The table below shows the mark distribution of candidates in an aptitude test for selection into the public service. Marks (in %) Freq 44 - 46 2 47 - 49 5 5...
Question 1 Report
The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.
| Marks (in %) | Freq |
| 44 - 46 | 2 |
| 47 - 49 | 5 |
| 50 - 52 | 11 |
| 53 - 55 | 20 |
| 56 - 61 | 42 |
| 62 - 64 | 46 |
| 65 - 67 | 36 |
| 68 - 70 | 9 |
| 71 - 73 | 3 |
(a) Make a cumulative frequency for the distribution
(b) Draw the cumulative frequency curve.
(c) From your graph, estimate the median mark.
(d) The cut-off mark was 63%. What percentage of the candidates was selected?
Answer Details
(a) Cumulative frequency table
Total frequency, \(N=174\).
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 44–46 | 43.5–46.5 | 2 | 2 |
| 47–49 | 46.5–49.5 | 5 | 7 |
| 50–52 | 49.5–52.5 | 11 | 18 |
| 53–55 | 52.5–55.5 | 20 | 38 |
| 56–61 | 55.5–61.5 | 42 | 80 |
| 62–64 | 61.5–64.5 | 46 | 126 |
| 65–67 | 64.5–67.5 | 36 | 162 |
| 68–70 | 67.5–70.5 | 9 | 171 |
| 71–73 | 70.5–73.5 | 3 | 174 |
(b) Cumulative frequency curve (ogive)
Plot the upper class boundaries against the cumulative frequencies, beginning with \((43.5,0)\), and join the points with a smooth curve.
(c) Median mark
The median corresponds to \(\frac{N}{2}=\frac{174}{2}=87\) on the cumulative-frequency axis. From the ogive, this gives a mark of approximately \(\boxed{62\%}\).
(d) Percentage selected
At a mark of \(63\%\), the cumulative frequency read from the curve is approximately \(103\). Hence, the number selected is
\[174-103=71.\]
\[\text{Percentage selected}=\frac{71}{174}\times100\%=40.8\%\approx\boxed{41\%}.\]
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