The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. Assume that the distribution of taxi and takeoff times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after the other, and that they take off one at a time on a given runway. (a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (Round your answer to four decimal places.)(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes? (Round your answer to four decimal places.)(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal places.)

Answer:a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.Step-by-step explanation:The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].Problems of normally distributed samples can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.In this problem, we have that:The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that [tex]\mu = 8.3, \sigma = 3.3[/tex].(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? We are working with a sample mean of 37 jets. So we have that:[tex]s = \frac{3.3}{\sqrt{37}} = 0.5425[/tex]Total time of 320 minutes for 37 jets, so[tex]X = \frac{320}{37} = 8.65[/tex]This probability is the pvalue of Z when [tex]X = 8.65[/tex]. So[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{8.65 - 8.3}{0.5425}[/tex][tex]Z = 0.65[/tex][tex]Z = 0.65[/tex] has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes? Total time of 275 minutes for 37 jets, so[tex]X = \frac{275}{37} = 7.43[/tex]This probability is subtracted by the pvalue of Z when [tex]X = 7.43[/tex][tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{7.43 - 8.3}{0.5425}[/tex][tex]Z = -1.60[/tex][tex]Z = -1.60[/tex] has a pvalue of 0.0548.There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? Total time of 320 minutes for 37 jets, so[tex]X = \frac{320}{37} = 8.65[/tex]Total time of 275 minutes for 37 jets, so[tex]X = \frac{275}{37} = 7.43[/tex]This probability is the pvalue of Z when [tex]X = 8.65[/tex] subtracted by the pvalue of Z when [tex]X = 7.43[/tex].So:From a), we have that for [tex]X = 8.65[/tex], we have [tex]Z = 0.65[/tex], that has a pvalue of 0.7422.From b), we have that for [tex]X = 7.43[/tex], we have [tex]Z = -1.60[/tex], that has a pvalue of 0.0548.So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.