| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| 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: reading values off axes, finding percentiles (median, quartiles, 90th percentile), and calculating IQR. Part (iv) requires simple comparison statements. All techniques are routine S1 content 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(156 - 55 = 99\) | B1 | \(98 \leqslant \text{answer} < 100\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(90\%\) of \(160 = 144\) | M1 | 144 seen, may be marked on graph |
| \((L =)\ 22\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Median \(= 15.6\), \(UQ = 18.8\), \(LQ = 12.7\) | B1 | \(15.5 <\) median \(< 15.8\) |
| \(IQR = 18.8 - 12.7\) | M1 | \(18.5 < UQ < 19\), \(12.5 < LQ < 13\) |
| \(6.1\) | A1 | \(6.0 \leqslant IQR \leqslant 6.2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The Median higher for Ransha (1st set of data) | B1 | Any correct comparison of central tendency, must mention median |
| IQR lower for Ransha (1st set of data) | B1 | Any correct comparison of spread, must refer to IQR |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $156 - 55 = 99$ | B1 | $98 \leqslant \text{answer} < 100$ |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $90\%$ of $160 = 144$ | M1 | 144 seen, may be marked on graph |
| $(L =)\ 22$ | A1 | |
---
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Median $= 15.6$, $UQ = 18.8$, $LQ = 12.7$ | B1 | $15.5 <$ median $< 15.8$ |
| $IQR = 18.8 - 12.7$ | M1 | $18.5 < UQ < 19$, $12.5 < LQ < 13$ |
| $6.1$ | A1 | $6.0 \leqslant IQR \leqslant 6.2$ |
---
## Question 5(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The Median higher for Ransha (1st set of data) | B1 | Any correct comparison of central tendency, must mention median |
| IQR lower for Ransha (1st set of data) | B1 | Any correct comparison of spread, must refer to IQR |
---5 Ransha measured the lengths, in centimetres, of 160 palm leaves. His results are illustrated in the cumulative frequency graph below.\\
\includegraphics[max width=\textwidth, alt={}, center]{7ea494c0-5e1a-4da9-a189-30128654fa1d-08_1090_1424_404_356}\\
(i) Estimate how many leaves have a length between 14 and 24 centimetres.\\
(ii) $10 \%$ of the leaves have a length of $L$ centimetres or more. Estimate the value of $L$.\\
(iii) Estimate the median and the interquartile range of the lengths.\\
Sharim measured the lengths, in centimetres, of 160 palm leaves of a different type. He drew a box-and-whisker plot for the data, as shown on the grid below.\\
\includegraphics[max width=\textwidth, alt={}, center]{7ea494c0-5e1a-4da9-a189-30128654fa1d-09_540_1287_1181_424}\\
(iv) Compare the central tendency and the spread of the two sets of data.\\
\hfill \mbox{\textit{CAIE S1 2019 Q5 [8]}}