SLet the remaining two observations be x and y. Then we are given that

2+4+10+12+14+x+y

7

= 8
⇒ 42 + x + y = 56 ⇒ x + y = 14 ... (i)
Also , = 16
⇒ = 16
⇒ 460 + x2+y2 = 560 ⇒ x2+y2 = 100 ... (ii)
Now (x+y)2+(x–y)2 = 2(x2+y2)
⇒ (14)2+(x–y)2 = 2 × 100 [Using (i) & (ii)]
⇒ (x–y)2 = 200 – 196 ⇒ (x–y)2 = 4 ⇒ x – y = ± 2
When x – y = 2 and x + y = 14, we get x = 8 and y = 6
When x – y = –2 and x + y = 14 we get x = 6 and y = 8.
So, the remaining two observations are 6 and 8.