The marks obtained by 40 students in an examination are as follows : 85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 ...

Answer Details

(a) Completed frequency distribution

Sorting the 40 marks into the given class boundaries and tallying each mark:

(Check: \(2+3+6+11+8+6+2+2 = 40\) students.)

(b) Histogram

Since all class widths are equal (5 marks), the bar heights equal the frequencies. The class boundaries are plotted on the horizontal axis and frequency on the vertical axis, with the bars drawn adjoining (no gaps):

(c) Modal mark from the histogram

The modal class is 74.5 - 79.5, the tallest bar (frequency 11). To read the mode, straight lines are drawn inside the modal bar: one from its top-left corner to the top-left corner of the next bar, and one from its top-right corner to the top-right corner of the previous bar. The vertical through their point of intersection meets the mark axis at the mode (shown dashed above), giving mode \(\approx 77.6\).

This agrees with the standard formula \[\text{Mode} = L + \left(\frac{f_1 - f_0}{(f_1-f_0)+(f_1-f_2)}\right)c,\] where \(L = 74.5,\ f_1 = 11,\ f_0 = 6,\ f_2 = 8,\ c = 5\): \[\text{Mode} = 74.5 + \left(\frac{11-6}{(11-6)+(11-8)}\right)\times 5 = 74.5 + \frac{5}{8}\times 5 = 74.5 + 3.125 \approx 77.6.\] So the modal mark is approximately 77.6 (about 78 marks).

(d) Probability of a mark greater than 79

Marks greater than 79 fall in the four highest classes (79.5 and above), giving \[8 + 6 + 2 + 2 = 18 \text{ students.}\] With 40 students in all, \[P(\text{mark} > 79) = \frac{18}{40} = \frac{9}{20} = 0.45.\]