$.
\item State why use of the Central Limit Theorem was not required when calculating this confidence interval.
\item Give a reason why the confidence interval is approximate rather than exact.
\end{enumerate}\item Give a reason in support of the claim that:
\begin{enumerate}[label=(\roman*)]
\item $\mu > 1$;
\item $\mathrm { P } ( 0.94 < X < 1.16 )$ is approximately 1 .
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{156f9453-ebc6-4406-b5bc-08d1918ebc62-10_2486_1714_221_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{156f9453-ebc6-4406-b5bc-08d1918ebc62-11_2486_1714_221_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2011 Q3 [13]}}