The weight of 60 boys are given in the following distribution table :

Weight (kg)	No. of boys
37	10
38	14
39	18
40	12
41	6

Find :

(i) Median

(ii) Lower quartile

(iii) Upper quartile

(iv) Inter quartile range.

Cumulative frequency distribution table :

(i) Here, n = 60, which is even.

By formula,

Median = $\dfrac{\dfrac{n}{2} \text{ th term} + \Big(\dfrac{n}{2} + 1\Big) \text{ th term}}{2}$

Substituting values we get :

Median = $\dfrac{\dfrac{60}{2} \text{ th term} + \Big(\dfrac{60}{2} + 1\Big)\text{ th term}}{2} = \dfrac{\text{30th term + 31st term}}{2}$

From table,

The weight of each boy from 25th to 42nd is 39 kg.

∴ 30th term and 31st term = 39

Substituting value to get median :

Median = $\dfrac{39 + 39}{2} = \dfrac{78}{2}$ = 39 kg.

(ii) Here, n = 60, which is even.

By formula,

Lower quartile = $\Big(\dfrac{n}{4}\Big)$ th term

= $\Big(\dfrac{60}{4}\Big)$ = 15th term.

From table,

The weight of each boy from 11th to 24th term is 38 kg.

Hence, lower quartile = 38.

(iii) Here, n = 60, which is even.

By formula,

Upper quartile = $\Big(\dfrac{3n}{4}\Big)$ th term

= $\Big(\dfrac{3 \times 60}{4}\Big)$ = 45th term.

From table,

The weight of each boy from 43rd to 54th is 40 kg.

Hence, upper quartile = 40.

(iv) Inter quartile range = Upper quartile - Lower quartile

= 40 - 38

= 2.

Hence, inter-quartile range = 2.