Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.
\(Mean\,\bar{x}\frac{6+7+10+12+13+x+y}{8}=9\)
\(⇒60+x+y+72\)
\(⇒x+y=12\)…….(1)
\(varience=9.25=\frac{1}{n}\sum_{i=1}^8(x_i-\bar{x})^2\)
\(9.25=\frac{1}{8}[(-3)^2+(-2)^2+(1)^1+(3)^2+(3)^2+(4)^2+x^2+y^2-2×9(x+y)+2×(9)^2]\)
\(9.25=\frac{1}{8}[9+4+1+9+9+16+x^2+y^2-18(12)+162].........[Using(1)]\)
\(9.25=\frac{1}{8}[48+x^2+y^2-216+162]\)
\(9.25=\frac{1}{8}[x^2+y^2-6]\)
\(⇒x^2+y^2=80\) ……(2)
From (1), we obtain
\(x ^2 + y^2 + 2xy = 144 -(3)\)
From (2) and (3), we obtain \(2xy = 64 -(4)\)
Subtracting (4) from (2), we obtain
\(x 2 + y 2 -2xy = 80-64 = 16\)
\(⇒ x-y = ± 4(5)\)
Therefore, from (1) and (5), we obtain
x = 8 and y = 4, when x-y = 4
x = 4 and y = 8, when x-y = 4
Thus, the remaining observations are 4 and 8.