The problem can be modeled using a binomial distribution where the probability of success (owning a cell phone) is given by \(p = 0.75\), and the number of trials \(n = 8\).

We need to find the probability that the number of successes (adults owning a cell phone) is between 4 and 6 inclusive, i.e., \(P(4 \leq X \leq 6)\).

This can be calculated as:

\(P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)\)

Using the binomial probability formula \(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\), we calculate each term:

\(P(X = 4) = \binom{8}{4} (0.75)^4 (0.25)^4\)

\(P(X = 5) = \binom{8}{5} (0.75)^5 (0.25)^3\)

\(P(X = 6) = \binom{8}{6} (0.75)^6 (0.25)^2\)

Calculating each:

\(P(X = 4) = 70 \times (0.75)^4 \times (0.25)^4\)

\(P(X = 5) = 56 \times (0.75)^5 \times (0.25)^3\)

\(P(X = 6) = 28 \times (0.75)^6 \times (0.25)^2\)

Summing these probabilities gives:

\(P(4 \leq X \leq 6) = 0.606\)