| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Known variance confidence intervals |
| Difficulty | Moderate -0.3 This is a straightforward application of known variance confidence intervals with standard bookwork parts. Part (a) requires calculating x̄ and applying the z-interval formula (routine), (b) is interpretation, (c) is direct recall (answer is 5%), and (d) tests understanding that unknown σ requires a t-distribution. All parts are standard textbook exercises with no problem-solving or novel insight required, though the multi-part structure and need to articulate reasoning in (d) elevates it slightly above pure recall. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mean \(= 504\) | B1 | 504 may be seen in part (b) |
| \(1.96\) | B1 | For realising a normal distribution must be used as a model and finding the correct value 1.96 |
| \(504 \pm \frac{5.4}{\sqrt{8}} \times "1.96"\) | M1 | For \(504 \pm \frac{5.4}{\sqrt{8}} \times "z \text{ value}"\), \( |
| \((500.258, 507.742)\) | A1 | awrt 500.26 and 507.74. NB using \(t\) gives 500.29 and 507.71 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 505 is in the confidence interval therefore there is evidence that the machine is working properly | B1ft | Drawing a correct inference (ft) using their answer to part (a) and the 505 from the question. Reason must be given. Ignore incorrect non-contextual |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5\%\) oe | B1 | \(5\%\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s\) needs to be used instead of \(\sigma\) and a \(t\)-value instead of the \(z\) value | B1 | Create new model by using \(s\) and \(t\). Allow if state use CI \(\mu \pm \frac{s}{\sqrt{n}} \times "t"\) or use \(s=4.44\) and \(t=2.365\) |
| Since the sample is small therefore you can't use the normal distribution | B1 | For recognising that the sample is small |
## Question 1:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mean $= 504$ | B1 | 504 may be seen in part (b) |
| $1.96$ | B1 | For realising a normal distribution must be used as a model and finding the correct value 1.96 |
| $504 \pm \frac{5.4}{\sqrt{8}} \times "1.96"$ | M1 | For $504 \pm \frac{5.4}{\sqrt{8}} \times "z \text{ value}"$, $|z|>1$. May be implied by a correct CI |
| $(500.258, 507.742)$ | A1 | awrt 500.26 and 507.74. NB using $t$ gives 500.29 and 507.71 |
**(4 marks)**
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **505** is in the confidence **interval** therefore there is evidence that the machine is **working** properly | B1ft | Drawing a correct inference (ft) using their answer to part (a) and the 505 from the question. Reason must be given. Ignore incorrect non-contextual |
**(1 mark)**
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5\%$ oe | B1 | $5\%$ |
**(1 mark)**
### Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s$ needs to be used instead of $\sigma$ and a $t$-value instead of the $z$ value | B1 | Create new model by using $s$ and $t$. Allow if state use CI $\mu \pm \frac{s}{\sqrt{n}} \times "t"$ or use $s=4.44$ and $t=2.365$ |
| Since the sample is small therefore you can't use the normal distribution | B1 | For recognising that the sample is small |
**(2 marks)**
---1 A machine is set to fill pots with yoghurt such that the mean weight of yoghurt in a pot is 505 grams.
To check that the machine is working properly, a random sample of 8 pots is selected. The weight of yoghurt, in grams, in each pot is as follows
$$\begin{array} { l l l l l l l l }
508 & 510 & 500 & 500 & 498 & 503 & 508 & 505
\end{array}$$
Given that the weights of the yoghurt delivered by the machine follow a normal distribution with standard deviation 5.4 grams,
\begin{enumerate}[label=(\alph*)]
\item find a $95 \%$ confidence interval for the mean weight, $\mu$ grams, of yoghurt in a pot. Give your answers to 2 decimal places.
\item Comment on whether or not the machine is working properly, giving a reason for your answer.
\item State the probability that a $95 \%$ confidence interval for $\mu$ will not contain $\mu$ grams.
\item Without carrying out any further calculations, explain the changes, if any, that would need to be made in calculating the confidence interval in part (a) if the standard deviation was unknown. Give a reason for your answer.\\
You may assume that the weights of the yoghurt delivered by the machine still follow a normal distribution.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 2019 Q1 [8]}}