Let the random variable \(X\) represent the number of cars made by Ford in a sample of 50 cars. \(X\) follows a binomial distribution with parameters \(n = 50\) and \(p = 0.28\).

To use a normal approximation, we calculate the mean \(\mu\) and variance \(\sigma^2\) of the distribution:

\(\mu = 50 \times 0.28 = 14\)

\(\sigma^2 = 50 \times 0.28 \times 0.72 = 10.08\)

We apply a continuity correction to find \(P(X > 18)\), which is approximated by \(P(X > 18.5)\).

Standardize the variable:

\(Z = \frac{18.5 - 14}{\sqrt{10.08}} = 1.417\)

Using the standard normal distribution table, find \(\Phi(1.417) \approx 0.9218\) or \(0.9217\).

Thus, \(P(X > 18) = 1 - \Phi(1.417) = 1 - 0.9218 = 0.0782\) or \(1 - 0.9217 = 0.0783\).