Find :

(i) Median

(ii) Lower quartile (Q₁)

(iii) Upper quartile (Q₃)

(iv) Interquartile range

(v) Semi-interquartile range for the following series :

5, 23, 9, 16, 0, 14, 19, 8, 2, 26, 13, 18

By arranging data in ascending order, we get:

0, 2, 5, 8, 9, 13, 14, 16, 18, 19, 23, 26

Number of observations, n = 12, which is even.

(i) By formula,

$\Rightarrow \text{Median} = \dfrac{\Big(\dfrac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\dfrac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\Big(\dfrac{12}{2}\Big) \text{th} \text{ term} + \Big(\dfrac{12}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{6 \text{th} \text{ term} + (6 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{6 \text{th} \text{ term} + 7\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{13 + 14}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{27}{2} \\[1em] \Rightarrow \text{Median} = 13.5$

Hence, Median = 13.5.

(ii) By formula,

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

= $\Big(\dfrac{12}{4}\Big)$ th term

= 3 rd term

= 5.

Hence, lower quartile (Q₁) = 5.

(iii) By formula,

Upper Quartile (Q₃) = $\Big(\dfrac{3\text{n}}{4}\Big)$ th term

= $\Big(\dfrac{3 \times 12}{4}\Big)$ th term

= $\Big(\dfrac{36}{4}\Big)$ th term

= 9 th term

= 18.

Hence, Upper Quartile (Q₃) = 18.

(iv) By formula,

Inter quartile range = Upper quartile - Lower quartile

= 18 - 5

= 13.

Hence, the inter-quartile range is 13.

(v) By formula,

Semi-interquartile range = $\dfrac{1}{2}$ × Inter quartile range

= $\dfrac{1}{2} \times 13$

= 6.5

Hence, semi-interquartile range = 6.5.