WAEC 2010 · Paper 2 · Q6

The table gives the distribution of heights, in metres, of 100 students of the same age group.

Height (m)	1.40–1.42	1.43–1.45	1.46–1.48	1.49–1.51	1.52–1.54	1.55–1.57	1.58–1.60	1.61–1.63
Frequency	2	4	19	30	24	14	6	1

(a)
Calculate the mean height of the distribution.
(b)
What is the probability that the height of a student selected at random is greater than the mean height of the distribution?

Worked solution (try it first)

(a)

Take the class mark (middle value) of each class: $1.41, 1.44, 1.47, 1.50, 1.53, 1.56, 1.59, 1.62$ .
Multiply each by its frequency: $2.82, 5.76, 27.93, 45.00, 36.72, 21.84, 9.54, 1.62$ .
Add them: $\sum fx = 151.23$ , with $\sum f = 100$ .
Mean $= \dfrac{151.23}{100} = 1.5123$ m.

(b)

The classes from $1.52$ m upwards lie above the mean: $24 + 14 + 6 + 1 = 45$ students.
Probability $= \dfrac{45}{100} = 0.45$ .

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