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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)
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Solution
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.
