Question: Your flight has been delayed: At Denver International Airport, 81% of recent flights have arrived…

Your flight has been delayed: At Denver International Airport, 81% of recent flights have arrived on time. A sample of 12 flights is studied. Round the probabilities to four decimal places.

Part 1 of 4 (a) Find the probability that all 12 of the flights were on time. The probability that all 12 of the flights were on time is \(\boxed{\phantom{0}}\).

Part 2 of 4 (b) Find the probability that exactly 10 of the flights were on time. The probability that exactly 10 of the flights were on time is \(\boxed{\phantom{0}}\).

Part 3 of 4 (c) Find the probability that 10 or more of the flights were on time. The probability that 10 or more of the flights were on time is \(\boxed{\phantom{0}}\).

Part 4 of 4 (d) Would it be unusual for 11 or more of the flights to be on time? It \(\boxed{\text{(Choose one)}}\) be unusual for 11 or more of the flights to be on time since the probability is \(\boxed{\phantom{0}}\).

Solution

The problem involves calculating probabilities using the binomial distribution. Given that 81% of recent flights have arrived on time, and a sample of 12 flights is studied, let’s solve each part: Part (a): Find the probability that all 12 flights were on time. The probability of a single flight being on time is \( p = 0.81 \). The probability that all 12 flights were on time is calculated as: \[ P(X = 12) = (0.81)^{12} \] Calculate: \[ P(X = 12) = 0.81^{12} \approx 0.0835 \] Part (b): Find the probability that exactly 10 of the flights were on time. We use the binomial probability formula: \[ P(X = k) = \binom{n}{k} (p^k) (1-p)^{n-k} \] For this problem, \( n = 12 \), \( k = 10 \), and \( p = 0.81 \). \[ P(X = 10) = \binom{12}{10} (0.81)^{10} (1-0.81)^2 \] Calculate: \[ P(X = 10) = \binom{12}{10} (0.81)^{10} (0.19)^2 \approx 0.2319 \] Part (c): Find the probability that 10 or more flights were on time. This is the cumulative probability: \[ P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) \] Calculate each term: \[ P(X = 11) = \binom{12}{11} (0.81)^{11} (0.19)^1 \approx 0.2842 \] Using the result from Part (a): \[ P(X = 12) \approx 0.0835 \] Add them: \[ P(X \geq 10) \approx 0.2319 + 0.2842 + 0.0835 = 0.5996 \] Part (d): Would it be unusual for 11 or more of the flights to be on time? Compute \( P(X \geq 11) \): \[ P(X = 11) \approx 0.2842 \] From Part (a): \[ P(X = 12) \approx 0.0835 \] \[ P(X \geq 11) \approx 0.2842 + 0.0835 = 0.3677 \] Since the probability is greater than 0.05, it would not be unusual for 11 or more flights to be on time.