The frequency distribution table shows the marks obtained by 100 students in a Mathematics test. Marks (%) 1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81...
Question 1 Report
The frequency distribution table shows the marks obtained by 100 students in a Mathematics test.
| Marks (%) | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
| Frequency | 2 | 3 | 5 | 13 | 19 | 31 | 13 | 9 | 4 | 1 |
(a) Draw the cumulative curve for the distribution.
(b) Use the graph to find the : (i) 60th percentile ; (ii) probability that a student passed the test if the pass mark was fixed at 35%.
Answer Details
(a) Cumulative frequency table
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 1–10 | 0.5–10.5 | 2 | 2 |
| 11–20 | 10.5–20.5 | 3 | 5 |
| 21–30 | 20.5–30.5 | 5 | 10 |
| 31–40 | 30.5–40.5 | 13 | 23 |
| 41–50 | 40.5–50.5 | 19 | 42 |
| 51–60 | 50.5–60.5 | 31 | 73 |
| 61–70 | 60.5–70.5 | 13 | 86 |
| 71–80 | 70.5–80.5 | 9 | 95 |
| 81–90 | 80.5–90.5 | 4 | 99 |
| 91–100 | 90.5–100.5 | 1 | 100 |
Plot cumulative frequency against the upper class boundaries and join the points with a smooth increasing curve.
(b)(i) 60th percentile
The 60th percentile corresponds to cumulative frequency 60. It lies in the class 51–60:
\[P_{60}=50.5+\left(\frac{60-42}{73-42}\right)\times 10=50.5+\frac{18}{31}\times10\approx56.3\%\]
Thus, the 60th percentile is approximately \(56.3\%\).
(b)(ii) Probability of passing at 35%
From the ogive, the cumulative frequency below 35% is approximately 16. Hence the number passing is approximately
\[100-16=84.\]
Therefore,
\[P(\text{student passes})=\frac{84}{100}=\boxed{0.84}.\]
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