Let the random variable \(X\) represent the number of households with no broadband service out of 120. \(X\) follows a binomial distribution with parameters \(n = 120\) and \(p = 0.35\).

To approximate, we use a normal distribution with mean \(\mu = np = 120 \times 0.35 = 42\) and variance \(\sigma^2 = np(1-p) = 120 \times 0.35 \times 0.65 = 27.3\).

We apply a continuity correction to approximate \(P(X > 32)\) as \(P(X > 32.5)\).

Standardize using \(Z = \frac{X - \mu}{\sigma}\):

\(P(X > 32.5) = P\left( Z > \frac{32.5 - 42}{\sqrt{27.3}} \right) = P(Z > -1.818)\)

Using the standard normal distribution table, \(\Phi(1.818) \approx 0.966\).

Thus, \(P(X > 32) = 1 - \Phi(-1.818) = \Phi(1.818)\).

Therefore, the probability is approximately \(0.965 < p < 0.966\).