Step 1 :First, we need to find the mean and standard deviation of the binomial distribution. The mean (expected value) of a 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 2 :In this case, \(n = 153\) and \(p = 0.8\). So, the mean is \(np = 153 \times 0.8 = 122.4\) and the standard deviation is \(\sqrt{np(1-p)} = \sqrt{153 \times 0.8 \times 0.2} = 6.4807\).
Step 3 :For part (a), we want to find the probability that exactly 114 flights are on time. To use the normal approximation, we need to convert this to a z-score, which is given by \((x - \mu) / \sigma\), where \(x\) is the value we're interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 4 :The z-score for 114 is \((114 - 122.4) / 6.4807 = -1.296\).
Step 5 :We can then use a standard normal distribution table to find the probability associated with this z-score. The table gives the probability that a value is less than the given z-score, so to find the probability that a value is exactly equal to the z-score, we need to subtract the probability that the value is one less than the z-score.
Step 6 :The z-score for 113 is \((113 - 122.4) / 6.4807 = -1.45\).
Step 7 :Looking up these z-scores in the standard normal distribution table, we find that the probability associated with a z-score of -1.296 is 0.0975 and the probability associated with a z-score of -1.45 is 0.0735.
Step 8 :So, the probability that exactly 114 flights are on time is \(0.0975 - 0.0735 = 0.024\).
Step 9 :Therefore, \(P(114) = 0.024\).