The number of days the flight arrives early can be modeled by a binomial distribution with parameters \(n = 60\) and \(p = 0.15\).

First, calculate the mean and variance of the distribution:

Mean \(= np = 60 \times 0.15 = 9\)

Variance \(= np(1-p) = 60 \times 0.15 \times 0.85 = 7.65\)

Since \(n\) is large and \(p\) is small, we can use a normal approximation. The normal distribution has mean \(9\) and variance \(7.65\).

We need to find \(P(X \geq 12)\). Using continuity correction, this is approximated by \(P(X > 11.5)\).

Standardize using the normal distribution:

\(P\left(Z > \frac{11.5 - 9}{\sqrt{7.65}}\right)\)

Calculate the standardized value:

\(Z = \frac{11.5 - 9}{\sqrt{7.65}} \approx 0.9039\)

Find the probability from the standard normal distribution table:

\(1 - \Phi(0.9039) = 1 - 0.8169 = 0.183\)

Thus, the probability that Richard’s flight arrives early at least 12 times is approximately 0.183.