(i) Here, N=200
Now, `(N)/(2)=(200)/(2)=100`, which lies in the interval 15-20.
Lower limit, l=15, h=5, f=80 and cf=55
`"Median"=l+(((N)/(2)-cf)/(f))xxh=15+((100-55)/(80))xx5`
`=15+((45)/(16))=15+2.81=17.81"hec"`
(ii) In a given table 80 is the highest frequency So, the modal class is 15-20
Here, `l=15,f_(m)=80,f_(1)=30,f_(2)=40 and h=5`
`"Mode"=l+((f_(m)-f_(1))/(2f_(m)-f_(1)-f_(2)))xx5`
`=15+((50)/(160-70))xx5`
`=15+((50)/(90))xx5=15+(25)/(9)`
=15+2.77=17.77 hec