At Denver International Airport, 82% of recent flights have arrived on time. A sample of 14 flights is studied. Round the probabilities to four decimal places.Part 1 of 4(a) Find the probability that all 14 of the flights were on time.The probability that all 14 of the flights were on time is .Part 2 of 4(b) Find the probability that exactly 12 of the flights were on time.The probability that exactly 12 of the flights were on time is .Part 3 of 4(c) Find the probability that 12 or more of the flights were on time.The probability that 12 or more of the flights were on time is .Part 4 of 4(d) Would it be unusual for 13 or more of the flights to be on time?It ▼(Choose one) be unusual for 13 or more of the flights to be on time since the probability is
Question
At Denver International Airport, 82% of recent flights have arrived on time. A sample of 14 flights is studied. Round the probabilities to four decimal places.Part 1 of 4(a) Find the probability that all 14 of the flights were on time.The probability that all 14 of the flights were on time is .Part 2 of 4(b) Find the probability that exactly 12 of the flights were on time.The probability that exactly 12 of the flights were on time is .Part 3 of 4(c) Find the probability that 12 or more of the flights were on time.The probability that 12 or more of the flights were on time is .Part 4 of 4(d) Would it be unusual for 13 or more of the flights to be on time?It ▼(Choose one) be unusual for 13 or more of the flights to be on time since the probability is
Solution
Part 1 of 4(a) The probability that all 14 of the flights were on time can be calculated by raising the probability of a single flight being on time to the power of the number of flights. In this case, that's 0.82^14 = 0.0523.
Part 2 of 4(b) The probability that exactly 12 of the flights were on time can be calculated using the binomial probability formula, which is P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)). Here, n is the number of trials (14 flights), k is the number of successful trials (12 flights), p is the probability of success (0.82), and C(n, k) is the binomial coefficient. Plugging in these values, we get P(X=12) = C(14, 12) * (0.82^12) * ((1-0.82)^(14-12)) = 0.2785.
Part 3 of 4(c) The probability that 12 or more of the flights were on time can be calculated by adding the probabilities of exactly 12, 13, and 14 flights being on time. Using the binomial probability formula as before, we get P(X>=12) = P(X=12) + P(X=13) + P(X=14) = 0.2785 + 0.1558 + 0.0523 = 0.4866.
Part 4 of 4(d) It would not be unusual for 13 or more of the flights to be on time since the probability is 0.2081 (the sum of the probabilities for exactly 13 and 14 flights being on time), which is not a low probability.
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