| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate single values from cumulative frequency graph |
| Difficulty | Moderate -0.8 This is a straightforward cumulative frequency graph reading exercise requiring standard techniques: reading quartiles, percentiles, and frequencies from a given curve, plus a simple binomial probability calculation. Part (v) requires minimal interpretation about distribution assumptions. All parts are routine S1 skills with no problem-solving or novel insight required. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | \(S_{xx} = 20400 - \frac{360^2}{8} = 4200\); \(S_{yy} = 6.88 - \frac{6.8^2}{8} = 1.1\) | M1 (both) |
| \(S_{xy} = 241 - \frac{360 \times 6.8}{8} = -65\) | M1 | Correct sub in 3 correct S formulae and a correct r formula |
| \(r = \frac{-65}{\sqrt{4200 \times 1.1}} = -0.956\) (3 sf) | A1 | Correct ans with no working M2A1; Ignore comment about –1 < r < –0.9 |
| 5(ii) | eg As you move further away, prices drop | B1 |
| 5(iii) | None | B1 |
| 5(iv) | \(b = \frac{"-65"}{4200} = -0.0154762\); \(y - \frac{6.8}{8} = "\)-0.0154762"\((x - \frac{360}{8})\) oe; \(y = -0.0155x + 1.55\) (3 sf) oe or \(y = \frac{433}{280} - \frac{13}{840}x\) oe | M1, M1, A1 |
| 5(v) | Values of x are chosen beforehand or x is independent or controlled | B1 |
**5(i)** | $S_{xx} = 20400 - \frac{360^2}{8} = 4200$; $S_{yy} = 6.88 - \frac{6.8^2}{8} = 1.1$ | M1 (both) | Correct sub in a correct S formula |
| $S_{xy} = 241 - \frac{360 \times 6.8}{8} = -65$ | M1 | Correct sub in 3 correct S formulae and a correct r formula |
| $r = \frac{-65}{\sqrt{4200 \times 1.1}} = -0.956$ (3 sf) | A1 | Correct ans with no working M2A1; Ignore comment about –1 < r < –0.9 |
**5(ii)** | eg As you move further away, prices drop | B1 | High prices go with short distances oe; Allow "Strong (or high or good or equiv) neg corr'n between price and distance" |
**5(iii)** | None | B1 | Ignore all else, even if incorrect |
**5(iv)** | $b = \frac{"-65"}{4200} = -0.0154762$; $y - \frac{6.8}{8} = "$-0.0154762"$(x - \frac{360}{8})$ oe; $y = -0.0155x + 1.55$ (3 sf) oe or $y = \frac{433}{280} - \frac{13}{840}x$ oe | M1, M1, A1 | ft their $S_y$ & $S_{xx}$ from (i) for M-marks only; or $a = \frac{6.8}{8} + "$0.0154762"$ × $\frac{360}{8}$ oe; allow $y = -0.015x + 1.5$ (figs which round to correct figures to either 3 sf or 2 sf, even if result from arith errors. Must have "$y =$" |
**5(v)** | Values of x are chosen beforehand or x is independent or controlled | B1 | Not "x is constant." Not just "y depends on x" Ignore all other, even if incorrect |5 The examination marks obtained by 1200 candidates are illustrated on the cumulative frequency graph, where the data points are joined by a smooth curve.\\
\includegraphics[max width=\textwidth, alt={}, center]{11316ea6-3999-4003-b77d-bee8b547c1da-04_1335_1319_404_413}
Use the curve to estimate\\
(i) the interquartile range of the marks,\\
(ii) $x$, if $40 \%$ of the candidates scored more than $x$ marks,\\
(iii) the number of candidates who scored more than 68 marks.
Five of the candidates are selected at random, with replacement.\\
(iv) Estimate the probability that all five scored more than 68 marks.
It is subsequently discovered that the candidates' marks in the range 35 to 55 were evenly distributed - that is, roughly equal numbers of candidates scored $35,36,37 , \ldots , 55$.\\
(v) What does this information suggest about the estimate of the interquartile range found in part (i)?
\section*{June 2005}
\hfill \mbox{\textit{OCR S1 Q5 [13]}}