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 provided data lacks continuous class intervals. As the difference between consecutive class intervals is 1, \( \frac{1}{2} = 0.5 \) must be added to the upper class limits and subtracted from the lower class limits to establish continuous intervals. The continuous class intervals with their corresponding cumulative frequencies are presented below.

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

The cumulative frequency immediately greater than \( \frac{n}{2} \) (which is \( \frac{40}{2} = 20 \)) is 29. This corresponds to the class interval 144.5 - 153.5.
The median class is 144.5 - 153.5.
The lower limit (\(l\)) of the median class is 144.5.
The frequency (\(f\)) of the median class is 12.
The cumulative frequency (\(cf\)) of the class preceding the median class is 17.
The class size (\(h\)) is 9.

The median is calculated using the formula: Median = \(l + (\frac{\frac{n}{2} - cf}{f} \times h)\)

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

Median = 144.5 + \( \frac{27}{12} \)
Median = 144.5 + \( \frac{9}{4} \)
Median = 146.75

Therefore, the median length of the leaves is 146.75 mm.