The following table shows the distribution of marks obtained by some students 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 | 50 | 50 | 40 | 60 | 100 | 100 | 50 | 25 | 15 | 10 |
(a) Construct a cumulative frequency table for the distribution
(b) Draw an ogive for the distribution
(c) Use your graph in (b) to determine : (i) semi- interquartile range ; (ii) number of students who failed, if the pass mark for the examination is 37 ; (iii) probability that a student selected at random scored between 20% and 60%.
Explanation
(a)
|
Marks (x) |
No of students (f) |
Class boundaries |
Cum Freq ((sum f)) |
| 0-9 | 50 | -0.5 – 9.5 | 50 |
| 10-19 | 50 | 9.5 – 19.5 | 100 |
| 20-29 | 40 | 19.5 – 29.5 | 140 |
| 30-39 | 60 | 29.5 – 39.5 | 200 |
| 40-49 | 100 | 39.5 – 49.5 | 300 |
| 50-59 | 100 | 49.5 – 59.5 | 400 |
| 60-69 | 50 | 59.5 – 69.5 | 450 |
| 70-79 | 25 | 69.5 – 79.5 | 475 |
| 80-89 | 15 | 79.5 – 89.5 | 490 |
| 90-99 | 10 | 89.5 – 99.5 | 500 |
(c)(i) Position of lower quartile, (Q_{1} = frac{sum f + 1}{4} = frac{500 + 1}{4})
= (frac{501}{4} = 125.25th) position.
(therefore Q_{1} = 26.0)
Position of upper quartile, (Q_{3} = 3(frac{sum f + 1}{4}) = 3(125.25) = 375.75th) position.
(therefore Q_{3} = 56.3)
Semi-interquartile range = (frac{Q_{3} – Q_{1}}{2} = frac{56.3 – 26}{2})
= (frac{30.3}{2} = 15.15)
(ii) If the pass mark is 37, then 190 students failed.
(iii) Those who scored between 0% and 20% = 100
Those who scored between 0% and 60% = 400
Those who scored between 20% and 60% = 400 – 100 = 300.
P(20% < x < 60%) = (frac{300}{500} = frac{3}{5})