The heights (in nearest cm) of 63 students of a certain school are given in the following frequency distribution table:

Height (in cm)	Number of students
150	9
151	12
152	10
153	8
154	11
155	7
156	6

Find :

(i) Median

(ii) Lower quartile (Q₁)

(iii) Upper quartile (Q₃)

(iv) Interquartile range from the above data.

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

Here number of observations, n = 63, which is odd.

(i) By formula,

$\Rightarrow \text{Median} = \dfrac{\text{n} + 1}{2} \text{th} \text{observation} \\[1em] \Rightarrow \text{Median} = \dfrac{63 + 1}{2} \text{th} \text{observation} \\[1em] \Rightarrow \text{Median} = \dfrac{64}{2} \text{th} \text{observation} \\[1em] \Rightarrow \text{Median} = 32 \text{th} \text{observation}$

From table,

32 nd term is 153.

Hence, median = 153.

(ii) By formula,

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

= $\Big(\dfrac{63 + 1}{4}\Big) = \dfrac{64}{4}$ th term

= 16 th term

From table,

16 th term is 151.

Hence, lower quartile = 151.

(iii) By formula,

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

= $\Big(\dfrac{3 \times (63 + 1)}{4}\Big)$ th term

= $\Big(\dfrac{3 \times 64}{4}\Big) = \dfrac{192}{4}$ th term

= 48th term

From table,

48 th term is 154.

Hence, Upper Quartile = 154.

(iv) By formula,

Inter quartile range = Upper quartile - Lower quartile

= 154 - 151

= 3.

Hence, the inter-quartile range is 3.