The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Length (in mm)	Number of leaves
118 - 126	3
127 - 135	5
136 - 144	9
145 - 153	12
154 - 162	5
163 - 171	4
172 - 180	2

Find the median length of leaves.

Here the given data is discontinuous.

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\dfrac{127 - 126}{2} = \dfrac{1}{2}$ = 0.5

∴ 0.5 has to be added to the upper-class limit and 0.5 has to be subtracted from the lower-class limit of each interval.

Cumulative frequency distribution table is as follows :

Here, n = 40, $\dfrac{n}{2} = \dfrac{40}{2} = 20$.

Cumulative frequency just greater than 20 is 29, belonging to class-interval 144.5 − 153.5

Therefore, median class = 144.5 - 153.5

⇒ Class size (h) = 9

⇒ Lower limit of median class (l) = 144.5

⇒ Frequency of median class (f) = 12

⇒ Cumulative frequency of class preceding median class (cf) = 17

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\Rightarrow \text{Median } = 144.5 + \dfrac{20 - 17}{12} \times 9 \\[1em] = 144.5 + \dfrac{27}{12} \\[1em] = 144.5 + 2.25 \\[1em] = 146.75$

Hence, median length = 146.75 mm.