The mean and variance of 8 observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

#### Solution

Let *x* and *y* be the remaining two observations.

\[n = 8 \]

\[\text{ Variance } = 9 . 25\]

\[ \bar{X} = \text{ Mean } = 9 \]

\[ \Rightarrow \frac{6 + 7 + 10 + 12 + 12 + 13 + x + y}{8} = 9\]

\[ \Rightarrow 60 + x + y = 72\]

\[ \Rightarrow x + y = 12 . . . (1)\]

\[\text{ Variance } X = \frac{1}{n} \sum^8_{i = 1} {x_i}^2 - \left( \bar{X} \right)^2 \]

\[ \Rightarrow 9 . 25 = \left( \frac{1}{8} \times \left( 6^2 + 7^2 + {10}^2 + {12}^2 + {12}^2 + {13}^2 + x^2 + y^2 \right) \right) - 9^2 \]

\[ \Rightarrow 9 . 25 = \frac{1}{8}\left( 642 + x^2 + y^2 \right) - 81\]

\[ \Rightarrow 9 . 25 \times 8 = 642 + x^2 + y^2 - 648\]

\[ \Rightarrow x^2 + y^2 = 80 . . . . (2)\]

\[ \text{ We know } , \]

\[ \left( x + y \right)^2 + \left( x - y \right)^2 = 2\left( x^2 + y^2 \right)\]

\[ \Rightarrow {12}^2 + \left( x - y \right)^2 = 2 \times 80 \left[ \text{ using equations (1) and (2) } \right]\]

\[ \Rightarrow 144 + \left( x - y \right)^2 = 160\]

\[ \Rightarrow \left( x - y \right)^2 = 16\]

\[ \Rightarrow x - y = \pm 4 \]

\[\text{ If x - y = 4, then x + y = 12 and x - y = 4 give x = 8 and } y = 4\]

\[\text{ If x - y = - 4, then x + y = 12 and x - y = 4 give x = 4 and } y = 8\]

Thus, the remaining two observations are 8 and 4.