Answer:

mark my answer as brain-list

Solution:

We know that,

Median = l + [(n/2 - cf)/f] × h

Class size, h

Number of observations, n

Lower limit of median class, l

Frequency of median class, f

Cumulative frequency of class preceding median class, cf

Let's construct a continuous class data.

you have to write the table that i have given

then,

next step

From the table, it can be observed that n = 40 ⇒ n/2 = 20

Cumulative frequency (cf) just greater than 20 is 29, belonging to class 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

Median = l + [(n/2 - cf)/f] × h

= 144.5 + [(20 - 17)/12] × 9

= 144.5 + (3/12) × 9

= 144.5 + 9/4

= 144.5 +2.25

= 146.75

Therefore, median length of leaves is 146.75 mm.