The table shows the marks obtained by a group of students in a class test. Marks 40 - 44 45 - 49 50 - 54 55 - 59 60 - 64 65 - 69 No of students 4 9 18 23 10...
Question 1 Report
The table shows the marks obtained by a group of students in a class test.
| Marks | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 |
| No of students |
4 | 9 | 18 | 23 | 10 | 6 |
(a) Draw a histogram for the distribution ;
(b) Use your histogram to estimate the median of the distribution.
Answer Details
(a) Histogram
Convert the class limits to continuous class boundaries. Since every class has width 5, the bar heights are the frequencies.
| Marks | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 40–44 | 39.5–44.5 | 4 | 4 |
| 45–49 | 44.5–49.5 | 9 | 13 |
| 50–54 | 49.5–54.5 | 18 | 31 |
| 55–59 | 54.5–59.5 | 23 | 54 |
| 60–64 | 59.5–64.5 | 10 | 64 |
| 65–69 | 64.5–69.5 | 6 | 70 |
The required histogram is:
(b) Median
The total frequency is
\[N=70,\qquad \frac{N}{2}=35.\]
The 35th observation lies in the class \(55\text{–}59\), since the cumulative frequency before this class is 31 and the cumulative frequency at the end is 54.
Using \(L=54.5\), \(c_f=31\), \(f=23\) and class width \(c=5\),
\[\begin{aligned}\text{Median} &=L+\frac{\left(\frac{N}{2}-c_f\right)}{f}\times c\\&=54.5+\frac{35-31}{23}\times5\\&=54.5+0.87\\&\approx55.4.\end{aligned}\]
Therefore, the estimated median mark is \(55.4\).
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