For two data sets, each of size $$5$$, the variances are given to be $$4$$ and $$5$$ and the corresponding means are given to be $$2$$ and $$4$$, respectively. The variance of the combined data set is

Let the sizes of the two samples be $$n_1 = 5$$ and $$n_2 = 5$$.

Their means are $$\mu_1 = 2$$ and $$\mu_2 = 4$$, while their variances are $$\sigma_1^{2} = 4$$ and $$\sigma_2^{2} = 5$$, respectively.

Step 1: Combined mean
The mean of the combined sample is obtained by the weighted average formula: $$\mu = \frac{n_1\mu_1 + n_2\mu_2}{n_1 + n_2}$$

Substituting the values,
$$\mu = \frac{5 \times 2 + 5 \times 4}{5 + 5} = \frac{10 + 20}{10} = 3$$

Step 2: Formula for combined variance
For two groups, the variance of the combined data set is

$$\sigma^{2} = \frac{n_1\bigl(\sigma_1^{2} + (\mu_1 - \mu)^{2}\bigr) + n_2\bigl(\sigma_2^{2} + (\mu_2 - \mu)^{2}\bigr)}{n_1 + n_2}$$

This comes from expanding the total sum of squared deviations from the overall mean.

Step 3: Deviations of individual means from the combined mean
$$\mu_1 - \mu = 2 - 3 = -1 \;\;\Rightarrow\;\; (\mu_1 - \mu)^{2} = 1$$
$$\mu_2 - \mu = 4 - 3 = 1 \;\;\Rightarrow\;\; (\mu_2 - \mu)^{2} = 1$$

Step 4: Substitute into the variance formula
$$\sigma^{2} = \frac{5\bigl(4 + 1\bigr) + 5\bigl(5 + 1\bigr)}{10} = \frac{5 \times 5 + 5 \times 6}{10} = \frac{25 + 30}{10} = \frac{55}{10}$$

Step 5: Simplify
$$\sigma^{2} = \frac{55}{10} = \frac{11}{2} = 5.5$$

Hence, the variance of the combined data set is $$\frac{11}{2}$$.

Option A which is: $$\frac{11}{2}$$