The following table shows the distribution of marks obtained by some students in an examination....
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 |
.jpg)
(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%) = \(\frac{300}{500} = \frac{3}{5}\)
Post an Explanation Or Report an Error
Your email address will not be published. Required fields are marked *