From the following frequency distribution find :

(i) Median

(ii) Lower quartile (Q₁)

(iii) Upper quartile (Q₃)

(iv) Interquartile range

Variate	Frequency
26	6
25	4
18	8
16	9
30	5
28	11
20	13
23	4

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 = 20

31st term = 23

$\Rightarrow \text{Median} = \dfrac{30 \text{ th} \text{ term} + 31 \text { st} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{20 + 23}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{43}{2} \\[1em] \Rightarrow \text{Median} = 21.5$

Hence, median = 21.5.

(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

= 28.

Hence, Upper Quartile = 28.

(iv) By formula,

Interquartile range = Upper quartile - Lower quartile

= 28 - 18

= 10

Hence, interquartile range = 10.