| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Standard +0.8 This question requires understanding of confidence intervals, working backwards from interval width to find the confidence level (involving inverse normal distribution), and applying probability to multiple independent intervals. Part (ii) is non-routine as students typically calculate intervals given α, not vice versa, requiring algebraic manipulation and z-table work in reverse. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks |
|---|---|
| (i) \(\frac{14800}{50}\) or 296 | B1 |
| \(\sqrt{\frac{504390000}{50} - '296'^2} = \sqrt{\frac{504390000 - 4381200000}{50}} = \sqrt{187.755} = 188\) (3 sf) | M1 A1 |
| Answer | Marks |
|---|---|
| (ii) \(2 \times z \times \frac{\sqrt{187.755}}{\sqrt{50}} = 5.45\) or equivalent | M1 |
| \(z = 1.406\) or 1.405 | A1 |
| \(\Phi('1.406') = 0.92\) or 0.9199 | M1 |
| \(\alpha = 84\) (2 sf) allow 83.98 | A1 |
| Answer | Marks |
|---|---|
| (iii) \(\frac{0.96}{\sqrt{...}} = 0.849\) (3 sf) | M1 A1 |
(i) $\frac{14800}{50}$ or 296 | B1
$\sqrt{\frac{504390000}{50} - '296'^2} = \sqrt{\frac{504390000 - 4381200000}{50}} = \sqrt{187.755} = 188$ (3 sf) | M1 A1
**Total: 3**
(ii) $2 \times z \times \frac{\sqrt{187.755}}{\sqrt{50}} = 5.45$ or equivalent | M1
$z = 1.406$ or 1.405 | A1
$\Phi('1.406') = 0.92$ or 0.9199 | M1
$\alpha = 84$ (2 sf) allow 83.98 | A1
**Total: 4**
(iii) $\frac{0.96}{\sqrt{...}} = 0.849$ (3 sf) | M1 A1
**Total: 2**
If '2 $\times$' omitted: $z \times \frac{\sqrt{187.755}}{\sqrt{50}} = 5.45$ M1; $z = 2.812$ or 2.810 A0; $\Phi('2.812') = 0.9975$; $\alpha = 99.5$ or 99 or 100 M1 A0
For complete method to find $\alpha$
SR use of biased var(184) scores M1A1(1.4205); $\alpha = 84.5$ M1A1
**Total: 9**
---5 The volumes, $v$ millilitres, of juice in a random sample of 50 bottles of Cooljoos are measured and summarised as follows.
$$n = 50 \quad \Sigma v = 14800 \quad \Sigma v ^ { 2 } = 4390000$$
(i) Find unbiased estimates of the population mean and variance.\\
(ii) An $\alpha \%$ confidence interval for the population mean, based on this sample, is found to have a width of 5.45 millilitres. Find $\alpha$.
Four random samples of size 10 are taken and a $96 \%$ confidence interval for the population mean is found from each sample.\\
(iii) Find the probability that these 4 confidence intervals all include the true value of the population mean.
\hfill \mbox{\textit{CAIE S2 2015 Q5 [9]}}