Let the probability of choosing a rope of length 3 m be denoted as \(P(3)\) and the probability of choosing a rope of length 5 m be \(P(5)\).

Given that there are four times as many ropes of length 3 m as there are ropes of length 5 m, we have:

\(P(3) = \frac{4}{5} \quad \text{and} \quad P(5) = \frac{1}{5}\)

The expectation \(E(X)\) is calculated as:

\(E(X) = 3 \times P(3) + 5 \times P(5) = 3 \times \frac{4}{5} + 5 \times \frac{1}{5} = \frac{12}{5} + \frac{5}{5} = \frac{17}{5} = 3.4\)

The variance \(\text{Var}(X)\) is calculated as:

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

\(E(X^2) = 3^2 \times P(3) + 5^2 \times P(5) = 9 \times \frac{4}{5} + 25 \times \frac{1}{5} = \frac{36}{5} + \frac{25}{5} = \frac{61}{5}\)

Then, use the formula for variance:

\(\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{61}{5} - \left(\frac{17}{5}\right)^2 = \frac{61}{5} - \frac{289}{25}\)

Convert \(\frac{61}{5}\) to \(\frac{305}{25}\) and subtract:

\(\text{Var}(X) = \frac{305}{25} - \frac{289}{25} = \frac{16}{25} = 0.64\)