The table shows the frequency distribution of 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 |
| Freq | 2 | 5 | 8 | 18 | 20 | 15 | 5 | 4 | 2 | 1 |
(a) Draw the cumulative frequency curve of the distribution.
(b) Use your graph to estimate the :
(i) semi-interquartile range of the distribution; (ii) percentage of candidates who passed with distinction if the least mark for distinction was 72.
Explanation
| Marks | Class boundaries | Freq | Cum Freq |
| 0-9 | 0 – 9.5 | 2 | 2 |
| 10 – 19 | 9.5 – 19.5 | 5 | 7 |
| 20 – 29 | 19.5 – 29.5 | 8 | 15 |
| 30 – 39 | 29.5 – 39.5 | 18 | 33 |
| 40 – 49 | 39.5 – 49.5 | 20 | 53 |
| 50 – 59 | 49.5 – 59.5 | 15 | 68 |
| 60 – 69 | 59.5 – 69.5 | 5 | 73 |
| 70 – 79 | 69.5 – 79.5 | 4 | 77 |
| 80 – 89 | 79.5 – 89.5 | 2 | 79 |
| 90 – 99 | 89.5 – 99.5 | 1 | 80 |
(a)
(b)(i) Lower quartile, (Q_{1} = 32.5)
Upper quartile, (Q_{3} = 53.0)
Semi-interquartile range = (frac{1}{2} (Q_{3} – Q_{1}))
= (frac{1}{2} (53.0 – 32.5) = frac{1}{2} (20.5))
= (10.25)
(ii) Number of students that passed with distinction = (80 – 74)
= 6 students
Percentage = (frac{6}{80} times 100% = 7.5%)