Question

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 the leaves. 
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Answer 1

The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, \(\frac{1}2 = 0.5\) has to be added and subtracted to upper class limits and lower class limits respectively.  Continuous class intervals with respective cumulative frequencies can be represented as follows. 
The cumulative frequencies with their respective class intervals are as follows.

Length (in mm)	Number of leaves	Cumulative frequency
117.5 - 126.5	3	3
126.5 - 135.5	5	3 + 5 = 8
135.5 - 144.5	9	8 + 9 = 17
144.5 - 153.5	12	17 + 12 = 29
153.5 - 162.5	5	29 + 5 = 34
162.5 - 171.5	4	34 + 4 = 38
171.5 - 180.5	2	38 + 2 = 40
Total (n)	40

Cumulative frequency just greater \(\frac{n}2 ( i.e., \frac{40}2 = 20)\) than is 29, belonging to class interval 144.5 - 153.5.
Median class = 144.5 - 153.5
Lower limit (\(l\)) of median class = 144.5
Frequency (\(f\)) of median class = 12
Cumulative frequency (\(cf\)) of median class = 17
Class size (\(h\)) = 9

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

Median =  \(144.5 + (\frac{20  - 17}{12} )\times 9\)

Median = 144.5 +\(\frac9{4}\)
Median = 146.75

Therefore, median length of leaves is 146.75 mm.