The table below shows the marks obtained by forty pupils in a Mathematics test. Marks 0 - 9 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59 No of pupils 4 5 6 12 8 ...
Question 1 Report
The table below shows the marks obtained by forty pupils in a Mathematics test.
| Marks | 0 - 9 | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 |
| No of pupils | 4 | 5 | 6 | 12 | 8 | 5 |
(a) Draw a histogram for the mark distribution ;
(b) Use your histogram to estimate the mode ;
(c) Calculate the median of the distribution.
Answer Details
(a) Histogram
Use continuous class boundaries. Since all class widths are 10 marks, the bar heights are the frequencies.
| Marks | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 0 - 9 | -0.5 - 9.5 | 4 | 4 |
| 10 - 19 | 9.5 - 19.5 | 5 | 9 |
| 20 - 29 | 19.5 - 29.5 | 6 | 15 |
| 30 - 39 | 29.5 - 39.5 | 12 | 27 |
| 40 - 49 | 39.5 - 49.5 | 8 | 35 |
| 50 - 59 | 49.5 - 59.5 | 5 | 40 |
(b) Mode
The modal class is \(30-39\), with class boundaries \(29.5-39.5\).
\[\text{Mode}=L+\frac{d_1}{d_1+d_2}\times c\]
\[=29.5+\frac{12-6}{(12-6)+(12-8)}\times10\]
\[=29.5+\frac{6}{10}\times10=\boxed{35.5\text{ marks}}\]
(c) Median
\[N=40,\qquad \frac{N}{2}=20\]
The 20th value lies in the class \(30-39\), since the cumulative frequency rises from \(15\) to \(27\) in this class.
\[\text{Median}=L+\frac{\left(\frac{N}{2}-\text{cumulative frequency before median class}\right)}{f_m}\times c\]
\[=29.5+\frac{20-15}{12}\times10\]
\[=29.5+4.1667=\boxed{33.67\text{ marks}}\]
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