Let x and y be the remaining two observations Then X = mean =4.4 ⇒
1+2+6+x+y
5
=4.4 ⇒9+x+y=22 ⇒x+y=22−9=13....(i) and variance =8.24
⇒
1
5
(12+22+62+x2+y2)−Mean2=8.24 ⇒
1
5
(1+4+36+x2+y2)−(4.4)2=8.24
⇒
1
5
(41+x2+y2)−19.36=8.24
⇒
1
5
(41+x2+y2)=8.24+19.36=27.60
⇒41+x2+y2=27.60×5=138 ⇒x2+y2=138−41=97 Now, (x+y)2+(x–y)2=2(x2+y2) ⇒169+(x−y)2=2(97) ⇒(x−y)2=194−169=25 ⇒x−y=±5....(ii) from (i) and (ii), we get, if x−y=5 Then,2x=18⇒x=9 and 2y=8⇒y=4 If x−y=−5∴2x=8⇒x=4 and 2y=18⇒y=9 Hence, the remaining two observations are 9 and 4 or 4 and 9