Following is the distribution of marks in Economics obtained by 50 students :
`{:("Marks (more than)",0,10,20,30,40,50),("Number of Students",50,46,40,20,10,3):}`
Calculate the median marks . If 60% of the students pass this test , find the minimum marks obtained by a pass candidate.

Explanation

Let us first of all convert cumulative frequencies into simple frequencies .

(i) M = Size of `((N)/(2))` th item
= Size of `((50)/(2))` th item = Size of 25th item
Median lies class 20-30.
`M=l_(1)+(N/(2)-c.f.)/(f)xxi`
`=20+(50/(2)-10)/(20)xx10=20+(25-10)/(20)xx10`
`=20+(15)/(20)xx10=20+7.5=27.5`
(ii) Since , 60% candidates are passed and 40% have failed , the minimum pass marks are given by `P_(40)` . In other words , we have to find the value of `P_(40)`.
`P_(40)` = Size of `40((N)/(100))` th item
Size of `40((50)/(100))`th item
= Size of 20 th item
`P_(40)` lies in the class 20-30.
`P_(40)=l_(1)+(40(N/(100))-c.f.)/(f) xxi =20+(40(50/(100))-10)/(20)xx10`
`=20+(20-10)/(20)xx10=20+(10)/(20)xx10`
`=20+5=25`
Median = 27.5.
Minimum marks obtained by a pass candedate =25.