| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Proportion confidence interval |
| Difficulty | Standard +0.3 This is a straightforward capture-recapture question with standard confidence interval calculations. Part (i) uses basic proportion reasoning, part (ii) applies the standard normal approximation formula for proportions (taught material), and part (iii) requires simple algebraic manipulation of the interval. While it involves multiple steps, each is routine application of S3 techniques with no novel insight required, making it slightly easier than average A-level difficulty. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(24/250 = 500/n\) | M1 | AEF |
| \(n = 500 \times 250/24 \approx 5208\) | A1 | Or 5210 or 5200. Must be integer. |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(p_s = 24/250\ (0.096)\) | ||
| \(s^2 = 0.096 \times 0.904/250 = 0.000347\) or \(s = 0.0186\) | B1 | or \(/249\) |
| Use \(p_s \pm zs\) | M1 | \(p_s + s\) M0. Incorrect \(n\), eg 5208, B0M1B1A0 |
| \(z = 1.96\) | B1 | |
| \((0.05948,\ 0.13252) \sim (0.0595,\ 0.133)\) | A1 | or \((0.0594, 0.1326)\). Allow 3DP answers. Must be interval, indicated somewhere. Allow limits reversed. |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(500/n_1 = \text{"0.0595"}\) and \(500/n_2 = \text{"0.1325"}\) | M1 | |
| \(n_1 = 8406\) or \(8410\) (3SF) \([8400, 8420]\) | A1 | Allow M1A1A0 if both values in range, but neither integers |
| \(n_2 = 3774\) or \(3770\) (3SF) \([3759, 3774]\) | A1 | |
| [3] |
# Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $24/250 = 500/n$ | M1 | AEF |
| $n = 500 \times 250/24 \approx 5208$ | A1 | Or 5210 or 5200. Must be integer. |
| | **[2]** | |
---
# Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $p_s = 24/250\ (0.096)$ | | |
| $s^2 = 0.096 \times 0.904/250 = 0.000347$ or $s = 0.0186$ | B1 | or $/249$ |
| Use $p_s \pm zs$ | M1 | $p_s + s$ M0. Incorrect $n$, eg 5208, B0M1B1A0 |
| $z = 1.96$ | B1 | |
| $(0.05948,\ 0.13252) \sim (0.0595,\ 0.133)$ | A1 | or $(0.0594, 0.1326)$. Allow 3DP answers. Must be interval, indicated somewhere. Allow limits reversed. |
| | **[4]** | |
---
# Question 2(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $500/n_1 = \text{"0.0595"}$ and $500/n_2 = \text{"0.1325"}$ | M1 | |
| $n_1 = 8406$ or $8410$ (3SF) $[8400, 8420]$ | A1 | Allow M1A1A0 if both values in range, but neither integers |
| $n_2 = 3774$ or $3770$ (3SF) $[3759, 3774]$ | A1 | |
| | **[3]** | |
---2 In order to estimate the total number of rabbits in a certain region, a random sample of 500 rabbits is captured, marked and released. After two days a random sample of 250 rabbits is captured and 24 are found to be marked. It may be assumed that there is no change in the population during the two days.\\
(i) Estimate the total number of rabbits in the region.\\
(ii) Calculate an approximate $95 \%$ confidence interval for the population proportion of marked rabbits.\\
(iii) Using your answer to part (ii), estimate a 95\% confidence interval for the total number of rabbits in the region.
\hfill \mbox{\textit{OCR S3 2013 Q2 [9]}}