According to flightstats.com, American Airlines flights from Dallas to Chicago are on time $80 \%$ of the time, Suppose 10 fights are randomly selected, and the number of on-time fights is recorded. (a) Explain why this is a binomial experiment. (b) Determine the values of $n$ and $p$. (c) Find and interpret the probability that exactly 7 fights are on time. (d) Find and interpret the probability that fewer than 7 flights are on time. (e) Find and interpret the probability that at least 7 flights are on time. (f) Find and interpret the probability that between 5 and 7 fights, inclusive, are on time. (c) Using the binomial distribution, the probability that exactly 7 flights are on time is $\square$. (Round to four decimal places as needed.) Interpret the probability. In 100 trials of this experiment, it is expected that about $\square$ will result in exactly 7 flights being on time. (Round to the nearest whole number as needed.) (d) Using the binomial distribution, the probability that fewer than 7 flights are on time is $\square$. (Round to four decimal places as needed.) Interpret the probability. In 100 trials of this experiment, it is expected that about $\square$ will result in fewer than 7 fights being on time. (Ft) und to the nearest whole number as needed.) (e) Using the binomial distribution, the probability that at least 7 fights are on time is $\square$. (Round to four decimal places as needed.) Interpret the probability. In 100 trials of this experiment, it is expected that about $\Pi$ will result in at least 7 fights being on time. Next
Solution
Step 1 ：This is a binomial experiment because it meets the four conditions of a binomial experiment: 1. The experiment consists of a fixed number of trials (10 flights). 2. Each trial is independent of the others. 3. Each trial has only two possible outcomes (the flight is on time or it is not). 4. The probability of success (the flight being on time) is the same for each trial (80%).
Step 2 ：The values of n and p are: n = 10 (the number of trials, which is the number of flights) p = 0.80 (the probability of success, which is the probability of a flight being on time)
Step 3 ：The probability that exactly 7 flights are on time can be calculated using the binomial probability formula: \(P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k))\)
Step 4 ：Substitute the values into the formula: \(P(X = 7) = C(10, 7) * (0.80^7) * ((1-0.80)^(10-7)) = 120 * (0.2097152) * (0.008)\)
Step 5 ：\(\boxed{P(X = 7) = 0.2013}\) (rounded to four decimal places)
Step 6 ：The probability that fewer than 7 flights are on time is the sum of the probabilities that exactly 0, 1, 2, 3, 4, 5, and 6 flights are on time. This can be calculated using the binomial probability formula and then summing the results. The final result is \(\boxed{0.3222}\) (rounded to four decimal places)
Step 7 ：The probability that at least 7 flights are on time is 1 minus the probability that fewer than 7 flights are on time. This is \(1 - 0.3222 = \boxed{0.6778}\) (rounded to four decimal places)
Step 8 ：The probability that between 5 and 7 flights, inclusive, are on time is the sum of the probabilities that exactly 5, 6, and 7 flights are on time. This can be calculated using the binomial probability formula and then summing the results. The final result is \(\boxed{0.8791}\) (rounded to four decimal places)
