| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| 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: part (a) uses the sampling distribution of the mean with known variance (routine CLT application), and part (b) involves constructing a confidence interval with known variance and making a simple interpretation. Both parts require only direct formula application with no problem-solving insight, making it slightly easier than average but not trivial due to the two-part structure and need for careful interpretation. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(L \sim N(1005, 15^2)\) | ||
| \(V(\text{pack}) = \dfrac{15^2}{12}\) or \(\dfrac{225}{12}\) or \(\dfrac{75}{4}\) or \(18.7\) to \(18.8\) OR \(SD(\text{pack}) = \dfrac{15}{\sqrt{12}}\) or \(\dfrac{15}{2\sqrt{3}}\) or \(\dfrac{5\sqrt{3}}{2}\) or \(4.3\) to \(4.4\) | B1 | CAO; AWFW \((18.75)\); CAO; OE; AWFW \((4.33013)\) |
| \(P(L < 1000) = P\!\left(\dfrac{1000 - 1005}{15/\sqrt{12}}\right)\) | M1 | Standardising 1000 using 1005 and \(\dfrac{15}{\sqrt{12}}\) OE; allow \((1005 - 1000)\) |
| \(P(Z < -1.1547) = 1 - P(Z < 1.1547)\) | m1 | Correct area change; may be implied by correct answer or answer \(< 0.5\) |
| \(1 - (0.87698\) to \(0.87493) = \mathbf{0.123}\) to \(\mathbf{0.126}\) | A1 | AWFW \((0.12411)\); \((1 - \text{answer}) \Rightarrow\) B1 M1 max |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(99\%\ (0.99) \Rightarrow z = 2.57\) to \(2.58\) | B1 | AWFW \((2.5758)\) |
| CI for \(\mu\) is \(\bar{x} \pm z \times \dfrac{\sigma}{\sqrt{n}}\) | M1 | Used with \(z\) \((2.05\) to \(2.58)\), \(\bar{x}\ (4.65)\), \(\sigma\ (0.15)\), and \(\div\sqrt{n}\) with \(n > 1\) |
| \(4.65 \pm 2.5758 \times \dfrac{0.15}{\sqrt{24}}\) | A1 | \(z\) \((2.05\) to \(2.06\) or \(2.32\) to \(2.33\) or \(2.57\) to \(2.58)\), \(\bar{x}\ (4.65)\), \(\sigma\ (0.15)\), and \(\div\sqrt{24}\) or \(23\) or \(12\) or \(11\) |
| \(4.65 \pm 0.08\) OR \((4.57,\ 4.73)\) | A1 | CAO/AWRT; AWRT |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Clear correct comparison of \(4.5\) with LCL or CI (eg \(4.5 <\) LCL or its value, or \(4.5 <\) CI or its limits) | BF1 | F on CI only providing \(\text{LCL} > 4.5\) (ie whole of \(\text{CI} > 4.5\)); quoting values for LCL or CI not required; BF0 for '\(4.5\) is outside CI'; OE |
| so Agree with manufacturer's specification | Bdep1 | OE; dependent on previous BF1 |
| Total | 2 |
# Question 6:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $L \sim N(1005, 15^2)$ | | |
| $V(\text{pack}) = \dfrac{15^2}{12}$ or $\dfrac{225}{12}$ or $\dfrac{75}{4}$ **or** $18.7$ to $18.8$ **OR** $SD(\text{pack}) = \dfrac{15}{\sqrt{12}}$ or $\dfrac{15}{2\sqrt{3}}$ or $\dfrac{5\sqrt{3}}{2}$ **or** $4.3$ to $4.4$ | B1 | CAO; AWFW $(18.75)$; CAO; OE; AWFW $(4.33013)$ |
| $P(L < 1000) = P\!\left(\dfrac{1000 - 1005}{15/\sqrt{12}}\right)$ | M1 | Standardising 1000 using 1005 and $\dfrac{15}{\sqrt{12}}$ OE; allow $(1005 - 1000)$ |
| $P(Z < -1.1547) = 1 - P(Z < 1.1547)$ | m1 | Correct area change; may be implied by correct answer or answer $< 0.5$ |
| $1 - (0.87698$ to $0.87493) = \mathbf{0.123}$ **to** $\mathbf{0.126}$ | A1 | AWFW $(0.12411)$; $(1 - \text{answer}) \Rightarrow$ B1 M1 max |
| **Total** | **4** | |
## Part (b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $99\%\ (0.99) \Rightarrow z = 2.57$ to $2.58$ | B1 | AWFW $(2.5758)$ |
| CI for $\mu$ is $\bar{x} \pm z \times \dfrac{\sigma}{\sqrt{n}}$ | M1 | Used with $z$ $(2.05$ to $2.58)$, $\bar{x}\ (4.65)$, $\sigma\ (0.15)$, and $\div\sqrt{n}$ with $n > 1$ |
| $4.65 \pm 2.5758 \times \dfrac{0.15}{\sqrt{24}}$ | A1 | $z$ $(2.05$ to $2.06$ or $2.32$ to $2.33$ or $2.57$ to $2.58)$, $\bar{x}\ (4.65)$, $\sigma\ (0.15)$, and $\div\sqrt{24}$ or $23$ or $12$ or $11$ |
| $4.65 \pm 0.08$ **OR** $(4.57,\ 4.73)$ | A1 | CAO/AWRT; AWRT |
| **Total** | **4** | |
## Part (b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Clear correct comparison of $4.5$ with LCL or CI (eg $4.5 <$ LCL or its value, or $4.5 <$ CI or its limits) | BF1 | F on CI only providing $\text{LCL} > 4.5$ (ie whole of $\text{CI} > 4.5$); quoting values for LCL or CI **not** required; BF0 for '$4.5$ is outside CI'; OE |
| so **Agree** with manufacturer's specification | Bdep1 | OE; dependent on previous BF1 |
| **Total** | **2** | |
---6
\begin{enumerate}[label=(\alph*)]
\item The length of one-metre galvanised-steel straps used in house building may be modelled by a normal distribution with a mean of 1005 mm and a standard deviation of 15 mm .
The straps are supplied to house builders in packs of 12, and the straps in a pack may be assumed to be a random sample.
Determine the probability that the mean length of straps in a pack is less than one metre.
\item Tania, a purchasing officer for a nationwide house builder, measures the thickness, $x$ millimetres, of each of a random sample of 24 galvanised-steel straps supplied by a manufacturer. She then calculates correctly that the value of $\bar { x }$ is 4.65 mm .
\begin{enumerate}[label=(\roman*)]
\item Assuming that the thickness, $X \mathrm {~mm}$, of such a strap may be modelled by the distribution $\mathrm { N } \left( \mu , 0.15 ^ { 2 } \right)$, construct a $99 \%$ confidence interval for $\mu$.
\item Hence comment on the manufacturer's specification that the mean thickness of such straps is greater than 4.5 mm .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2013 Q6 [10]}}