Let the remaining two observations be x and y
The observations are 2, 4, 10, 12, 14, x, y.
\(Mean\,\bar{x}\frac{2+4+10+12+14+x+y}{7}=8\)
\(⇒56=42+x+y\)
\(⇒x+y=14\)…….(1)
\(varience=16=\frac{1}{n}\sum_{i=1}^7(x_i-\bar{x})^2\)
\(16=\frac{1}{7}[(-6)^2+(-4)^2+(2)^1+(4)^2+(6)^2+x^2+y^2-2×8(x+y)+2×(8)^2]\)
\(16=\frac{1}{7}[36+16+4+16+36+x^2+y^2-16(14)+2(64)].........[Using(1)]\)
\(16=\frac{1}{7}[108+x^2+y^2-224+128]\)
\(16=\frac{1}{7}[12+x^2+y^2]\)
\(⇒x^2+y^2=100\) ……(2)
From (1), we obtain
\(x ^2 + y^2 + 2xy = 196 (3)\)
From (2) and (3), we obtain
2xy = 196-100
⇒ 2xy = 96-(4)
Subtracting (4) from (2), we obtain
\(x^2+ y^2 2xy = 100-96\)
\(⇒ (x -y)^2 = 4\)
\(⇒ x-y = ± 2(5)\)
Therefore, from (1) and (5), we obtain
x = 8 and y = 6 when x-y = 2
x = 6 and y = 8 when x-y = -2
Thus, the remaining observations are 6 and 8.