Question :
| Length (in mm) | 118 – 126 | 127 – 135 | 136 – 144 | 145 – 153 | 154 – 162 | 163 – 171 | 172 – 180 |
| Number of Leaves | 3 | 5 | 9 | 12 | 5 | 4 | 2 |
Solution :
Here the frequency table is given in the inclusive form. So, we first convert it into exclusive form by subtracting and adding \(h\over 2\) to the lower and upper limits respectively of each class, where h denotes the difference of lower limit of a class and the upper limit of the previous class.
Converting the given table into exclusive form and preparing the cumulative frequency table, we get
Length (in mm)
Number of leaves (frequency)
Cumulative frequency
117.5 – 126.5
3
3
126.5 – 135.5
5
8
135.5 – 144.5
9
17
144.5 – 153.5
12
29
153.5 – 162.5
5
34
162.5 – 171.5
4
38
171.5 – 180.5
2
40
We have : n = 40 So, \(n\over 2\) = 20
The cumulative frequency just greater than \(n\over 2\) is 29 and the corresponding class is (144.5 – 153.5).
So, it is the median class.
Here, l = 144.5, h = 9, f = 12 and cf = 17
Substituting these values in the formula
Median = l + (\({n\over 2} – cf\over f\) \(\times\) h,
Median = 144.5 + \(3\over 12\) \(\times\) 9 = 144.5 + 2.25 = 146.75 mm
