Let the random variable \(X\) represent the number of households satisfied with their wifi connection. \(X\) follows a binomial distribution with parameters \(n = 150\) and \(p = 0.66\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(np, npq)\), where \(q = 1 - p\).

Calculate the mean and variance:

\(np = 0.66 \times 150 = 99\)

\(npq = 0.66 \times (1 - 0.66) \times 150 = 33.66\)

Standard deviation \(\sigma = \sqrt{33.66} \approx 5.8017\)

We need to find \(P(X > 84)\). Using continuity correction, this is \(P(X > 84.5)\).

Standardize using the normal distribution:

\(P(X > 84.5) = P\left( Z > \frac{84.5 - 99}{\sqrt{33.66}} \right)\)

\(= P(Z > -2.499)\)

Using standard normal distribution tables or a calculator, \(P(Z > -2.499) = 0.994\).