The table shows the distribution of marks scored by students in a test.
|
Mark (%) |
10 – 19 | 20 – 29 | 30 – 39 | 40 – 49 | 50 – 59 | 60 – 69 | 70 – 79 | 80 – 89 | 90 – 99 |
| Frequency | 4 | 7 | 12 | 18 | 20 | 14 | 9 | 4 | 2 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw a cumulative frequency curve for the distribution.
(c) Use the curve to estimate the:
(i) median;
(ii) probability that a student selected at random obtained distinction, if the lowest mark for distinction is 75%.
Explanation
(a)
| Marks |
Class Boundaries |
Frequency |
Cumulative Frequency |
| 10 – 19 | 9.5 – 19.5 | 4 | 4 |
| 20 – 29 | 19.5 – 29.5 | 7 | 11 |
| 30 – 39 | 29.5 – 39.5 | 12 | 23 |
| 40 – 49 | 39.5 – 49.5 | 18 | 41 |
| 50 – 59 | 49.5 – 59.5 | 20 | 61 |
| 60 – 69 | 59.5 – 69.5 | 14 | 75 |
| 70 – 79 | 69.5 – 79.5 | 9 | 84 |
| 80 – 89 | 79.5 – 89.5 | 4 | 88 |
| 90 – 99 | 89.5 – 99.5 | 2 | 90 |
(b) The cumulative frequency curve
(c)(i) using the curve, the median was 51.5.
(ii) Using the curve, the estimate of tbe number of students who scored at least 75% was 90 – 80 = 10
Therefore, Prob(that a student selected at random obtained distinction) = (frac{10}{90} = frac{1}{9})