We are given, the mean and variance of eight observations are $8$ and $16$ respectively.
Also, we have five observations $2$, $4$, $10$, $12$ and $14$.
Suppose that the remaining two observations to be $a$ and $b$.
Now the observations are $2$, $4$, $10$, $12$, $14$, $a$ and $b$.
The formula to calculate mean of the observations is
$\overline X = \frac{{2 + 4 + 10 + 12 + 14 + a + b}}{7}$
$8 = \frac{{42 + a + b}}{7}$
$42 + a + b = 56$
$a + b = 56 – 42$
$a + b = 14$ …… (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,
$16 = \frac{1}{7}\left[ {{{( – 6)}^2} + {{( – 4)}^2} + {{(2)}^2} + {{(4)}^2} + {{(6)}^2} + {a^2} + {b^2} – 2 \times 8(x + y)} \right.\left. { + 2 \times {{(8)}^2}} \right]$
Substituting equation (1) to get,
$16 = \frac{1}{7}\left[ {36 + 16 + 4 + 16 + 36 + {a^2} + {b^2} – 16(14) + 2(64)} \right]$
$16 = \frac{1}{7}\left[ {12 + {a^2} + {b^2}} \right]$
${a^2} + {b^2} = 112 – 12$
${a^2} + {b^2} = 100$ …… (2)
Squaring on both sides of equation (1) gives,
${\left( {a + b} \right)^2} = {14^2}$
${a^2} + {b^2} + 2ab = 196$ …… (3)
Substituting equation (2) to get,
$100 + 2ab = 196$
$2ab = 196 – 100$
$2ab = 96$ …… (4)
Let us subtract equation (4) from (2) to get,
${a^2} + {b^2} – 2ab = 100 – 96$
${\left( {a – b} \right)^2} = 4$
$a – b = \pm 2$ …… (5)
Solving equations (1) and (5) gives,
When $a – b = 4$ then
$a = 8$ and $b = 6$
And when $a – b = – 4$then
$a = 6$ and $b = 8$
Thus, the remaining two observations are $6$ and $8$.