Let the random variable \(X\) represent the number of families owning a dishwasher in a sample of 145 families. \(X\) follows a binomial distribution with parameters \(n = 145\) and \(p = 0.22\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(\mu, \sigma^2)\).

Calculate the mean \(\mu\) and variance \(\sigma^2\):

\(\mu = 145 \times 0.22 = 31.9\)

\(\sigma^2 = 145 \times 0.22 \times 0.78 = 24.882\)

We use a continuity correction for \(P(X > 26)\), which becomes \(P(X > 26.5)\).

Standardize the variable:

\(P(X > 26.5) = P\left( z > \frac{26.5 - 31.9}{\sqrt{24.882}} \right) = P(z > -1.08255)\)

Using the standard normal distribution table, \(\Phi(1.08255) = 0.861\).

Thus, the probability that more than 26 families own a dishwasher is \(0.861\).