Final answer:
To answer the given questions, we can use the normal approximation to the binomial distribution. We calculate the mean (μ) and standard deviation (σ) using the number of trials (125 flights) and the probability of success (0.88). We then use the z-score formula to find the probabilities for each scenario.
Step-by-step explanation:
To answer the given questions, we can use the normal approximation to the binomial distribution. The probability of success is 88% or 0.88, and the number of trials is 125 flights.
1) To find the probability that exactly 115 flights are on time, we can use the normal approximation. We calculate the mean (μ) as n * p, which is 125 * 0.88 = 110. The standard deviation is sqrt(n * p * (1 - p)), which is sqrt(125 * 0.88 * 0.12) = 3.01. We then use the z-score formula (z = (x - μ) / σ), where x is 115, to find the probability using a standard normal distribution table or a calculator.
2) To find the probability that at least 115 flights are on time, we need to calculate the probability that 115, 116, 117, ..., 125 flights are on time and sum them up. We can use the normal approximation as described above for each probability and add them together.
3) To find the probability that fewer than 106 flights are on time, we calculate the probability that 1, 2, 3, ..., 105 flights are not on time and sum them up. Again, we can use the normal approximation for each probability and add them together.
4) To find the probability that between 106 and 115 flights (inclusive) are on time, we calculate the probability that 106, 107, 108, ..., 115 flights are on time and sum them up. We can use the normal approximation for each probability and add them together.