The problem involves a binomial distribution where the probability of obtaining a 5 on a single spin is \(p = \frac{1}{5} = 0.2\), and the probability of not obtaining a 5 is \(1 - p = 0.8\). George spins the spinner 10 times, so \(n = 10\).

We need to find the probability of obtaining a 5 more than 4 times but fewer than 8 times, i.e., \(P(5 \leq X \leq 7)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate \(P(X = 5)\), \(P(X = 6)\), and \(P(X = 7)\):

\(P(X = 5) = \binom{10}{5} (0.2)^5 (0.8)^5 = 0.02642\)

\(P(X = 6) = \binom{10}{6} (0.2)^6 (0.8)^4 = 0.005505\)

\(P(X = 7) = \binom{10}{7} (0.2)^7 (0.8)^3 = 0.0007864\)

Sum these probabilities:

\(P(5 \leq X \leq 7) = 0.02642 + 0.005505 + 0.0007864 = 0.0327\)