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प्रश्न
The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mode 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 |
बेरीज
उत्तर
Mode
Grouping Table
Class Interval | I | II | III | IV | V | VI |
50 - 53 | 3 | |||||
53 - 56 | 8 | 11 | 22 | 25 | 52 | 80 |
56 - 59 | 14 | |||||
59 - 62 | 30 | 44 | 66 | 94 | ||
62 - 65 | 36 | |||||
65 - 68 | 28 | 64 | 44 | 80 | 54 | |
68 - 71 | 16 | |||||
71 - 74 | 10 | 26 | 15 | 31 | ||
74 - 77 | 5 |
Analysis Table
Column | 50 - 53 | 53 - 56 | 56 - 59 | 59 - 62 | 62 - 65 | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 |
I | √ | ||||||||
II | √ | √ | |||||||
III | √ | √ | |||||||
IV | √ | √ | √ | ||||||
V | √ | √ | √ | ||||||
VI | √ | √ | √ | ||||||
Total | - | - | 1 | 3 | 6 | 3 | 1 | - | - |
Modal class = 62 - 65
Mode = `L + (f_1 - f_0)/(2f_1 - f_0 - f_2) xx i`
= `62 + (36 - 30)/(2 xx 36 - 30 - 28) xx 3`
= `62 + 6/(72 - 30 - 28) xx 3`
= `62 + 6/14 xx 3`
= `62 + 18/14`
= 63.28 kg per hectare
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Measures of Central Tendency - Mode
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