| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Implementing simple random or systematic sampling |
| Difficulty | Moderate -0.8 This is a straightforward conceptual question about systematic sampling requiring understanding of basic sampling properties. Part (i) is simple arithmetic (900รท12=75), part (ii) tests standard definitions (equal probability: yes, independence: no), and part (iii) requires recognizing that dice sums aren't uniformly distributed. No calculations beyond division are needed, and the concepts are fundamental S2 material with no problem-solving insight required. |
| Spec | 2.01a Population and sample: terminology2.01c Sampling techniques: simple random, opportunity, etc |
| \cline { 3 - 8 } \multicolumn{2}{c|}{} | Score on green dice | ||||||||
| \cline { 3 - 8 } \multicolumn{2}{c|}{} | 1 | 2 | 3 | 4 | 5 | 6 | |||
| 1,2 or 3 | 1 | 2 | 3 | 4 | 5 | 6 | ||
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{900 \div 12}{75}\) | B1 | 75 only |
| (ii) (a) True, first choice is random | B1 | True stated with reason based on first choice |
| (b) False, chosen by pattern | B1 | False stated, with any non-invalidating reason |
| (iii) Not equally likely | M1 | "Not equally likely" or "Biased" stated |
| e.g. \(P(1) = 0\), or triangular | Non-invalidating reason |
(i) $\frac{900 \div 12}{75}$ | B1 | 75 only
(ii) (a) True, first choice is random | B1 | True stated with reason based on first choice
(b) False, chosen by pattern | B1 | False stated, with any non-invalidating reason
(iii) Not equally likely | M1 | "Not equally likely" or "Biased" stated
e.g. $P(1) = 0$, or triangular | | Non-invalidating reason2 A school has 900 pupils. For a survey, Jan obtains a list of all the pupils, numbered 1 to 900 in alphabetical order. She then selects a sample by the following method. Two fair dice, one red and one green, are thrown, and the number in the list of the first pupil in the sample is determined by the following table.
\begin{center}
\begin{tabular}{ | l | c | c c c c c c | }
\cline { 3 - 8 }
\multicolumn{2}{c|}{} & \multicolumn{6}{c|}{Score on green dice} \\
\cline { 3 - 8 }
\multicolumn{2}{c|}{} & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\begin{tabular}{ l }
Score on \\
red dice \\
\end{tabular} & 1,2 or 3 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\end{tabular}
\end{center}
For example, if the scores on the red and green dice are 5 and 2 respectively, then the first member of the sample is the pupil numbered 8 in the list.
Starting with this first number, every 12th number on the list is then used, so that if the first pupil selected is numbered 8 , the others will be numbered $20,32,44 , \ldots$.\\
(i) State the size of the sample.\\
(ii) Explain briefly whether the following statements are true.
\begin{enumerate}[label=(\alph*)]
\item Each pupil in the school has an equal probability of being in the sample.
\item The pupils in the sample are selected independently of one another.\\
(iii) Give a reason why the number of the first pupil in the sample should not be obtained simply by adding together the scores on the two dice. Justify your answer.
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2007 Q2 [5]}}