| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unbiased estimator from summary statistics |
| Difficulty | Standard +0.3 This is a straightforward S1 question testing standard procedures: calculating mean and unbiased variance from summary statistics, applying the Central Limit Theorem to justify normality of sample means, constructing a confidence interval using given formulas, and making basic statistical comments. All steps are routine applications of taught methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{2290}{50} = 45.8\) or \(45800\) | B1 | |
| \(\left(s^2 =\right)\frac{28225.5}{49 \text{ or } 50}\) or \(\left(s =\right)\sqrt{\frac{28225.5}{49 \text{ or } 50}}\) | M1 | |
| \(s = 24(.0)\) or \(24000\) to \(24001\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} - ns = (45.8 - n \times 24.0) < 0\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| and negative salaries are impossible | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Large sample or \(n > 25\) or \(30\) or \(n = 50\) so CLT applies | B1, Bdep1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(99\% (0.99) \Rightarrow z = 2.57\) to \(2.58\) | B1 | |
| CI for \(\mu\) is \(\bar{x} \pm z \times \frac{s}{\sqrt{n}}\) | M1 | |
| Thus \(45.8 \pm 2.5758 \times \frac{24.0}{\sqrt{50}}\) | AF1 | |
| Hence \(45.8 \pm (8.7\) to \(8.8)\) or \(45800 \pm (8700\) to \(8800)\) OR \((37(.0)\) to \(37.1, 54.5\) to \(54.6)\) or \((37000\) to \(37100, 54500\) to \(54600)\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Clear correct comparison of \(55\) or \(55000\) with c's UCL or CI | B1 | |
| \((6/50\) or \(0.12\) or \(12\%) | B1 | |
| Reject both/each of the two claims | Bdep1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 0 \mid N(45.8, 24.0^2)) = P(Z < -1.91)\) | M1 | Standardising 0 using 45.8 & 24.0 |
| \(= 0.027 \text{ to } 0.03\) | A1 [2] | In addition to probability within range, must state that negative salaries are impossible |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 60 \mid N(45.8, 24.0^2)) = P(Z > 0.59)\) | M1 | Standardising 60 using 45.8 & 24.0 |
| \(= 0.27 \text{ to } 0.28\) | A1 [2] | In addition to probability within range, must compare calculated value to \(\frac{6}{50} = 0.12\) OE |
| Answer | Marks | Guidance |
|---|---|---|
| It/(claimed) mean/(claimed) value \(>\) UCL/CI | B0 | Must indicate 55 or 55000 |
| 99% have (mean) weights between CLs so ... | B0 | |
| Any comparison of 60 (£60 000) with UCL/CI | B0 | Value of 60 does not refer to mean |
| \(P(X > 60 \mid N(45.8, 24.0^2)) = P(Z > 0.59) = (0.27 \text{ to } 0.28) > \frac{6}{50} = 0.12\) | B0 | Assumes salaries \(\sim N\); cf (a)(ii) |
# Question 7:
## Part (a)(i):
$\bar{x} = \frac{2290}{50} = 45.8$ or $45800$ | B1 | | CAO
$\left(s^2 =\right)\frac{28225.5}{49 \text{ or } 50}$ **or** $\left(s =\right)\sqrt{\frac{28225.5}{49 \text{ or } 50}}$ | M1 | | Ignore notation
$s = 24(.0)$ or $24000$ to $24001$ | A1 | 3 | AWRT/AWFW (24.00064); $(\sigma = 23.75942)$
SCs: (for no seen working) M1 A1 for $24.0$ or $24000$ to $24001$; M1 A0 for $24$ or $23700$ to $23800$
## Part (a)(ii):
$\bar{x} - ns = (45.8 - n \times 24.0) < 0$ | M1 | | Allow (45 to 47) and any multiple of (23.5 to 24.5) which gives value $< 0$; Must clearly state the value of a numerical expression
SC: Accept quoted values of $(-4$ to $-1)$ $(n=2)$ or $(-28.5$ to $-23.5)$ $(n=3)$ (both AWFW)
and negative salaries are impossible | A1 | 2 | OE; must be in context; Negative values impossible $\Rightarrow$ A0
