JAMB 1984 · UME · Q48

The diameters in centimetres of 20 copper spheres are distributed as shown. What is the mean diameter?

Class boundary (cm)	3.35–3.45	3.45–3.55	3.55–3.65	3.65–3.75
Frequency	3	6	7	4

Worked solution (try it first)

Use the mid-point of each class: 3.40, 3.50, 3.60 and 3.70.
Multiply by the frequencies and add: $3(3.40) + 6(3.50) + 7(3.60) + 4(3.70) = 10.2 + 21 + 25.2 + 14.8$ , which is 71.2.
Divide by the total frequency: $\frac{71.2}{20} = 3.56$ cm, option C.