| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question with standard parts: calculating a CI with known σ, explaining why CLT isn't needed (population already normal), commenting on a claim, and finding a probability for sample mean. All parts are routine S3 techniques with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-part structure. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \frac{806.4}{36} = 22.4\) | B1 | For 22.4 |
| \(22.4 \pm 2.3263 \times \frac{0.4}{\sqrt{36}}\) | M1 B1 | M1 for use of \(\bar{x} \pm z\) value \(\times\frac{\sigma}{\sqrt{n}}\) with \(1.2 < z < 2.6\); B1 for \(z\) value \(= 2.3263\) or better |
| \((22.24\ldots, 22.55\ldots)\) awrt \((22.2, 22.6)\) | A1 | awrt (22.2, 22.6). This does not imply the B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| [The Central Limit Theorem is not required as] the original population is normally distributed | B1 | For reference to data modelled by normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 22.5 is within the confidence interval | B1ft | ft their CI. Comment on whether 22.5 is or is not in their CI. Allow e.g. range for CI: "22.24" \(< 22.5 <\) "22.6" |
| So no reason to doubt the manufacturers claim | dB1ft | Dependent on B1ft. Correct comment ft their CI. Answer must be compatible with their CI |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{Y} \sim N\!\left(850, \left(\frac{5}{\sqrt{10}}\right)^2\right)\) | B1 | For \(\bar{Y} \sim N(850,\ldots)\) or \(\bar{Y} < \frac{848-850}{5}\). Must have \(\bar{Y}\) or \(N\!\left(850,\left(\frac{5}{\sqrt{10}}\right)^2\right)\) or \(N(850, 2.5)\) seen or used. Or N(8500, 250) seen or used. Both implied by correct standardisation |
| \(P(\bar{Y}<848) = P\!\left(Z < \frac{848-850}{\frac{5}{\sqrt{10}}}\right) = [P(Z < -1.26)]\) | M1 | For \(\pm\)(a correct standardisation) implied by a correct answer |
| \(= 0.1038\) (Calculator gives 0.10295…) awrt 0.103/0.104 | A1 | awrt 0.103 to 0.104 |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \frac{806.4}{36} = 22.4$ | B1 | For 22.4 |
| $22.4 \pm 2.3263 \times \frac{0.4}{\sqrt{36}}$ | M1 B1 | M1 for use of $\bar{x} \pm z$ value $\times\frac{\sigma}{\sqrt{n}}$ with $1.2 < z < 2.6$; B1 for $z$ value $= 2.3263$ or better |
| $(22.24\ldots, 22.55\ldots)$ awrt $(22.2, 22.6)$ | A1 | awrt (22.2, 22.6). This does not imply the B1 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| [The Central Limit Theorem is not required as] the original population is **normally distributed** | B1 | For reference to data modelled by **normal distribution** |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 22.5 is within the confidence interval | B1ft | ft their CI. Comment on whether 22.5 is or is not in their CI. Allow e.g. range for CI: "22.24" $< 22.5 <$ "22.6" |
| So no reason to doubt the manufacturers claim | dB1ft | Dependent on B1ft. Correct comment ft their CI. Answer must be compatible with their CI |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{Y} \sim N\!\left(850, \left(\frac{5}{\sqrt{10}}\right)^2\right)$ | B1 | For $\bar{Y} \sim N(850,\ldots)$ or $\bar{Y} < \frac{848-850}{5}$. Must have $\bar{Y}$ or $N\!\left(850,\left(\frac{5}{\sqrt{10}}\right)^2\right)$ or $N(850, 2.5)$ seen or used. Or N(8500, 250) seen or used. Both implied by correct standardisation |
| $P(\bar{Y}<848) = P\!\left(Z < \frac{848-850}{\frac{5}{\sqrt{10}}}\right) = [P(Z < -1.26)]$ | M1 | For $\pm$(a correct standardisation) implied by a correct answer |
| $= 0.1038$ (Calculator gives 0.10295…) awrt 0.103/0.104 | A1 | awrt 0.103 to 0.104 |
---6 A garden centre sells bags of stones and large bags of gravel.\\
The weight, $X$ kilograms, of stones in a bag can be modelled by a normal distribution with unknown mean $\mu$ and known standard deviation 0.4
The stones in each of a random sample of 36 bags from a large batch is weighed. The total weight of stones in these 36 bags is found to be 806.4 kg
\begin{enumerate}[label=(\alph*)]
\item Find a 98\% confidence interval for the mean weight of stones in the batch.
\item Explain why the use of the Central Limit theorem is not required to answer part (a)
The manufacturer of these bags of stones claims that bags in this batch have a mean weight of 22.5 kg
\item Using your answer to part (a), comment on the claim made by the manufacturer.
The weight, $Y$ kilograms, of gravel in a large bag can be modelled by a normal distribution with mean 850 kg and standard deviation 5 kg
A builder purchases 10 large bags of gravel.
\item Find the probability that the mean weight of gravel in the 10 large bags is less than 848 kg
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2023 Q6 [10]}}