The table shows the marks scored by some candidates in an examination. Marks (%) 0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99 Frequency 7 11 17...
Question 1 Report
The table shows the marks scored by some candidates in an examination.
| Marks (%) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| Frequency | 7 | 11 | 17 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
(a) Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve.
(b) Use the curve to estimate, correct to one decimal place, the :
(i) Lowest mark for distinction if 5% of the candidates passed with distinction ; (ii) probability of selecting a candidate who scored at most 45%.
Answer Details
(a) Cumulative frequency table
| Marks (%) | Frequency, \(f\) | Upper class boundary | Cumulative frequency |
|---|---|---|---|
| 0–9 | 7 | 9.5 | 7 |
| 10–19 | 11 | 19.5 | 18 |
| 20–29 | 17 | 29.5 | 35 |
| 30–39 | 20 | 39.5 | 55 |
| 40–49 | 29 | 49.5 | 84 |
| 50–59 | 34 | 59.5 | 118 |
| 60–69 | 30 | 69.5 | 148 |
| 70–79 | 25 | 79.5 | 173 |
| 80–89 | 21 | 89.5 | 194 |
| 90–99 | 6 | 99.5 | 200 |
Total number of candidates \(=200\).
Plot the upper class boundaries against the cumulative frequencies, including \((-0.5,0)\), and draw a smooth cumulative frequency curve.
(b)(i) The number who passed with distinction is
\[5\%\text{ of }200=10.\]
Hence the lowest distinction mark has cumulative frequency
\[200-10=190.\]
From the cumulative frequency curve, a cumulative frequency of \(190\) gives a mark of approximately \(87.5\%\).
Lowest mark for distinction \(=87.5\%\).
(b)(ii) From the curve, the cumulative frequency at \(45\%\) is approximately \(69\).
\[P(\text{score at most }45\%)=\frac{69}{200}=0.345\approx 0.3\text{ (to 1 d.p.)}.\]
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