Since , we are given the mid- value , we should first find out the upper and lower limits of the various classes . As the mid - values are 5,15,25,35,45,55,65,75, so , class size is 10(=15-5) . For determining limits of different classes , applying the formula :
Lower Limit : `l_(1)=m-(1)/(2)xxiand`
Upper limit : `l_(2)=m-(1)/(2)xxi`
where , i = difference between two mid - values. In this case , the upper and the lower limtis of the first mid - value are : `l_(1)=5-(1)/(2)xx10=0and l_(1)=5+(1)/(2)xx10=10`
i.e., 0-10 class intervals for other values are obtained as :
m= Size of `((N)/(2))`th item
= Size of `((100)/(2))` th item = Size of 50th item
Hence ,median lies in the class 40-50.
`M=l_(1)+(N/(2)-c.f.)/(f)xxi`
`=40+(100/(2)-43)/(13)xx10=40+(50-43)/(13)xx10`
`=40+(7)/(13)xx10=40+5.38=45.38`
Median = 45.38.
