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

The mean and variance of 5 observations are 5 and 8 respectively. Three observations are 1, 3, 5. We need to find the sum of cubes of the remaining two observations.

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

Mean $$= \frac{1 + 3 + 5 + a + b}{5} = 5$$

$$9 + a + b = 25$$

$$a + b = 16$$ ... (i)

Variance $$= \frac{\sum x_i^2}{n} - \bar{x}^2 = 8$$

$$\frac{\sum x_i^2}{5} = 8 + 25 = 33$$

$$\sum x_i^2 = 165$$

$$1^2 + 3^2 + 5^2 + a^2 + b^2 = 165$$

$$1 + 9 + 25 + a^2 + b^2 = 165$$

$$a^2 + b^2 = 130$$ ... (ii)

From (i): $$(a + b)^2 = 256$$

$$a^2 + 2ab + b^2 = 256$$

$$130 + 2ab = 256$$

$$ab = 63$$ ... (iii)

$$a^3 + b^3 = (a + b)(a^2 - ab + b^2) = (a + b)(a^2 + b^2 - ab)$$

$$= 16 \times (130 - 63)$$

$$= 16 \times 67$$

$$= 1072$$

The answer is Option A: $$1072$$.