| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for confidence intervals |
| Difficulty | Standard +0.3 This is a straightforward S3 question requiring standard CLT explanation, routine confidence interval calculation (σ known, large n=100), and basic interpretation. The CLT application is explicit and the calculations are mechanical with no conceptual challenges beyond typical A-level statistics. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((X_1, X_2, X_3, \ldots, X_n\) is a random) sample of size \(n\), for \(n\) is large | B1 | |
| (from a population with mean \(\mu\) and variance \(\sigma^2\)) then \(\bar{X}\) is (approximately) Normal | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = \frac{1740000}{100} = 17400\) | B1 | |
| \(\bar{x} \pm z\frac{\sigma}{\sqrt{n}} = 17400 \pm 1.96 \times \frac{5000}{\sqrt{100}}\) | M1, B1 | Recognisable \(z\) value required for method; 2nd B1: 1.96 or better seen |
| \([16420, 18380]\) | A1A1 | Accept 3sf if correct expression seen; 5/5 for \([16420, 18380]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{X}\): Normal (approx) by CLT, and normal needed to find CI | B1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 20000 above upper confidence limit (not just outside) | B1ft | |
| Complaint justified | dB1ft |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(X_1, X_2, X_3, \ldots, X_n$ is a random) **sample** of size $n$, for $n$ is **large** | B1 | |
| (from a population with mean $\mu$ and variance $\sigma^2$) then $\bar{X}$ is (approximately) Normal | B1 | |
**(2 marks)**
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = \frac{1740000}{100} = 17400$ | B1 | |
| $\bar{x} \pm z\frac{\sigma}{\sqrt{n}} = 17400 \pm 1.96 \times \frac{5000}{\sqrt{100}}$ | M1, B1 | Recognisable $z$ value required for method; 2nd B1: 1.96 or better seen |
| $[16420, 18380]$ | A1A1 | Accept 3sf if correct expression seen; 5/5 for $[16420, 18380]$ |
**(5 marks)**
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{X}$: Normal (approx) by CLT, and normal needed to find CI | B1, B1 | |
**(2 marks)**
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| 20000 **above** upper confidence limit (**not** just outside) | B1ft | |
| Complaint justified | dB1ft | |
**(2 marks)**3.\\
\begin{enumerate}[label=(\alph*)]
\item Explain what you understand by the Central Limit Theorem.
A garage services hire cars on behalf of a hire company. The garage knows that the lifetime of the brake pads has a standard deviation of 5000 miles. The garage records the lifetimes, $x$ miles, of the brake pads it has replaced. The garage takes a random sample of 100 brake pads and finds that $\sum x = 1740000$
\item Find a 95\% confidence interval for the mean lifetime of a brake pad.
\item Explain the relevance of the Central Limit Theorem in part (b).
Brake pads are made to be changed every 20000 miles on average.\\
The hire car company complain that the garage is changing the brake pads too soon.
\item Comment on the hire company's complaint. Give a reason for your answer.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2012 Q3 [11]}}