Final answer:
To find the probability of at least 102 out of 120 flights being on time, we use the normal approximation of the binomial distribution checking that np and n(1-p) are both greater than 5. After calculating the mean and standard deviation, we look for the probability that the z-score is above the calculated value for 101.5 using the standard normal distribution.
Step-by-step explanation:
The question asks to find the probability of at least 102 out of 120 domestic flights arriving on time given a historical on-time arrival rate of 78.73%. To tackle this problem, we can use the binomial distribution, but due to the large sample size, we approximate it using the normal distribution. This approximation is valid under the conditions that both np and n(1-p) are greater than 5, where n is the sample size and p is the probability of success (on-time arrival).
First, we need to check these conditions:
- np = 120 * 0.7873 > 5
- n(1-p) = 120 * (1 - 0.7873) > 5
Since both conditions are met, we can proceed with the normal approximation. Next, we calculate the mean (µ) and standard deviation (σ) for the normal distribution:
- µ = np = 120 * 0.7873
- σ = √(np(1-p)) ≈ √(120 * 0.7873 * 0.2127)
With these calculated, we can find the z-score for 101.5 (since we are looking for at least 102, we use the continuity correction by subtracting 0.5 from 102) and use the standard normal distribution to find the probability of z being greater than this calculated z-score. This will give us the probability that at least 102 flights will arrive on time.