Question

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, 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, `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.

Length (in mm)	Number or leaves fi	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	4	38 + 2 = 40

From the table, it can be observed that the cumulative frequency just greater than `N/2 (40/2 = 20)` is 29, belonging to the class interval 144.5 − 153.5.

Median class = 144.5 − 153.5

Lower limit (l) of median class = 144.5

Class size (h) = 9

Frequency (f) of median class = 12

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

Median = `l + ((N/2 - F)/f) xx h`

= `144.5 + ((20-17)/12)xx9`

= `144.5+9/4=146.75`

Therefore, the median length of leaves is 146.75 mm.