From the following frequency distribution, find:

(i) Median

(ii) Lower quartile

(iii) Upper quartile

(iv) Semi-interquartile range

Variate	Frequency
13	6
15	4
18	11
20	9
22	16
24	12
25	2

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

Here number of observations, n = 60, which is even.

(i) By formula,

$\Rightarrow \text{Median} = \dfrac{\Big(\dfrac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\dfrac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\Big(\dfrac{60}{2}\Big) \text{th} \text{ term} + \Big(\dfrac{60}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{30 \text{th} \text{ term} + (30 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{30 \text{th} \text{ term} + 31\text{th} \text{ term}}{2}$

From table,

30th term is 20

31st term is 22 (All observations from 31st to 46th term = 22)

$\Rightarrow \text{Median} = \dfrac{30 \text{th} \text{ term} + 31 \text{st} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{20 + 22}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{42}{2} \\[1em] \Rightarrow \text{Median} = 21$

Hence, median = 21.

(ii) By formula,

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

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

= 15 th term

= 18.

Hence, lower quartile = 18.

(iii) By formula,

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

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

= $\Big(\dfrac{180}{4}\Big)$ th term

= 45 th term

= 22

Hence, Upper Quartile = 22.

(iv) By formula,

Semi-interquartile range = $\dfrac{1}{2}$ × (Upper quartile - Lower quartile)

= $\dfrac{1}{2} \times (22 - 18) = \dfrac{1}{2} \times 4$

= 2

Hence, semi-interquartile range = 2.