The mean and variance of the data $$4, 5, 6, 6, 7, 8, x, y$$ where $$x < y$$ are $$6$$ and $$\frac{9}{4}$$ respectively. Then $$x^4 + y^2$$ is equal to

We are given data: $$4, 5, 6, 6, 7, 8, x, y$$ where $$x < y$$, with mean = 6 and variance = $$\frac{9}{4}$$.

Use the mean condition

Mean $$= \frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$$

$$36 + x + y = 48$$

$$x + y = 12 \quad \cdots(1)$$

Use the variance condition

Variance $$= \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{9}{4}$$

$$\frac{\sum x_i^2}{8} - 36 = \frac{9}{4}$$

$$\frac{\sum x_i^2}{8} = 36 + \frac{9}{4} = \frac{153}{4}$$

$$\sum x_i^2 = \frac{153 \times 8}{4} = 306$$

Calculate $$\sum x_i^2$$

$$4^2 + 5^2 + 6^2 + 6^2 + 7^2 + 8^2 + x^2 + y^2 = 306$$

$$16 + 25 + 36 + 36 + 49 + 64 + x^2 + y^2 = 306$$

$$226 + x^2 + y^2 = 306$$

$$x^2 + y^2 = 80 \quad \cdots(2)$$

Solve the system

From (1): $$y = 12 - x$$

Substituting in (2): $$x^2 + (12-x)^2 = 80$$

$$x^2 + 144 - 24x + x^2 = 80$$

$$2x^2 - 24x + 64 = 0$$

$$x^2 - 12x + 32 = 0$$

$$(x-4)(x-8) = 0$$

$$x = 4$$ or $$x = 8$$

Since $$x < y$$: $$x = 4, y = 8$$

Calculate $$x^4 + y^2$$

$$x^4 + y^2 = 4^4 + 8^2 = 256 + 64 = 320$$

Therefore, $$x^4 + y^2 = 320$$.

The correct answer is Option A: $$320$$.