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%.
(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) = \(\frac{10}{90} = \frac{1}{9}\)
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