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) $P(114)=$ (Round to four decimal places as needed.)
Answer
Therefore, \(P(114) = 0.024\).
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, \(P(114) = 0.024\).
