Calculate median , quartiles , 6th decile and 70 th percentile from the following data :
`{:("Marks (less than)",80,70,60,50,40,30,20,10),("Number of Students",100,90,80,60,32,20,13,5):}`

Explanation

The data is given in the form of a cumulative frequency distribution . First we convert it into simple frequency distribution and write it ascending order :

M= Size of `((N)/(2))` th item = Size of `((100)/(2))` th item
= Size of 50 th item
Median lies in the class 40-50.
`M=l_(1)+(N/(2)-c.f.)/(f)xxi=40+(100(2)-32)/(28)xx10` `=40+(50-32)/(28)xx10=40+(18)/(28)xx10`
`=40+6.43=46.43`
`Q_(1)` = Size of `((N)/(4))`th item
= Size of `((100)/(4))`th item = Size of 25th item `Q_(1)` lies in the class 30-40.
`Q_(1)=l_(1)+(N/(4)-c.f.)/(f)xxi=30+(100/(4)-20)/(12)xx10`
`=30+(25-20)/(12)xx10=30+(5)/(12)xx10`
`=30+4.17=34.17`
`Q_(3)` = Size of `3((N)/(4))`th item
= Size of `3((100)/(4))`th item
=Size of 75 th item
`Q_(3)` lies in the class 50-60.
`Q_(3)=l_(1)+(3((N)/(4))-c.f.)/(f)xxi=50+(3((100)/(4))-60)/(20)xx10` `=50+(75-60)/(20)xx10`
`=50+(15)/(20)xx10`
`=50+7.5=57.5`
`D_(6)` = Size of`6((N)/(10))` th item
= Size of `6((100)/(10))` th item
= Size of 60 th item
`D_(6)` lies in the class 40-50.
`D_(6)=l_(1)+(6(N/(10))-c.f.)/(f)xxi=40+(6(100/(10))-32)/(28)xx10`
`=40+(60=32)/(28)xx10=40+(28)/(28)xx10`
`=40 +10=50` `P_(70)` = Size of `70((N)/(100))` th item
= Size of `70((100)/(100))` th item
Size of 70 th item
`P_(70)` lies in the class 50-60.
`P_(70)=l_(1)+(70(N/(100))-c.f.)/(f)xxi=50+(70(100/(100))-60)/(20)xx10`
`=50+(70=60)/(20)xx10`
`=50+(10)/(20)xx10`
`=50+5=55`
`:.` Median = 46.43, `Q_(1)=34.17,Q_(3)=57.5,D_(6)=50,P_(70)=55`.