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 9...
Question 1 Report
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%.
Answer Details
(a) Cumulative frequency table
| Marks | Frequency, \(f\) | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 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 |
Total number of students, \(N=500\).
(b) Ogive
The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\).
(c)
(i) Semi-interquartile range
\(Q_1\) is the \(\frac{N}{4}=125\)th value and \(Q_3\) is the \(\frac{3N}{4}=375\)th value. From the ogive,
\[Q_1\approx 26.0,\qquad Q_3\approx 56.3.\]
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{56.3-26.0}{2}=\boxed{15.15\text{ marks}}.\]
(ii) Number who failed
At a mark of \(37\), the cumulative frequency read from the ogive is approximately \(190\). Hence, \(\boxed{190\text{ students}}\) failed.
(iii) Probability of scoring between \(20\%\) and \(60\%\)
From the cumulative-frequency curve, the number scoring below \(20\%\) is \(100\), while the number scoring below \(60\%\) is \(400\).
\[\text{Number scoring between }20\%\text{ and }60\%=400-100=300.\]
\[P(20\%<X<60\%)=\frac{300}{500}=\boxed{\frac35}.\]
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