| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | CI with two different confidence levels same sample |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question requiring standard transformations between confidence levels and application of independence. Part (a) uses the relationship between critical values (1.645 for 90%, 2.326 for 98%), part (b) applies a linear transformation (multiply by π), and part (c) uses basic probability (0.98³). All steps are routine S3 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x = \frac{1}{2}(118.8 + 121.2) = 120\) | B1 | |
| 1.6449 (or better) | B1 M1 | |
| "their 1.6449" \(\frac{\sigma}{\sqrt{n}} = 121.2 - 120\) | ||
| 2.3263 (or better) | B1 dM1 | |
| "their 2.3263" \(\frac{\sigma}{\sqrt{n}} = 2.3263 \times (\frac{121.2 - 120}{1.6449})\) | ||
| So 98% C.I. = \(120 \pm 1.424...= (118.3028..., 121.699...)\) | ||
| awrt (118, 122) | A1 | (6) |
| (b) awrt \((118\pi, 122\pi)\) or \((371/372, 382/383)\) | B1ft | (1) |
| (c) P(All) = \((0.98)^3\) | M1 A1 | |
| = 0.941 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| 1st M1 for an attempt to find "width" or "half-width" of a 90% CI ft their z value (\( | z | > 1.5\)) e.g. for \(z E = 121.2 - 120\) (o.e.) N.B. \(E = 0.7295...\) Condone missing 2 here. |
| Answer | Marks | Guidance |
|---|---|---|
| 2nd dM1 for a correct attempt at "width" or "half-width" of a 98% CI ft their z value (\( | z | > 2\)). Dependent on 1st M1 and ft their value or expression for s.e. |
**(a)** $x = \frac{1}{2}(118.8 + 121.2) = 120$ | B1 |
1.6449 (or better) | B1 M1 |
"their 1.6449" $\frac{\sigma}{\sqrt{n}} = 121.2 - 120$ | |
2.3263 (or better) | B1 dM1 |
"their 2.3263" $\frac{\sigma}{\sqrt{n}} = 2.3263 \times (\frac{121.2 - 120}{1.6449})$ | |
So 98% C.I. = $120 \pm 1.424...= (118.3028..., 121.699...)$ | |
awrt **(118, 122)** | A1 | (6)
**(b)** awrt $(118\pi, 122\pi)$ or $(371/372, 382/383)$ | B1ft | (1)
**(c)** P(All) = $(0.98)^3$ | M1 A1 |
= 0.941 | | (2)
**[Total 9]**
**Notes:**
**(a)** NB in part (a) only lose one of the B1 marks for not using the percentage points table
1st B1 for $\bar{x} = 120$
2nd B1 for 1.6449 or better in an attempt (could be 1.6449$\sigma = k$ or even 1.6449 $\sigma^2 = k$). Condone strange notation for standard error (E) here if it is used correctly
1st M1 for an attempt to find "width" or "half-width" of a 90% CI ft their z value ($|z| > 1.5$) e.g. for $z E = 121.2 - 120$ (o.e.) N.B. $E = 0.7295...$ Condone missing 2 here.
3rd B1 for 2.3263 or better in an attempt at CI. If score 2nd B0 for using 1.64 or 1.645 allow 3rd B1 for 2.32 or 2.33 here
2nd dM1 for a correct attempt at "width" or "half-width" of a 98% CI ft their z value ($|z| > 2$). Dependent on 1st M1 and ft their value or expression for s.e.
A1 for lower limit in range $[118, 118.35)$ and upper limit in range $(121.65, 122]$. Answer only of awrt (118, 122) with no incorrect working seen scores 6/6 if 1.6449 and 2.3263 are seen and 5/6 (B1B1M1B0M1A1) otherwise.
**(c)** M1 for a correct expression i.e. $(0.98)^3$
A1 for awrt 0.941
---\begin{enumerate}
\item A manufacturer produces circular discs with diameter $D \mathrm {~mm}$, such that $D \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$. A random sample of discs is taken and, using tables of the normal distribution, a $90 \%$ confidence interval for $\mu$ is found to be\\
(118.8, 121.2)\\
(a) Find a 98\% confidence interval for $\mu$.\\
(b) Hence write down a 98\% confidence interval for the circumference of the discs.
\end{enumerate}
Using three different random samples, three $98 \%$ confidence intervals for $\mu$ are to be found.\\
(c) Calculate the probability that all the intervals will contain $\mu$.\\
\hfill \mbox{\textit{Edexcel S3 2013 Q5 [9]}}