To solve this problem, we first need to determine the probability distribution of \(X\), the number of girls in the team.

We have 3 boys and 5 girls, and we are choosing a team of 4. The possible values for \(X\) are 1, 2, 3, and 4.

For \(X = 1\):

We choose 1 girl and 3 boys. The probability is given by:

\(P(X=1) = \frac{\binom{5}{1} \cdot \binom{3}{3}}{\binom{8}{4}} = \frac{5 \cdot 1}{70} = \frac{1}{14}\)

For \(X = 2\):

We choose 2 girls and 2 boys. The probability is given by:

\(P(X=2) = \frac{\binom{5}{2} \cdot \binom{3}{2}}{\binom{8}{4}} = \frac{10 \cdot 3}{70} = \frac{3}{7}\)

For \(X = 3\):

We choose 3 girls and 1 boy. The probability is given by:

\(P(X=3) = \frac{\binom{5}{3} \cdot \binom{3}{1}}{\binom{8}{4}} = \frac{10 \cdot 3}{70} = \frac{3}{7}\)

For \(X = 4\):

We choose 4 girls. The probability is given by:

\(P(X=4) = \frac{\binom{5}{4} \cdot \binom{3}{0}}{\binom{8}{4}} = \frac{5 \cdot 1}{70} = \frac{1}{14}\)

Thus, the probability distribution table is:

Given \(E(X) = \frac{5}{2}\), we calculate \(\text{Var}(X)\) using the formula:

\(\text{Var}(X) = E(X^2) - (E(X))^2\)

First, calculate \(E(X^2)\):

\(E(X^2) = 1^2 \cdot \frac{1}{14} + 2^2 \cdot \frac{3}{7} + 3^2 \cdot \frac{3}{7} + 4^2 \cdot \frac{1}{14}\)

\(= \frac{1}{14} + \frac{12}{7} + \frac{27}{7} + \frac{16}{14}\)

\(= \frac{1}{14} + \frac{24}{14} + \frac{54}{14} + \frac{16}{14}\)

\(= \frac{95}{14}\)

Now, calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{95}{14} - \left(\frac{5}{2}\right)^2\)

\(= \frac{95}{14} - \frac{25}{4}\)

\(= \frac{95}{14} - \frac{87.5}{14}\)

\(= \frac{15}{28}\)

Thus, \(\text{Var}(X) = \frac{15}{28}\) (0.536).