| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate single values from cumulative frequency graph |
| Difficulty | Easy -1.2 This is a straightforward cumulative frequency graph reading exercise requiring only basic graph interpretation skills. Parts (i)-(iv) involve simple reading of values from the graph and one elementary probability calculation (raising a proportion to the 5th power). Part (v) requires minimal conceptual understanding about how uniform distribution affects quartile estimates. No complex calculations or problem-solving insight needed. |
| Spec | 2.01b Informal inferences: from samples2.02f Measures of average and spread |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Read at 300 or 300.25 and 900 or 900.75: 44.5 to 45.5 and 69 to 69.9; IQR 23.5 to 25.4 | M1 A1 A1 | 3 |
| (ii) 0.6 or 60%; CF 720; 63 to 64 | M1 M1 A1 | 3 |
| (iii) \(1200 - 860 = 340\) | M1 A1 | 2 |
| (iv) \(\frac{340}{1200}\); \(0.283^*\) or \(= 0.00183\) | M1 M1dep A1 | 3 |
| (v) Incorrect reason or ambiguity: B0B0. Otherwise: Too low, or should be 26 or 27 or 2 or 3 higher | B2 | 2 |
(i) Read at 300 or 300.25 and 900 or 900.75: 44.5 to 45.5 and 69 to 69.9; IQR 23.5 to 25.4 | M1 A1 A1 | 3 | dep A1 Must look back, see method. No wking, ans in range: M1A1A1
(ii) 0.6 or 60%; CF 720; 63 to 64 | M1 M1 A1 | 3 | Seen or implied; Seen or implied; 55.5 to 56: SC B1
(iii) $1200 - 860 = 340$ | M1 A1 | 2 | Allow $1200 - (850$ to $890)$; $310$ to $350$
(iv) $\frac{340}{1200}$; $0.283^*$ or $= 0.00183$ | M1 M1dep A1 | 3 | their (iii)/1200 exactly; Allow $0.00114$ to $0.00212 > 2$ sfs
(v) Incorrect reason or ambiguity: B0B0. Otherwise: Too low, or should be 26 or 27 or 2 or 3 higher | B2 | 2 | $\frac{340}{c^{1200}/c^{-35} = 20}$ or IQR $= $ value $>27$; eg IQR $= 55-35 = 20$ or IQR $=$ value $>27$ in range; or originally, majority in range 35–55 are at top; this range: B15 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]{5faf0d93-4037-4958-8665-1008477a79de-4_1344_1335_386_425}
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)?
\hfill \mbox{\textit{OCR S1 2005 Q5 [13]}}