The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Explanation

Let remaining two observations, are `x_(1)` and `x_(2)`
Given that
Mean `barx=9`
`implies (sumx_(i))/n=9`
`implies sumx_(i)=9xx9 ( :'n=8)`
`implies x_(1)+x_(2)+6+7+10+12+12+13=72`
`implies x_(1)+x_(2)=12`……………..1
Again variance `sigma^(2)=9.25`
`= (sumx_(i)^(2))/n-((sumx)/n)^(2)=9.25`
`implies (sumx_(i)^(2))/8=9.25+(9)^(2)`
`=9.25+81=90.25`
`implies sumx_(i)^(2)=90.25xx8`
`implies x_(1)^(2)+x_(2)^(2)+6^(2)+7^(2)+10^(2)+12^(2)+12^(2)+13^(2)=722`
`implies x_(1)^(2)+(12-x_(1))^(2)+36+49+100+144+144+169`
`=722` [From equation (1)]
`implies x^(2)+(144-24x_(1)+x_(1)^(2))+642=722`
`implies 2x_(1)^(2)-24x_(1)+64=0`
`implies x_(1)^(2)-12x_(1)+32=0`
`implies (x_(1)-4)(x_(1)-8)=0`
`implies x_(1)-4=0` or `x_(1)-8=0`
`implies x_(1)=4` or `x_(1)=8`
`x_(1)=4` then `x_(2)=12-4=8`
`x_(1)=8` then `x_(2)=12-8=4`
`:.` Remaining two observations `=4,8`