WAEC 2014 · Paper 2 · Q12

Statistics & correlation

The histogram represents the scores of some candidates in an examination.

(a)

Using the histogram, construct a frequency distribution table, indicating clearly the class intervals.

Model answer

Class interval	Class boundaries	Frequency
9 – 18	8.5 – 18.5	4
19 – 28	18.5 – 28.5	6
29 – 38	28.5 – 38.5	8
39 – 48	38.5 – 48.5	13
49 – 58	48.5 – 58.5	15
59 – 68	58.5 – 68.5	10
69 – 78	68.5 – 78.5	3
79 – 88	78.5 – 88.5	1
	Total	60

(b)
Draw a cumulative frequency curve of the distribution and use it to estimate the: (i) median; (ii) quartile deviation.
Separate values with commas, e.g. 3, −2
Model answer

Worked solution (try it first)

(a)

The bars stand on the class boundaries 8.5, 18.5, …, 88.5 (width 10), so the class intervals are 9–18, 19–28, 29–38, 39–48, 49–58, 59–68, 69–78 and 79–88.
Read the bar heights for the frequencies: 4, 6, 8, 13, 15, 10, 3 and 1, a total of 60.

(b)

Add up the frequencies for the cumulative frequencies: 4, 10, 18, 31, 46, 56, 59 and 60.
Plot each cumulative frequency at its upper class boundary (18.5, 28.5, …, 88.5), start at $(8.5, 0)$ , and join the points with a smooth curve.

(i)

The median is at the $\frac{60}{2} = 30$ th score: reading across at 30 gives about 47.7.

(ii)

$Q_1$ is at the $\frac{60}{4} = 15$ th score: about 34.8.
$Q_3$ is at the 45th score: about 57.8.
Quartile deviation $= \frac12(Q_3 - Q_1)$ , which is $\frac12(57.8 - 34.8) \approx 11.5$ .
The median is about 47.7 and the quartile deviation is about 11.5.

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