## Part (b)(i):
Large sample or $n > 25$ or $30$ or $n = 50$ so CLT applies | B1, Bdep1 | 2 | OE; Must indicate CLT; dependent on B1; Indication that other than sample mean is normally distributed $\Rightarrow$ Bdep0
## Part (b)(ii):
$99\% (0.99) \Rightarrow z = 2.57$ to $2.58$ | B1 | | AWFW (2.5758)
CI for $\mu$ is $\bar{x} \pm z \times \frac{s}{\sqrt{n}}$ | M1 | | Used with $(\bar{x}$ & $s)$ from (a)(i) and $z(1.64$ to $2.58)$ & $\div\sqrt{n}$ with $n > 1$
Thus $45.8 \pm 2.5758 \times \frac{24.0}{\sqrt{50}}$ | AF1 | | F on $(\bar{x}$ & $s)$ with $\div\sqrt{50}$ or $49$ & $z(1.64$ to $1.65$ or $2.32$ to $2.33$ or $2.57$ to $2.58)$
Hence $45.8 \pm (8.7$ to $8.8)$ or $45800 \pm (8700$ to $8800)$ OR $(37(.0)$ to $37.1, 54.5$ to $54.6)$ or $(37000$ to $37100, 54500$ to $54600)$ | A1 | 4 | CAO/AWFW (8.74); Ignore (absence of) quoted units; AWFW
## Part (c):
Clear correct comparison of $55$ or $55000$ with c's UCL or CI | B1 | | Accept $55000$ compared with c's $54.5$ to $54.6$ (ie different units)
$(6/50$ or $0.12$ or $12\%) </\neq 0.25$ or $25\%$ | B1 | | OE; correct comparison mentioning both $12\%$ and $25\%$
Reject both/each of the two claims | Bdep1 | 3 | Dependent on B1 B1
**Total: 14 marks**
## Question 7:
### Part (a)(ii) – Alternative Solutions
**Solution 1:**
$P(X < 0 \mid N(45.8, 24.0^2)) = P(Z < -1.91)$ | M1 | Standardising 0 using 45.8 & 24.0
$= 0.027 \text{ to } 0.03$ | A1 [2] | In addition to probability within range, must state that negative salaries are impossible
---
**Solution 2:**
$P(X > 60 \mid N(45.8, 24.0^2)) = P(Z > 0.59)$ | M1 | Standardising 60 using 45.8 & 24.0
$= 0.27 \text{ to } 0.28$ | A1 [2] | In addition to probability within range, must compare calculated value to $\frac{6}{50} = 0.12$ OE
---
### Part (c) – Additional Comment Illustrations
It/(claimed) mean/(claimed) value $>$ UCL/CI | B0 | Must indicate 55 or 55000
99% have (mean) weights between CLs so ... | B0 |
Any comparison of 60 (£60 000) with UCL/CI | B0 | Value of 60 does not refer to mean
$P(X > 60 \mid N(45.8, 24.0^2)) = P(Z > 0.59) = (0.27 \text{ to } 0.28) > \frac{6}{50} = 0.12$ | B0 | Assumes salaries $\sim N$; cf (a)(ii)7 A random sample of 50 full-time university employees was selected as part of a higher education salary survey.
The annual salary in thousands of pounds, $x$, of each employee was recorded, with the following summarised results.
$$\sum x = 2290.0 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 28225.50$$
Also recorded was the fact that 6 of the 50 salaries exceeded $\pounds 60000$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Calculate values for $\bar { x }$ and $s$, where $s ^ { 2 }$ denotes the unbiased estimate of $\sigma ^ { 2 }$.
\item Hence show why the annual salary, $X$, of a full-time university employee is unlikely to be normally distributed. Give numerical support for your answer.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Indicate why the mean annual salary, $\bar { X }$, of a random sample of 50 full-time university employees may be assumed to be normally distributed.
\item Hence construct a $99 \%$ confidence interval for the mean annual salary of full-time university employees.
\end{enumerate}\item It is claimed that the annual salaries of full-time university employees have an average which exceeds $\pounds 55000$ and that more than $25 \%$ of such salaries exceed £60000.
Comment on each of these two claims.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2012 Q7 [14]}}