| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 10 |
| 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 normal distribution and sampling distribution of the mean with known variance. Part (a) is routine standardization, part (b) applies CLT with σ/√n, and part (c) is a standard confidence interval calculation. All steps are textbook procedures requiring no problem-solving insight, though the multi-part structure and S3 level place it slightly below average difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E \sim N(0, 0.5^2)\) or \(X \sim N(w, 0.5^2)\) | ||
| \(P(\ | E\ | <0.6) = P\left(\ |
| \(= 2\times0.8849 - 1 = 0.7698\) | A1 | awrt \(0.770\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{E} \sim N\left(0, \frac{1}{64}\right)\) or \(\bar{X} \sim N\left(w, \frac{0.5^2}{16}\right)\) | M1 | Correct attempt to define \(\bar{E}\) or \(\bar{X}\); must attempt \(\frac{\sigma^2}{n}\) |
| \(P\left(\ | \bar{E}\ | <0.3\right) = P\left(\ |
| \(= P(\ | Z\ | <2.4) = 2\times0.9918 - 1 = 0.9836\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(35.6 \pm 2.3263 \times \frac{1}{8}\) | M1 B1 | For \(35.6 \pm z\times\frac{0.5}{\sqrt{16}}\); for 2.3263 or better (use of 2.33 loses B1) |
| \((35.3, 35.9)\) | A1, A1 | awrt 35.3; awrt 35.9 |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E \sim N(0, 0.5^2)$ or $X \sim N(w, 0.5^2)$ | | |
| $P(\|E\|<0.6) = P\left(\|Z\| < \frac{0.6}{0.5}\right) = P(\|Z\|<1.2)$ | M1 | Identifying correct probability with 0.6 and attempting to standardise; need $\|\cdot\|$ |
| $= 2\times0.8849 - 1 = 0.7698$ | A1 | awrt $0.770$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{E} \sim N\left(0, \frac{1}{64}\right)$ or $\bar{X} \sim N\left(w, \frac{0.5^2}{16}\right)$ | M1 | Correct attempt to define $\bar{E}$ or $\bar{X}$; must attempt $\frac{\sigma^2}{n}$ |
| $P\left(\|\bar{E}\|<0.3\right) = P\left(\|Z\| < \frac{0.3}{\frac{1}{8}}\right)$ | M1, A1 | Identifying correct probability using $\bar{E}$ or $\bar{X}$; must have 0.3 and $\|\cdot\|$; correct standardisation |
| $= P(\|Z\|<2.4) = 2\times0.9918 - 1 = 0.9836$ | A1 | awrt $0.984$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $35.6 \pm 2.3263 \times \frac{1}{8}$ | M1 B1 | For $35.6 \pm z\times\frac{0.5}{\sqrt{16}}$; for 2.3263 or better (use of 2.33 loses B1) |
| $(35.3, 35.9)$ | A1, A1 | awrt 35.3; awrt 35.9 |
---3. A woodwork teacher measures the width, $w \mathrm {~mm}$, of a board. The measured width, $X \mathrm {~mm}$, is normally distributed with mean $w \mathrm {~mm}$ and standard deviation 0.5 mm .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that $X$ is within 0.6 mm of $w$.
The same board is measured 16 times and the results are recorded.
\item Find the probability that the mean of these results is within 0.3 mm of $w$.
Given that the mean of these 16 measurements is 35.6 mm ,
\item find a 98\% confidence interval for $w$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q3 [10]}}