Let the mean and variance of 8 numbers x, y, 10, 12, 6, 12, 4, 8 be 9 and 9.25 respectively. If $$x > y$$, then $$3x - 2y$$ is equal to ______.

To solve for $$x$$ and $$y$$, we use the definitions of mean and variance for the 8 numbers: $$x, y, 10, 12, 6, 12, 4, 8$$.

1. Use the Mean

The mean $$\bar{x} = 9$$.

$$\frac{x + y + 10 + 12 + 6 + 12 + 4 + 8}{8} = 9$$

$$x + y + 52 = 72 \implies \mathbf{x + y = 20} \quad \dots \text{(Eq. 1)}$$

2. Use the Variance

The variance $$\sigma^2 = 9.25$$. The formula is $$\frac{\sum x_i^2}{n} - (\bar{x})^2 = \sigma^2$$.

$$\frac{x^2 + y^2 + 10^2 + 12^2 + 6^2 + 12^2 + 4^2 + 8^2}{8} - 9^2 = 9.25$$

$$\frac{x^2 + y^2 + 100 + 144 + 36 + 144 + 16 + 64}{8} - 81 = 9.25$$

$$\frac{x^2 + y^2 + 504}{8} = 90.25$$

$$x^2 + y^2 + 504 = 722 \implies \mathbf{x^2 + y^2 = 218} \quad \dots \text{(Eq. 2)}$$

3. Solve for $$x$$ and $$y$$

Using $$(x+y)^2 = x^2 + y^2 + 2xy$$:

$$20^2 = 218 + 2xy \implies 400 - 218 = 2xy \implies 2xy = 182 \implies \mathbf{xy = 91}$$

We need two numbers that add to $$20$$ and multiply to $$91$$. These are 13 and 7.

Since the problem states $$x > y$$, we have:

• $$x = 13$$
• $$y = 7$$

4. Final Calculation

Find the value of $$3x - 2y$$:

$$3(13) - 2(7) = 39 - 14 = \mathbf{25}$$