The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is

Given: 5 observations with mean = 5 and variance = 8. Three of the observations are 1, 3, 5.

Let the remaining two observations be $$a$$ and $$b$$.

Using the mean:

$$ \frac{1 + 3 + 5 + a + b}{5} = 5 $$ $$ 9 + a + b = 25 $$ $$ a + b = 16 \quad \ldots (1) $$

Using the variance:

$$ \text{Variance} = \frac{\sum x_i^2}{n} - \bar{x}^2 = 8 $$ $$ \frac{\sum x_i^2}{5} = 8 + 25 = 33 $$ $$ \sum x_i^2 = 165 $$ $$ 1^2 + 3^2 + 5^2 + a^2 + b^2 = 165 $$ $$ 1 + 9 + 25 + a^2 + b^2 = 165 $$ $$ a^2 + b^2 = 130 \quad \ldots (2) $$

From (1): $$(a+b)^2 = a^2 + 2ab + b^2$$

$$ 256 = 130 + 2ab $$ $$ ab = 63 \quad \ldots (3) $$

Now, the sum of cubes:

$$ a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)[(a^2+b^2) - ab] $$ $$ = 16 \times (130 - 63) = 16 \times 67 = 1072 $$

Therefore, the sum of cubes of the remaining two observations is 1072.