| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Confidence intervals for population mean |
| Difficulty | Moderate -0.3 This is a straightforward application of standard formulas for unbiased estimates and confidence intervals. Part (i) requires direct substitution into memorized formulas for sample mean and unbiased variance. Part (ii) is a routine confidence interval calculation using the t-distribution. The question involves no problem-solving or conceptual challenges—just careful arithmetic and formula recall, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{3420}{60} (= 57)\) | B1 | |
| \(\frac{60}{59}\left(\frac{195200}{60} - \text{`}57\text{'}^2\right) = 4.40678\) | M1 | Oe |
| \(= 4.41\) (3 sf) | A1 [3] | As final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{`}57\text{'} \pm z\sqrt{\frac{\text{`}4.40678\text{'}}{60}}\) | M1 | |
| \(z = 2.326\) | B1 | \(2.326 - 2.329\) (accept 2.33 if no better seen) |
| \([56.4 \text{ to } 57.6]\) (3 sf) | A1 [3] | NB: use of biased variance in (ii) can score in full |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{3420}{60} (= 57)$ | B1 | |
| $\frac{60}{59}\left(\frac{195200}{60} - \text{`}57\text{'}^2\right) = 4.40678$ | M1 | Oe |
| $= 4.41$ (3 sf) | A1 [3] | As final answer |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{`}57\text{'} \pm z\sqrt{\frac{\text{`}4.40678\text{'}}{60}}$ | M1 | |
| $z = 2.326$ | B1 | $2.326 - 2.329$ (accept 2.33 if no better seen) |
| $[56.4 \text{ to } 57.6]$ (3 sf) | A1 [3] | NB: use of biased variance in (ii) can score in full |
---3 Jagdeesh measured the lengths, $x$ minutes, of 60 randomly chosen lectures. His results are summarised below.
$$n = 60 \quad \Sigma x = 3420 \quad \Sigma x ^ { 2 } = 195200$$
(i) Calculate unbiased estimates of the population mean and variance.\\
(ii) Calculate a $98 \%$ confidence interval for the population mean.
\hfill \mbox{\textit{CAIE S2 2015 Q3 [6]}}