| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Critique of sampling methods |
| Difficulty | Moderate -0.8 This is a straightforward critique of a flawed sampling method requiring only arithmetic calculations (multiply by 853, add 1, round) and observation that pupil 89 appears twice. The improvement is standard textbook knowledge (reject and resample). No complex probability theory or novel insight needed—purely procedural with minimal problem-solving. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 89, 90, 91, 91, 92 | B2 | All correct: B2; one error (e.g. all \(-1\)): B1. Allow 088, etc |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Not all equally likely (91 more than 90 etc) | B1 | Imply different likelihood/probability. *Not* "same pupil is selected twice" |
| Multiply by 1000 and ignore if \(> 853\) | B1 | Or equivalent method. Not "ignore repeats". Ignore extras. Number students, use random numbers and ignore outside range: B1 |
| Total: 2 |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 89, 90, 91, 91, 92 | B2 | All correct: B2; one error (e.g. all $-1$): B1. Allow 088, etc |
| | **Total: 2** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Not all equally likely (91 more than 90 etc) | B1 | Imply different likelihood/probability. *Not* "same pupil is selected twice" |
| Multiply by 1000 and ignore if $> 853$ | B1 | Or equivalent method. Not "ignore repeats". Ignore extras. Number students, use random numbers and ignore outside range: B1 |
| | **Total: 2** | |
---1 It is required to select a random sample of 30 pupils from a school with 853 pupils. A student suggests the following method.\\
"Give each pupil sequentially a three-digit number from 001 to 853 . Use a calculator to generate random three-digit numbers from 0.000 to 0.999 inclusive, multiply the answer by 853 , add 1 and round off to the nearest whole number. Select the corresponding pupil, and repeat as necessary".\\
(i) Determine which pupil would be picked for each of the following calculator outputs:
$$0.103 , \quad 0.104 , \quad 0.105 , \quad 0.106 , \quad 0.107$$
(ii) Use your answers to part (i) to show that this method is biased, and suggest an improvement.
\hfill \mbox{\textit{OCR S2 2013 Q1 [4]}}