The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

Explanation

Let remaining tw observations are `x_(1)` and `x_(2)`.
Here `n=7`
Give that `barx=8implies(sumx_(i))/n=8`
`implies x_(1)+x_(2)+2+4+10+12+14=8xx7`
`implies x_(1)+x_(2)=14`…………1
and variance `=16`
`implies (sumx_(i)^(2))/n-((sumx)/n)^(2)=16implies(sumx_(i)^(2))/7=16+(8)^(2)=80`
`implies sumx_(i)^(2)=560`
`implies x_(1)^(2)+x_(2)^(2)+2^(2)+4^(2)+10^(2)+12^(2)+14^(2)=560`
`=x_(1)^(2)+14-x_(1)^(2)+4+16+100144+196=560`
`implies x_(1)^(2)+196-28x_(1)+x_(1)^(2)+460-560=0`
`implies 2x_(1)^(2)-28x_(1)+96=0`
`=x_(1)^(2)-14x_(1)+48=0`
`implies(x_(1)-6)(x_(1)-8)=0`
`implies x_(1)-6=0` or `x_(1)-8=0`
`implies x_(1)=6` or `x_(1)=8`
If `x_(1)=6` then `x_(2)=14-6=8`
`x_(1)=8` then `x_(2)=14-8=6`
`:.` Remaining two observations `=6,8`.