Consider 10 observations $$x_1, x_2, \ldots, x_{10}$$, such that $$\sum_{i=1}^{10}(x_i - \alpha) = 2$$ and $$\sum_{i=1}^{10}(x_i - \beta)^2 = 40$$, where $$\alpha, \beta$$ are positive integers. Let the mean and the variance of the observations be $$\frac{6}{5}$$ and $$\frac{84}{25}$$ respectively. Then $$\frac{\beta}{\alpha}$$ is equal to:

Given 10 observations with mean $$\bar{x} = \frac{6}{5}$$ and variance $$\sigma^2 = \frac{84}{25}$$. Also, $$\sum_{i=1}^{10} (x_i - \alpha) = 2$$ and $$\sum_{i=1}^{10} (x_i - \beta)^2 = 40$$, where $$\alpha$$ and $$\beta$$ are positive integers.

First, use the sum of deviations from $$\alpha$$ to find $$\alpha$$. The sum of deviations from a constant $$A$$ is given by $$\sum (x_i - A) = n(\bar{x} - A)$$. Here, $$n = 10$$, so:

$$\sum (x_i - \alpha) = 10 \left( \frac{6}{5} - \alpha \right) = 2$$

Solve for $$\alpha$$:

$$\frac{6}{5} - \alpha = \frac{2}{10} = \frac{1}{5}$$

$$\alpha = \frac{6}{5} - \frac{1}{5} = \frac{5}{5} = 1$$

Thus, $$\alpha = 1$$.

Next, use the variance to find $$\sum (x_i - \bar{x})^2$$. The variance formula is $$\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$$, so:

$$\sum (x_i - \bar{x})^2 = n \sigma^2 = 10 \times \frac{84}{25} = \frac{840}{25} = \frac{168}{5}$$

Now, use the given $$\sum (x_i - \beta)^2 = 40$$. The formula relating sum of squared deviations from $$\beta$$ to the sum of squared deviations from the mean is:

$$\sum (x_i - \beta)^2 = \sum (x_i - \bar{x})^2 + n (\bar{x} - \beta)^2$$

Substitute the known values:

$$40 = \frac{168}{5} + 10 \left( \frac{6}{5} - \beta \right)^2$$

Solve for $$\beta$$:

$$40 - \frac{168}{5} = 10 \left( \frac{6}{5} - \beta \right)^2$$

$$\frac{200}{5} - \frac{168}{5} = \frac{32}{5} = 10 \left( \frac{6}{5} - \beta \right)^2$$

$$\frac{32}{50} = \frac{16}{25} = \left( \frac{6}{5} - \beta \right)^2$$

Take the square root:

$$\left| \frac{6}{5} - \beta \right| = \frac{4}{5}$$

This gives two cases:

Case 1: $$\frac{6}{5} - \beta = \frac{4}{5}$$

$$\beta = \frac{6}{5} - \frac{4}{5} = \frac{2}{5}$$

But $$\beta$$ must be a positive integer, so $$\frac{2}{5}$$ is invalid.

Case 2: $$\frac{6}{5} - \beta = -\frac{4}{5}$$

$$\beta = \frac{6}{5} + \frac{4}{5} = \frac{10}{5} = 2$$

This is a positive integer, so $$\beta = 2$$.

Thus, $$\alpha = 1$$ and $$\beta = 2$$, so:

$$\frac{\beta}{\alpha} = \frac{2}{1} = 2$$

The ratio $$\frac{\beta}{\alpha}$$ is 2, which corresponds to option A.