A certain flight arrives on fime 87 percent of the time. Suppose 127 fights are randomly selected Use the normal approximation to the binomial to approximate the probability that
(a) exactly 109 fights are on time.
(b) at least 109 flights are on time.
(c) fewer than 116 flights are on time
(d) between 116 and 119 , inclusive are on time
Answer
Final Answer: The approximate probabilities are as follows: (a) The probability that exactly 109 flights are on time is approximately \(\boxed{0.097}\). (b) The probability that at least 109 flights are on time is approximately \(\boxed{0.700}\). (c) The probability that fewer than 116 flights are on time is approximately \(\boxed{0.907}\). (d) The probability that between 116 and 119 flights, inclusive, are on time is approximately \(\boxed{0.084}\).
Step 1 ：Given that the flight arrives on time 87 percent of the time and 127 flights are randomly selected, we are asked to find the probability of certain outcomes using the normal approximation to the binomial distribution.
Step 2 ：The mean of the normal approximation to the binomial distribution is given by \(np\), and the standard deviation is given by \(\sqrt{np(1-p)}\), where \(n\) is the number of trials and \(p\) is the probability of success on each trial.
Step 3 ：Substituting \(n = 127\) and \(p = 0.87\) into the formulas, we find that the mean is \(110.49\) and the standard deviation is approximately \(3.79\).
Step 4 ：For part (a), we need to find the probability that exactly 109 flights are on time. In the normal approximation, this is the probability that the number of successes is between 108.5 and 109.5. The probability is approximately \(0.097\).
Step 5 ：For part (b), we need to find the probability that at least 109 flights are on time. In the normal approximation, this is the probability that the number of successes is greater than or equal to 108.5. The probability is approximately \(0.700\).
Step 6 ：For part (c), we need to find the probability that fewer than 116 flights are on time. In the normal approximation, this is the probability that the number of successes is less than 115.5. The probability is approximately \(0.907\).
Step 7 ：For part (d), we need to find the probability that between 116 and 119 flights, inclusive, are on time. In the normal approximation, this is the probability that the number of successes is between 115.5 and 119.5. The probability is approximately \(0.084\).
Step 8 ：Final Answer: The approximate probabilities are as follows: (a) The probability that exactly 109 flights are on time is approximately \(\boxed{0.097}\). (b) The probability that at least 109 flights are on time is approximately \(\boxed{0.700}\). (c) The probability that fewer than 116 flights are on time is approximately \(\boxed{0.907}\). (d) The probability that between 116 and 119 flights, inclusive, are on time is approximately \(\boxed{0.084}\).
