| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Confidence interval interpretation |
| Difficulty | Moderate -0.3 Part (a) requires straightforward manipulation of confidence interval endpoints using the relationship between z-values (1.96 vs 1.645), a standard S3 technique. Part (b) is a direct binomial probability calculation with p=0.9, n=4. Both parts are routine applications of well-practiced methods with no conceptual challenges beyond standard S3 content. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(\bar{x} = \frac{1}{2}(11.52 + 13.75) = 12.635\) | B1 | 12.635 (may be implied by correct CI) |
| Use of 1.96 | B1 | |
| \(\frac{\sigma}{\sqrt{n}} = \frac{13.75 - 12.635}{1.96} (= 0.56887...)\) or \(\frac{\sigma}{\sqrt{n}} = \frac{13.75 - 11.52}{2 \times 1.96} (= 0.56887...)\) | M1 | For attempt at standard error (may be implied by awrt 0.569) |
| Use of 1.6449 or better (1.644853... from calc) | B1 | Use of 1.64 or 1.65 is B0 |
| \(12.635 \pm 1.6449 \times 0.56887...\) | M1 | For (their \(\bar{x}\)) \(\pm\) (their 1.6449)(their \(\frac{\sigma}{\sqrt{n}}\)); \(\frac{\sigma}{\sqrt{n}}\) must be numerical |
| 90% CI is \((11.699..., 13.5707...)\) | A1 | awrt (11.7, 13.6) from correct working. Correct answer with no working scores B1B1M1B1M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(4 \times 0.9^3 \times 0.1 = 0.2916\) | M1 | \(4p^3(1-p)\) where \(0 < p < 1\) |
| \(= 0.2916\) | A1 | awrt 0.292 |
# Question 3:
## Part 3(a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $\bar{x} = \frac{1}{2}(11.52 + 13.75) = 12.635$ | B1 | 12.635 (may be implied by correct CI) |
| Use of 1.96 | B1 | |
| $\frac{\sigma}{\sqrt{n}} = \frac{13.75 - 12.635}{1.96} (= 0.56887...)$ or $\frac{\sigma}{\sqrt{n}} = \frac{13.75 - 11.52}{2 \times 1.96} (= 0.56887...)$ | M1 | For attempt at standard error (may be implied by awrt 0.569) |
| Use of 1.6449 or better (1.644853... from calc) | B1 | Use of 1.64 or 1.65 is B0 |
| $12.635 \pm 1.6449 \times 0.56887...$ | M1 | For (their $\bar{x}$) $\pm$ (their 1.6449)(their $\frac{\sigma}{\sqrt{n}}$); $\frac{\sigma}{\sqrt{n}}$ must be numerical |
| 90% CI is $(11.699..., 13.5707...)$ | A1 | awrt (11.7, 13.6) from correct working. Correct answer with no working scores B1B1M1B1M1A1 |
## Part 3(b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $4 \times 0.9^3 \times 0.1 = 0.2916$ | M1 | $4p^3(1-p)$ where $0 < p < 1$ |
| $= 0.2916$ | A1 | awrt 0.292 |
---\begin{enumerate}
\item Components are manufactured such that their length in mm is normally distributed with mean $\mu$ and variance $\sigma ^ { 2 }$. Below is a 95\% confidence interval for $\mu$ calculated from a random sample of components.\\
(11.52, 13.75)
\end{enumerate}
Using the same random sample,\\
(a) find a $90 \%$ confidence interval for $\mu$.
Four 90\% confidence intervals are found from independent random samples.\\
(b) Calculate the probability that only 3 of these 4 intervals will contain $\mu$.
\hfill \mbox{\textit{Edexcel S3 2021 Q3 [8]}}