Question

7. The data show 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, 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.

Accepted Answer

(a) Frequency distribution table (class interval, frequency, cumulative frequency, class boundary):
0–9: f=1, cum=1, boundary −0.5−9.5
10–19: f=3, cum=4, boundary 9.5–19.5
20–29: f=3, cum=7, boundary 19.5–29.5
30–39: f=6, cum=13, boundary 29.5–39.5
40–49: f=8, cum=21, boundary 39.5–49.5
50–59: f=12, cum=33, boundary 49.5–59.5
60–69: f=6, cum=39, boundary 59.5–69.5
70–79: f=3, cum=42, boundary 69.5–79.5
80–89: f=4, cum=46, boundary 79.5–89.5
90–99: f=4, cum=50, boundary 89.5–99.5

(b) The cumulative frequency curve (ogive) is plotted with class boundary on the horizontal axis and cumulative frequency on the vertical axis, joining the plotted points with a smooth curve.

(c)(i) Median position = ½Nᵗʰ = ½(50) = 25ᵗʰ. Tracing this on the graph, median ≈ 53.5 marks.
(ii) Tracing the point 66 (class boundary 66.5) on the graph gives a cumulative frequency of 37, meaning 37 students scored below 66. Students who scored at least 66 = 50−37 = 13. Percentage = 13/50×100% = 26%.

WAEC Mathematics 2019 Theory — Question 9

Sign up free to unlock