The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1,2 and 6, then the other two are

Let the two unknown numbers be x and y, then mean =4=1+2+6+x+y5=4⇒x+y=11…………………(1)σ2=5.212+22+62+x2+y25−mean2=5.241+x2+y2=55.2+42x2+y2=65……………...(2)Solve (1) & (2) x=4,y=7