| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Proportion confidence interval |
| Difficulty | Moderate -0.3 This is a straightforward application of standard proportion confidence interval formulas with clear numerical values. Parts (i)-(ii) are routine calculations, (iii) tests definition recall, and (iv) requires simple comparison of a value to an interval. The sample size is adequate for normal approximation, and no unusual complications arise—slightly easier than average due to its mechanical nature. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s^2 = 0.2 \times 0.8/90\) | B1 | OR /89 |
| \(p_s \pm zs\) | M1 | |
| \(z = 1.645\) | B1 | |
| \(0.1306 < p_y < 0.2693\) | A1 4 | Art (0.131, 0.269) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.7306 < p_p < 0.8694\) | B1ft 1 | ft (i) Art (0.731, 0.869) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If a large number of such intervals were calculated from independent samples, approximately 90% of all such intervals would contain \(p\) | B2 2 | Or: Probability that such an interval contains \(p\) is 0.9; B1 for right idea |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0.131, 0.269)\) encloses \(0.25\) so Mendel's theory is supported | M1, A1\(\sqrt{}\) 2 [9] | Or equivalent Ft CI(i) |
## Question 5:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s^2 = 0.2 \times 0.8/90$ | B1 | OR /89 |
| $p_s \pm zs$ | M1 | |
| $z = 1.645$ | B1 | |
| $0.1306 < p_y < 0.2693$ | A1 **4** | Art (0.131, 0.269) |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.7306 < p_p < 0.8694$ | B1ft **1** | ft (i) Art (0.731, **0.869**) |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| If a large number of such intervals were calculated from independent samples, approximately 90% of all such intervals would contain $p$ | B2 **2** | Or: Probability that such an interval contains $p$ is 0.9; B1 for right idea |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0.131, 0.269)$ encloses $0.25$ so Mendel's theory is supported | M1, A1$\sqrt{}$ **2** [9] | Or equivalent Ft CI(i) |
---5 An experiment with hybrid corn resulted in yellow kernels and purple kernels. Of a random sample of 90 kernels, 18 were yellow and 72 were purple.\\
(i) Calculate an approximate $90 \%$ confidence interval for the proportion of yellow kernels produced in all such experiments.\\
(ii) Deduce an approximate $90 \%$ confidence interval for the proportion of purple kernels produced in all such experiments.\\
(iii) Explain what is meant by a $90 \%$ confidence interval for a population proportion.\\
(iv) Mendel's theory of inheritance predicts that $25 \%$ of all such kernels will be yellow. State, giving a reason, whether or not your calculations support the theory.
\hfill \mbox{\textit{OCR S3 2011 Q5 [9]}}