The marks scored by 15 students in a class test are:

14, 20, 8, 17, 25, 27, 20, 16, 25, 0, 5, 19, 17, 30, 6

Find :

(i) Median

(ii) Lower quartile (Q₁)

(iii) Upper quartile (Q₃)

(iv) Interquartile range

(v) Semi-interquartile range

By arranging data in ascending order, we get:

0, 5, 6, 8, 14, 16, 17 17, 19, 20, 20, 25, 25, 27, 30

Number of observations, n = 15, which is odd.

(i) By formula,

$\Rightarrow \text{Median} = \dfrac{\text{n} + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \dfrac{15 + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \dfrac{16}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 8 \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 17$

Hence, median = 17.

(ii) By formula,

Lower Quartile = $\Big(\dfrac{\text{n} + 1}{4}\Big)$ th term

= $\Big(\dfrac{15 + 1}{4}\Big) = \dfrac{16}{4}$ th term

= 4th term

= 8

Hence, lower quartile (Q₁) = 8.

(iii) By formula,

Upper Quartile (Q₃) = $\Big(\dfrac{3(\text{n} + 1)}{4}\Big)$ th term

= $\Big(\dfrac{3 \times (15 + 1)}{4}\Big)$ th term

= $\Big(\dfrac{3 \times 16}{4}\Big) = \dfrac{48}{4}$ th term

= 12th term

= 25.

Hence, Upper Quartile (Q₃) = 25.

(iv) By formula,

Inter quartile range = Upper quartile - Lower quartile

= 25 - 8

= 17

Hence, the inter-quartile range is 17.

(v) By formula,

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

= $\dfrac{1}{2} \times 17$

= 8.5

Hence, semi-interquartile range = 8.5.