WAEC 2018 · Paper 2 · Q13

Marks	10	20	30	40	50	60	70	80	90
Frequency	1	1	$x$	5	$y$	1	4	3	1

The frequency table shows the marks distribution of a class of 30 students in an examination. The mean mark of the distribution is 52.

(a)
Find the values of $x$ and $y$ .
Separate values with commas, e.g. 3, −2
(b)
Construct a grouped frequency distribution table starting with a lower class limit of 1 and a class interval of 10.
Model answer

Marks Frequency

1–10 1

11–20 1

21–30 2

31–40 5

41–50 12

51–60 1

61–70 4

71–80 3

81–90 1

Total 30

Each mark goes in the class that contains it (10 in 1–10, 20 in 11–20, …), using $x = 2$ and $y = 12$ .
(c)
Draw a histogram for the distribution.
Model answer
Draw bars on the class boundaries, not the class limits (0.5, 10.5, …, 90.5), with no gaps between them. The height of each bar is the frequency, and each axis is labelled.

To estimate the mode, take the tallest bar. Join its top-left corner to the top-left corner of the bar on its right, and its top-right corner to the top-right corner of the bar on its left. Read down from where the two lines cross: the mode is about 44.4.
(d)
Use the histogram to estimate the mode.

Try it on a graph

Histogram with the crossed lines that locate the mode.

Worked solution (try it first)

(a)

There are 30 students: $1 + 1 + x + 5 + y + 1 + 4 + 3 + 1 = 30$ , so $x + y = 14$ .
The mean is 52, so the marks add up to $30 \times 52 = 1560$ : $10 + 20 + 30x + 200 + 50y + 60 + 280 + 240 + 90 = 1560$ , which gives $30x + 50y = 660$ , or $3x + 5y = 66$ .
From the first equation $x = 14 - y$ .
Substitute: $3(14 - y) + 5y = 66$ , so $42 + 2y = 66$ , $y = 12$ and $x = 2$ .

(b)

Classes of width 10 starting at 1: each mark goes in the class that ends with it (10 is in 1–10, 20 in 11–20, and so on).
Marks 1–10 11–20 21–30 31–40 41–50 51–60 61–70 71–80 81–90

Frequency 1 1 2 5 12 1 4 3 1

(c)

Draw the histogram on the class boundaries $0.5, 10.5, 20.5, \ldots, 90.5$ , with touching bars of heights $1, 1, 2, 5, 12, 1, 4, 3, 1$ .

(d)

The tallest bar is 40.5–50.5.
Join its top-left corner to the top-left corner of the next bar, and its top-right corner to the top-right corner of the bar before.
Read down from where the lines cross: about 44.
By formula: $40.5 + \frac{12 - 5}{(12 - 5) + (12 - 1)} \times 10 = 40.5 + \frac{7}{18} \times 10$
$\approx 44.4$ .

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