| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown variance confidence intervals |
| Difficulty | Standard +0.3 This is a standard S1 question testing routine application of CLT and confidence intervals with grouped data. Parts (a)-(d) involve straightforward calculations using mid-interval values and standard formulas. Part (e) requires basic interpretation. The grouped data adds minor computational complexity but no conceptual challenge beyond typical A-level statistics. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.01a Permutations and combinations: evaluate probabilities5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Number of visitors per day | Number of days |
| 1-10 | 13 |
| 11-20 | 33 |
| 21-25 | 17 |
| 26-30 | 12 |
| 31-35 | 8 |
| 36-40 | 5 |
| 41-50 | 5 |
| 51-100 | 7 |
| Total | 100 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \((\bar{x}) = 24.7\) to \(25.7\) | B2 | AWFW (25.2) |
| Standard Deviation \((s_n, s_{n-1}) = 16.7\) to \(17.7\) | B2 | AWFW (17.1474 or 17.2338) |
| MPs \((x)\): 5.5, 15.5, 23, 28, 33, 38, 45.5, 75.5 | (B1) | At least 4 correct |
| Mean \((\bar{x}) = \dfrac{\sum fx}{100}\) | (M1) | Use of |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Data is skewed or not symmetric; Discrete data or counts; \((\text{Mean} - 2 \times \text{SD}) < 0 \Rightarrow\) negative counts | B1 | One valid reason |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Since sample size large \((n > 30)\) can use Central Limit Theorem | B1 | Either point |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(= \mu\) | B1 | CAO; not \(\bar{x}\) or its value |
| Variance \(= \dfrac{\sigma^2}{100}\) | B1 | Accept \(\dfrac{\sigma^2}{n}\) or \(\dfrac{(\text{their SD})^2}{100}\), etc |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(99\% \Rightarrow z = 2.57\) to \(2.58\) | B1 | AWFW (2.5758) |
| CI for \(\mu\) is \(\bar{x} \pm z \times \dfrac{(\sigma \text{ or } s)}{\sqrt{n}}\) | M1 | Use of. Must have \((\div\sqrt{n})\) with \(n > 1\) |
| Thus \(25.2 \pm 2.5758 \times \dfrac{17.1 \text{ or } 17.2}{\sqrt{100}}\) | A1\(\checkmark\) | \(\checkmark\) on \(\bar{x}\), \(z\) and \(s > 0\); not on \(n\) |
| \((20.8,\ 29.6)\) | A1 | AWRT |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{UCL} < 30\) | B1\(\checkmark\) | \(\checkmark\) on CI |
| so Reject claim that \(\mu > 30\) | \(\uparrow\)dep B1\(\checkmark\) | \(\checkmark\) on CI |
| \(7/100\) or \(7\%\) of \(X > 50\) (from table) | B1 | CAO |
| so Reject claim that often \(X > 50\) | \(\uparrow\)dep B1 | CAO |
| Total: 4 marks |
# Question 6:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $(\bar{x}) = 24.7$ to $25.7$ | B2 | AWFW (25.2) |
| Standard Deviation $(s_n, s_{n-1}) = 16.7$ to $17.7$ | B2 | AWFW (17.1474 or 17.2338) |
| MPs $(x)$: 5.5, 15.5, 23, 28, 33, 38, 45.5, 75.5 | (B1) | At least 4 correct |
| Mean $(\bar{x}) = \dfrac{\sum fx}{100}$ | (M1) | Use of |
| **Total: 4 marks** | | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Data is skewed or not symmetric; Discrete data or counts; $(\text{Mean} - 2 \times \text{SD}) < 0 \Rightarrow$ negative counts | B1 | One valid reason |
| **Total: 1 mark** | | |
## Part (c)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Since sample size large $(n > 30)$ can use Central Limit Theorem | B1 | Either point |
| **Total: 1 mark** | | |
## Part (c)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= \mu$ | B1 | CAO; not $\bar{x}$ or its value |
| Variance $= \dfrac{\sigma^2}{100}$ | B1 | Accept $\dfrac{\sigma^2}{n}$ or $\dfrac{(\text{their SD})^2}{100}$, etc |
| **Total: 2 marks** | | |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $99\% \Rightarrow z = 2.57$ to $2.58$ | B1 | AWFW (2.5758) |
| CI for $\mu$ is $\bar{x} \pm z \times \dfrac{(\sigma \text{ or } s)}{\sqrt{n}}$ | M1 | Use of. Must have $(\div\sqrt{n})$ with $n > 1$ |
| Thus $25.2 \pm 2.5758 \times \dfrac{17.1 \text{ or } 17.2}{\sqrt{100}}$ | A1$\checkmark$ | $\checkmark$ on $\bar{x}$, $z$ and $s > 0$; not on $n$ |
| $(20.8,\ 29.6)$ | A1 | AWRT |
| **Total: 4 marks** | | |
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{UCL} < 30$ | B1$\checkmark$ | $\checkmark$ on CI |
| so Reject claim that $\mu > 30$ | $\uparrow$dep B1$\checkmark$ | $\checkmark$ on CI |
| $7/100$ or $7\%$ of $X > 50$ (from table) | B1 | CAO |
| so Reject claim that often $X > 50$ | $\uparrow$dep B1 | CAO |
| **Total: 4 marks** | | |6 On arrival at a business centre, all visitors are required to register at the reception desk. An analysis of the register, for a random sample of 100 days, results in the following information on the number, $X$, of visitors per day.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Number of visitors per day & Number of days \\
\hline
1-10 & 13 \\
\hline
11-20 & 33 \\
\hline
21-25 & 17 \\
\hline
26-30 & 12 \\
\hline
31-35 & 8 \\
\hline
36-40 & 5 \\
\hline
41-50 & 5 \\
\hline
51-100 & 7 \\
\hline
Total & 100 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate an estimate of:
\begin{enumerate}[label=(\roman*)]
\item $\mu$, the mean number of visitors per day;
\item $\sigma$, the standard deviation of the number of visitors per day.
\end{enumerate}\item Give a reason, based upon the data provided, why $X$ is unlikely to be normally distributed.
\item \begin{enumerate}[label=(\roman*)]
\item Give a reason why $\bar { X }$, the mean of a random sample of 100 observations on $X$, may be assumed to be normally distributed.
\item State, in terms of $\mu$ and $\sigma$, the mean and variance of $\bar { X }$.
\end{enumerate}\item Hence construct a $99 \%$ confidence interval for $\mu$.
\item The receptionist claims that she registers on average more than 30 visitors per day, and frequently registers more than 50 visitors on any one day.
Comment on each of these two claims.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2005 Q6 [12]}}