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The following table production yield in ...

The following table production yield in kg per hectare of wheat of 150 farms in a village . Culculate the mean , median and mode of the production yeild.
`{:("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):}`

Text Solution

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`overline(X)=A+(sumfd)/(sumf)xxC`
`=63.5+(16)/(150)xx3`
`63.5+0.32=63.82`
M= size of `((N)/(2))`th item
=Size of `((150)/(2))`th item
= Size of 75th item
This lies in 91th cumulative frequency and the corresponding class is 62-65.
`M=l_(2)+(N/(2)-c.f.)/(f)xxi`
`=62+(150/(2)-55)/(36)xx3`
`=62+(75-55)/(36)xx3`
`=62+(20)/(36)xx3`
`=62+1.67`
`=63.67`
A glance at the series reveals that 62-65 is the modal class because it has the maximum frequency , i.e., 36 Following formula is used to calculate modal value.
`Z=l_(1)+(f_(1)-f_(0))/(2f_(1)-f_(0)-f_(2))xxi`
`=62+(36-30)/(2(36)-30-28)xx3`
`=62+(6)/(72-58)xx3`
`=62+(6)/(14)xx3`
`=62+1.29`
`=63.29
`:.` Mean = 63. 82 kg per hectare.
Median = 63.67 kg per hectare.
Mode = 63.29 kg per hectare.]
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