We are given, the mean and variance of eight observations are $9$ and $9.25$ respectively.
Also, we have six observations $6$, $7$, $10$, $12$, $12$ and $13$.
Suppose that the remaining two observations to be $a$ and $b$.
Now the observations are $6$, $7$, $10$, $12$, $12$, $13$, $a$ and $b$.
The formula to calculate mean of the observations is
$\overline X = \frac{{6 + 7 + 10 + 12 + 12 + 13 + a + b}}{8}$
$9 = \frac{{60 + a + b}}{8}$
$60 + a + b = 72$
$a + b = 72 – 60$
$a + b = 12$ …… (1)
The formula for variance is given by,
Variance $ = \frac{1}{n}{\sum\limits_{i = 1}^8 {\left( {{{\rm X}_i} – {\rm X}} \right)} ^2}$
Substituting the given values to get,
$9.25 = \frac{1}{8}\left[ {{{( – 3)}^2} + {{( – 2)}^2} + {1^2} + {3^2} + {4^2} + {a^2} + {b^2} – 18(a + b) + 2 \times {9^2}} \right]$
Substituting equation (1) to get,
$9.25 = \frac{1}{8}\left[ {9 + 4 + 1 + 9 + 9 + 16 + {a^2} + {b^2} – 18 \times 12 + 162} \right]$
$9.25 = \frac{1}{8}\left[ {48 + {a^2} + {b^2} – 216 + 162} \right]$
$9.25 = \frac{1}{8}\left[ {{a^2} + {b^2} – 6} \right]$
${a^2} + {b^2} = 80$ …… (2)
Squaring on both sides of equation (1) gives,
${\left( {a + b} \right)^2} = {12^2}$
${a^2} + {b^2} + 2ab = 144$ …… (3)
Substituting equation (2) to get,
$80 + 2ab = 144$
$2ab = 144 – 80$
$2ab = 64$ …… (4)
Let us subtract equation (4) from (2) to get,
${a^2} + {b^2} – 2ab = 80 – 64$
${\left( {a – b} \right)^2} = 16$
$a – b = \pm 4$ …… (5)
Solving equations (1) and (5) gives,
When $a – b = 4$ then
$a = 8$ and $b = 4$
And when $a – b = – 4$then
$a = 4$ and $b = 8$
Thus, the remaining two observations are $4$ and $8$.