| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Single tail probability P(X < a) or P(X > a) |
| Difficulty | Standard +0.8 This is a confidence interval question requiring knowledge of normal distribution sampling theory with known σ, hypothesis testing interpretation, and understanding of when to use t-distribution. While the calculations are straightforward, it tests conceptual understanding of statistical inference at Further Maths level, including the distinction between z and t procedures, making it moderately harder than average A-level questions. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Uses formula \(\bar{x} \pm z\sqrt{\dfrac{\sigma^2}{n}}\) with \(z\) = AWRT 1.96 or 1.64 or 1.645 | M1 | PI |
| \(= 36 \pm 1.96\dfrac{6}{\sqrt{5}}\) | ||
| \(= (30.7,\ 41.3)\) | A1 | AWRT; condone poor notation |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The confidence interval does not support Alan's claim as 30 minutes is outside the confidence interval. | E1F | Infers no as 30 is outside the CI; follow through on their confidence interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A \(t\) distribution will be used instead of a normal distribution | E1 | Explains that \(z\) value would be replaced by a \(t\) value or gives correct formula; condone "use a \(t\)-test rather than a \(z\)-test"; ignore values given following a correct statement; implied by AWRT 1.96 replaced by AWRT 2.78 |
| Total: 4 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Uses formula $\bar{x} \pm z\sqrt{\dfrac{\sigma^2}{n}}$ with $z$ = AWRT 1.96 or 1.64 or 1.645 | M1 | PI |
| $= 36 \pm 1.96\dfrac{6}{\sqrt{5}}$ | | |
| $= (30.7,\ 41.3)$ | A1 | AWRT; condone poor notation |
| **Total: 2** | | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The confidence interval does not support Alan's claim as 30 minutes is outside the confidence interval. | E1F | Infers no as 30 is outside the CI; follow through on their confidence interval |
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| A $t$ distribution will be used instead of a normal distribution | E1 | Explains that $z$ value would be replaced by a $t$ value or gives correct formula; condone "use a $t$-test rather than a $z$-test"; ignore values given following a correct statement; implied by AWRT 1.96 replaced by AWRT 2.78 |
| **Total: 4** | | |3 Alan's journey time to work can be modelled by a normal distribution with standard deviation 6 minutes.
Alan measures the journey time to work for a random sample of 5 journeys. The mean of the 5 journey times is 36 minutes.
3
\begin{enumerate}[label=(\alph*)]
\item Construct a 95\% confidence interval for Alan's mean journey time to work, giving your values to one decimal place.\\
3
\item Alan claims that his mean journey time to work is 30 minutes.\\
State, with a reason, whether or not the confidence interval found in part (a) supports Alan's claim.\\
3
\item Suppose that the standard deviation is not known but a sample standard deviation is found from Alan's sample and calculated to be 6
Explain how the working in part (a) would change.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2019 Q3 [4]}}