The probability of selecting a sample of 50 flights on a particular day and finding the proportion is more than 0.40 is approximately 0.059 or 5.9%.
We can use the central limit theorem to approximate the probability of this event. The central limit theorem states that the distribution of the sample means will be approximately normal, regardless of the distribution of the individual flight delays, as long as the sample size is large enough (usually n > 30).
In this case, the sample size is 50, which is large enough to apply the central limit theorem. The mean of the sample proportions will be 0.35, and the standard deviation will be sqrt(0.35(1-0.35)/50) = 0.032.
We want to find the probability that the sample proportion is more than 0.40. This is equivalent to finding the probability that the z-score is greater than 1.56. The z-score is calculated as (x - μ) / σ, where x is the sample proportion, μ is the mean, and σ is the standard deviation.
In this case, the z-score is (0.40 - 0.35) / 0.032 = 1.56. Using a standard normal distribution table, we can find that the probability of a z-score greater than 1.56 is 0.059.