Let the random variable \(X\) represent the number of days the train arrives late out of 142 days. \(X\) follows a binomial distribution with parameters \(n = 142\) and \(p = 0.35\).

To approximate, we use a normal distribution with mean \(\mu\) and variance \(\sigma^2\).

Calculate the mean: \(\mu = np = 142 \times 0.35 = 49.7\).

Calculate the variance: \(\sigma^2 = npq = 142 \times 0.35 \times 0.65 = 32.305\).

Standard deviation: \(\sigma = \sqrt{32.305}\).

We approximate \(X\) by a normal distribution \(N(49.7, 32.305)\).

We need to find \(P(X > 40)\). Using continuity correction, this becomes \(P(X > 40.5)\).

Standardize: \(Z = \frac{40.5 - 49.7}{\sqrt{32.305}} \approx -1.619\).

Find \(P(Z > -1.619)\) using standard normal distribution tables or calculator.

\(P(Z > -1.619) = 0.947\).