Since the given data is very large, we construct a grouped frequency distribution table of class size 5. , class interval will be 0-5, 5-10, 10-15, 15-20 and so on. The data is represented in the grouped frequency distribution table as:

In the given table the classes do not overlap. Also we find that, the houses of 36 out of 40 engineers are below 20 km of distance

Primary data: when the information was collected by the investigator herself or himself with a definite objective in her or his mind, the data obtained is called primary data.

Primary data; (i), (ii) and (iii)

Secondary data; when the information was gathered from a source which already had the information stored, the data obtained is called secondary data

Secondary data; (iv) and (v)

(i) Diameter = 28 cm

Radius, r = 28/2 cm = 14cm

Volume of the solid spherical ball = (4/3) πr³

Volume of the ball = (4/3)×(22/7)×14³ = 34496/3

Hence, volume of the ball is 34496/3 cm³

(ii) Diameter = 0.21 m

Radius of the ball =0.21/2 m= 0.105 m

Volume of the ball = (4/3 )πr³

Volume of the ball = (4/3)× (22/7)×0.105³ m³

Hence, volume of the ball = 0.004851 m³

Let the diameter of earth be “d”. Therefore, the radius of earth will be will be d/2

Diameter of moon will be d/4 and the radius of moon will be d/8

Find the volume of the moon :

Volume of the moon = (4/3) πr³ = (4/3) π (d/8)³ = 4/3π(d³/512)

Find the volume of the earth :

Volume of the earth = (4/3) πr³= (4/3) π (d/2)³ = 4/3π(d³/8)

Fraction of the volume of the earth is the volume of the moon

Answer: Volume of moon is of the 1/64 volume of earth.