Question

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

Answer 1

Let the remaining two observations be x and y. Then,
`"mean"=8rArr(2+4+10+12+14+x+y)/(7)=8`
`rArr" "42+x+y=56`
`rArr" "x+y=14." ...(i)"`
Also, variance =16
`rArr" "(1)/(7)(2^(2)+4^(2)+10^(2)+12^(2)+14^(2)+x^(2)+y^(2))-8^(2)=16" "[because sigma^(2)=(Sigmax_(i)^(2))/(n)-(barx)^(2)]`
`rArr" "(1)/(7)(460+x^(2)+y^(2))=80`
`rArr" "460+x^(2)+y^(2)=560`
`rArr" "x^(2)+y^(2)=100." ...(ii)"`
`"Now, "(x+y)^(2)+(x-y)^(2)=2(x^(2)+y^(2))`
`rArr" "(x-y)^(2)=2(x^(2)+y^(2))-(x+y)^(2)`
`rArr" "(x-y)^(2)=(2xx100)-(14)^(2)=(200-196)=4`
`rArr" "x-y=pm2.`
`"Now, "x+y=14,x-y=2rArrx=8,y=6,`
`x+y =14, x-y=-2rArrx=6,y=8.`
Hence, the remainig two observations are 6 and 8.