A certain flight arrives on time 80 percent of the time. Suppose 153 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that
(a) exactly 114 flights are on time.
(b) at least 114 flights are on time.
(c) fewer than 112 flights are on time.
(d) between 112 and 113, inclusive are on time.
(a) $\mathrm{P}(114)=0.0191$ (Round to four decimal places as needed.)
(b) $P(X \geq 114)=\square$ (Round to four decimal places as needed.)

Step 1 ：We are given that the probability of a flight arriving on time is 0.8. We are also given that we are selecting 153 flights. This is a binomial distribution problem, but we are asked to use the normal approximation to the binomial.

Step 2 ：The mean of a binomial distribution is np and the standard deviation is sqrt(np(1-p)). Here, n is the number of trials (153 in this case) and p is the probability of success (0.8 in this case).

Step 3 ：For part (b), we need to find the probability that at least 114 flights are on time. In terms of the normal distribution, this is equivalent to finding the probability that the number of successes is greater than or equal to 114.

Step 4 ：We can use the z-score formula to convert 114 to a z-score. The z-score is given by \((X - \text{mean}) / \text{standard deviation}\).

Step 5 ：Once we have the z-score, we can use a z-table to find the probability that the z-score is less than or equal to our calculated z-score. However, since we want the probability that the z-score is greater than or equal to our calculated z-score, we need to subtract this probability from 1.

Step 6 ：Calculating the mean and standard deviation, we get \(\text{mean} = 122.4\) and \(\text{std_dev} = 4.947726750741192\).

Step 7 ：Calculating the z-score, we get \(z = -1.6977493752543322\).

Step 8 ：Using the z-table, we find that the probability that the z-score is less than or equal to our calculated z-score is approximately 0.9552224627931788.

Step 9 ：Subtracting this probability from 1, we get the probability that at least 114 flights are on time is approximately 0.9552.

Step 10 ：Final Answer: \(P(X \geq 114)=\boxed{0.9552}\)