| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown variance confidence intervals |
| Difficulty | Moderate -0.3 This is a straightforward application of standard S1 techniques: calculating mean/SD from grouped data, applying the Central Limit Theorem with standard justification (n=36 is large enough), and finding a probability using normal distribution. All steps are routine textbook exercises requiring no problem-solving insight, though the multi-part structure and calculation burden place it slightly below average difficulty. |
| Spec | 2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.01a Permutations and combinations: evaluate probabilities |
| Time (seconds) | Number of customers |
| \(20 < x \leqslant 30\) | 2 |
| \(30 < x \leqslant 40\) | 7 |
| \(40 < x \leqslant 60\) | 18 |
| \(60 < x \leqslant 80\) | 27 |
| \(80 < x \leqslant 100\) | 23 |
| \(100 < x \leqslant 120\) | 13 |
| \(120 < x \leqslant 150\) | 7 |
| \(150 < x \leqslant 180\) | 3 |
| Total | 100 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a) | ||
| \(\sum fx = 8025\), \(\sum fx^2 = 739975\) | ||
| Mean \((\bar{x}) = 80.2\) to \(80.3\) | B2 | AWFW, 80.25 |
| Standard Deviation \((s_n, s_{n-1}) = 30.9\) to \(31.2\) | B2 | AWFW, 30.97882 or 31.13489 |
| MPs (x): 25, 35, 50, 70, 90, 110, 135, 165 | B1 | At least 4 correct |
| Mean \((\bar{x}) = \frac{\sum fx}{100}\) | M1 | Use of |
| 4 | ||
| 4(b)(i) | ||
| Large (\(n > 30\)) sample or Central Limit Theorem | B1 | OE |
| 4(b)(ii) | ||
| Mean \((\bar{Y}) = 80.2\) to \(80.3\) | B1\(\checkmark\) | \(\checkmark\) on (a) |
| Standard error \((\bar{Y}) = \frac{30.9 \text{ to } 31.2}{\sqrt{36}} = 5.1\) to \(5.25\) | M1 | \(\sqrt{s^2} > 0\) in (a) \(\checkmark\) \(\sqrt{36}\) or 6 |
| 2 | ||
| 4(b)(iii) | ||
| \(\mathbb{P}\left(\bar{Y} < 90\right) = \mathbb{P}\left(Z < \frac{90 - (80.2 \text{ to } 80.3)}{(5.1 \text{ to } 5.25)}\right)\) | M1 | Standardising 90 using values from (b)(ii) with \(\sqrt{s^2/36} > 0\) or \(\sqrt{s^2/100} > 0\) |
| \(= \mathbb{P}(Z < 1.84\) to \(1.93)\) | M1 | |
| \(= 0.967\) to \(0.974\) | A1 | AWFW |
| 3 |
| **4(a)** |
|----------|
| $\sum fx = 8025$, $\sum fx^2 = 739975$ | | |
| Mean $(\bar{x}) = 80.2$ to $80.3$ | B2 | AWFW, 80.25 |
| Standard Deviation $(s_n, s_{n-1}) = 30.9$ to $31.2$ | B2 | AWFW, 30.97882 or 31.13489 |
| MPs (x): 25, 35, 50, 70, 90, 110, 135, 165 | B1 | At least 4 correct |
| Mean $(\bar{x}) = \frac{\sum fx}{100}$ | M1 | Use of |
| | | 4 |
| **4(b)(i)** |
|----------|
| Large ($n > 30$) sample or Central Limit Theorem | B1 | OE |
| **4(b)(ii)** |
|----------|
| Mean $(\bar{Y}) = 80.2$ to $80.3$ | B1$\checkmark$ | $\checkmark$ on (a) |
| Standard error $(\bar{Y}) = \frac{30.9 \text{ to } 31.2}{\sqrt{36}} = 5.1$ to $5.25$ | M1 | $\sqrt{s^2} > 0$ in (a) $\checkmark$ $\sqrt{36}$ or 6 |
| | | 2 |
| **4(b)(iii)** |
|----------|
| $\mathbb{P}\left(\bar{Y} < 90\right) = \mathbb{P}\left(Z < \frac{90 - (80.2 \text{ to } 80.3)}{(5.1 \text{ to } 5.25)}\right)$ | M1 | Standardising 90 using values from (b)(ii) with $\sqrt{s^2/36} > 0$ or $\sqrt{s^2/100} > 0$ |
| $= \mathbb{P}(Z < 1.84$ to $1.93)$ | M1 | |
| $= 0.967$ to $0.974$ | A1 | AWFW |
| | | 3 |
**Total for Q4: 10**
---4 The time, $x$ seconds, spent by each of a random sample of 100 customers at an automatic teller machine (ATM) is recorded. The times are summarised in the table.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Time (seconds) & Number of customers \\
\hline
$20 < x \leqslant 30$ & 2 \\
\hline
$30 < x \leqslant 40$ & 7 \\
\hline
$40 < x \leqslant 60$ & 18 \\
\hline
$60 < x \leqslant 80$ & 27 \\
\hline
$80 < x \leqslant 100$ & 23 \\
\hline
$100 < x \leqslant 120$ & 13 \\
\hline
$120 < x \leqslant 150$ & 7 \\
\hline
$150 < x \leqslant 180$ & 3 \\
\hline
Total & 100 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate estimates for the mean and standard deviation of the time spent at the ATM by a customer.
\item The mean time spent at the ATM by a random sample of $\mathbf { 3 6 }$ customers is denoted by $\bar { Y }$.
\begin{enumerate}[label=(\roman*)]
\item State why the distribution of $\bar { Y }$ is approximately normal.
\item Write down estimated values for the mean and standard error of $\bar { Y }$.
\item Hence estimate the probability that $\bar { Y }$ is less than $1 \frac { 1 } { 2 }$ minutes.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2006 Q4 [10]}}