The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $$\sigma^2$$ respectively. If the variance of all the 30 numbers in the two sets is 13, then $$\sigma^2$$ is equal to

Given: Set 1: $$n_1 = 15$$, $$\bar{x}_1 = 12$$, $$\sigma_1^2 = 14$$. Set 2: $$n_2 = 15$$, $$\bar{x}_2 = 14$$, $$\sigma_2^2 = \sigma^2$$. Combined variance = 13.

The combined mean is:

$$ \bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{15(12) + 15(14)}{30} = \frac{180 + 210}{30} = 13 $$

The deviations of each set's mean from the combined mean:

$$ d_1 = \bar{x}_1 - \bar{x} = 12 - 13 = -1, \quad d_2 = \bar{x}_2 - \bar{x} = 14 - 13 = 1 $$

The combined variance formula is:

$$ \sigma_c^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2} $$

Substituting:

$$ 13 = \frac{15(14 + 1) + 15(\sigma^2 + 1)}{30} $$

$$ 390 = 15(15) + 15(\sigma^2 + 1) = 225 + 15\sigma^2 + 15 $$

$$ 390 = 240 + 15\sigma^2 $$

$$ 15\sigma^2 = 150 \implies \sigma^2 = 10 $$

The correct answer is $$\sigma^2 = \mathbf{10}$$.