| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question requiring standard formulas for unbiased estimates and normal distribution critical values. Part (iii) tests basic understanding that 98% of intervals contain μ (expected value = 0.98 × 50 = 49), which is conceptual but routine. Slightly below average difficulty as it's purely procedural with no problem-solving or novel insight required. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4200/80\ (= 52.5)\) | B1 | |
| \(= \frac{80}{79}\!\left(\frac{229\,000}{80} - 52.5^2\right)\ (= 107.595)\) | M1 A1 [3] | |
| \(= 108\) (3 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(52.5 \pm z\sqrt{\frac{107.595}{80}}\) | M1 | Correct form – must be \(z\)-value – allow one side only |
| \(z = 2.326\) | B1 | Seen |
| 49.8 to 55.2 | A1f [3] | ft their 52.5 and 107.595. Must be an interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 49 | B1 [1] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4200/80\ (= 52.5)$ | B1 | |
| $= \frac{80}{79}\!\left(\frac{229\,000}{80} - 52.5^2\right)\ (= 107.595)$ | M1 A1 [3] | |
| $= 108$ (3 sf) | | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $52.5 \pm z\sqrt{\frac{107.595}{80}}$ | M1 | Correct form – must be $z$-value – allow one side only |
| $z = 2.326$ | B1 | Seen |
| 49.8 to 55.2 | A1f [3] | ft their 52.5 and 107.595. Must be an interval |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 49 | B1 [1] | |
---5 The masses, $m$ grams, of a random sample of 80 strawberries of a certain type were measured and summarised as follows.
$$n = 80 \quad \Sigma m = 4200 \quad \Sigma m ^ { 2 } = 229000$$
(i) Find unbiased estimates of the population mean and variance.\\
(ii) Calculate a 98\% confidence interval for the population mean.
50 random samples of size 80 were taken and a $98 \%$ confidence interval for the population mean, $\mu$, was found from each sample.\\
(iii) Find the number of these 50 confidence intervals that would be expected to include the true value of $\mu$.
\hfill \mbox{\textit{CAIE S2 2015 Q5 [7]}}