The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations is:

Let other two observations be x, y
∴Mean = (3+4+4+x+y)/5
∴4 = (11+x+y)/5
∴20 = 11+x+y
∴x+y = 9 ——— (1)
Variance = (1/5)[(3-4)²+(4-4)²+(4-4)²+(x-4)²+(y-4)²]
5.20 = (1/5)[1+0+0+x²-8x+16+y²-8y+16]
26 = 33+x²+y²-8(x+y)
26 = 33+x²+y²-8(9) [From (1)]
∴x²+y² = 65 ——— (2)
x+y = 9 [From (1)]
x²+y²+2xy = 81 [Squaring both sides]
65+2xy = 81 [From (2)]
2xy = 16
xy = 8
∴x = 8/y ——— (3)
Substituting (3) in (1),
8/y + y = 9
y²+8 = 9y
y²-9y+8 = 0
∴y = 8 and y = 1
For y = 8, x+8 = 9 ∴x = 1
∴We get x = 1 and y = 8
∴Difference between other two observations = 8-1 = 7