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Question
Compute the mode of the following data:
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
| Frequency | 25 | 16 | 28 | 20 | 5 |
Solution
Here, the maximum class frequency is 28, and the class corresponding to this frequency is 40 – 60. So, the modal class is 40 – 60.
Modal class = 40 – 60, lower limit (l) of modal class = 40, class size (h) = 20,
frequency `(f_1)` of the modal class = 28,
frequency `(f_0`) of class preceding the modal class = 16,
frequency `(f_2)` of class succeeding the modal class = 20
Now, let us substitute these values in the formula:
Mode = `l + ((f_1− f_0)/(2f_1− f_0− f_2)) × h`
`= 40 + ((28−16)/(56−16−20)) × 20`
`= 40 + ((12)/(20)) × 20`
= 40 + 12
= 52
Hence, the mode is 52.
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