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Question
The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the median values.
| Production yield (kg. per hectare) | 50 - 53 | 53 - 56 | 56 - 59 | 59 - 62 | 62 - 65 | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 |
| Number of farms | 3 | 8 | 14 | 30 | 36 | 28 | 16 | 10 | 5 |
Solution
Median
| Class Interval | Frequency (f) | CF |
| 50 - 53 | 3 | 3 |
| 53 - 56 | 8 | 11 |
| 56 - 59 | 14 | 25 |
| 59 - 62 | 30 | 55 |
| 62 - 65 | 36 | 91 |
| 65 - 68 | 28 | 119 |
| 68 - 71 | 16 | 135 |
| 71 - 74 | 10 | 145 |
| 74 - 77 | 5 | 150 |
| Total | ∑f = 150 |
Median class = Size of `(N/2)^{th}` item
= Size of `(150/2)^{th}` item = 75th item
75th item lies in the 91st cumulative frequency and the corresponding class interval is 62 - 65.
Median = `L + (N/2 - CF)/f xx i`
= `62 + (75 - 55)/36 xx 3`
= `62 + 20/36 xx 3`
= 63.67 kg per hectare
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