Here `n=100impliesN/2=100/2=50`
`:.` Median clas `=35.5-40.5`
Now `l_(1)=35.5,l_(2)=40.5,C=37,f=26`
Median `M=l_(1)+((N/2-C)(l_(2)-l_(1)))/f`
`=35.5+((50-37)xx(40.5xx35.5))/26`
`=35.5+2.5=38`
Now mean deviation `=(sumf_(i)|x_(i)-M|)/(sumf_(i))=735/100`
`=7.35`