Question 1 Report
The data shows the marks obtained by students in a biology test
| 52 | 56 | 25 | 56 | 68 | 73 | 66 | 64 | 56 | 48 |
| 20 | 39 | 9 | 50 | 46 | 54 | 54 | 40 | 50 | 96 |
| 36 | 44 | 18 | 97 | 65 | 21 | 60 | 44 | 54 | 32 |
| 92 | 49 | 37 | 94 | 72 | 88 | 89 | 35 | 59 | 34 |
| 15 | 88 | 53 | 16 | 84 | 52 | 72 | 46 | 60 | 42 |
(a) Construct a frequency distribution table using the class interval 0 - 9, 10 - 19, 19, 20, 29...
(b) Draw a cumulative frequency curve for the distribution
(c) Use the graph to estimate the;
(i) median
(ii) Percentage of students who scored at least 66 marks, correct to the nearest whole number.
(a) Frequency distribution
| Class interval | Class boundaries | Frequency, \(f\) | Cumulative frequency, \(cf\) |
|---|---|---|---|
| 0 - 9 | -0.5 - 9.5 | 1 | 1 |
| 10 - 19 | 9.5 - 19.5 | 3 | 4 |
| 20 - 29 | 19.5 - 29.5 | 3 | 7 |
| 30 - 39 | 29.5 - 39.5 | 6 | 13 |
| 40 - 49 | 39.5 - 49.5 | 8 | 21 |
| 50 - 59 | 49.5 - 59.5 | 12 | 33 |
| 60 - 69 | 59.5 - 69.5 | 6 | 39 |
| 70 - 79 | 69.5 - 79.5 | 3 | 42 |
| 80 - 89 | 79.5 - 89.5 | 4 | 46 |
| 90 - 99 | 89.5 - 99.5 | 4 | 50 |
| Total | 50 | ||
(b) Cumulative frequency curve (less-than ogive)
The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\), and joined by a smooth curve.
(c)(i) Median
There are \(50\) students, so the median corresponds to cumulative frequency \(\frac{50}{2}=25\). Reading from \(cf=25\) on the ogive gives a mark of approximately \(53\).
Therefore, the estimated median mark is \(53\) marks.
(c)(ii) Percentage scoring at least 66 marks
From the ogive, the cumulative frequency at \(66\) marks is approximately \(37\). Hence, the estimated number scoring at least \(66\) marks is
\[50-37=13.\]
\[\text{Percentage}=\frac{13}{50}\times100\%=26\%.\]
Therefore, the percentage is \(26\%\), correct to the nearest whole number.
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