The mean and variance of 7 observations are 8 and 16, respectively.

Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

\( Mean,\overline{X}=\dfrac{2+4+10+12+14+x+y}{7}\)

= 8

= 56 + 42 + x + y

= x + y =14 ….............(1)

Variance = \( \dfrac{1}{n} \displaystyle\sum_{i=1}^{n} f_i (x_i-\overline {x} )^2 \)

16 = \( \dfrac{1}{7}[(-6)^2 + (-4)^2+(2)^2+(4)^2+(6)^2+x^2\)

\(+y^2 -2\times 8(x+y)+2\times (8)^2]\)

By using equation (i) substiute 14 instead of (x+y)

16=\( \dfrac{1}{7}[36 + 16+4+(16)+36+x+y\)

\( -16(14)+2(64)]\)

16 = \( \dfrac{1}{7}[12+x^2+y^2]\)

= x²+y² = 112 - 12

= x²+y² = 100 ....................(2)

From (1), we obtain

x²+y² +2xy = 196 ....................(3)

From (2) and (3), we obtain
2xy = 196 – 100

2xy = 96 ............................(4)

Subtracting (4) from (2), we obtain

x²+y²-2xy = 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.