Problem 7 [18 points]: The number of cracks in a section of highway that is significant enough to require repair is assumed to follow a Poisson distribution. Let X be the number of cracks in 20km, and we have the information that the probability of 4 cracks in 20km is equal to the probability of 5 cracks in 20km (Pr(X = 4) = Pr(X = 5)) (a) [3 points] Please find the Expect value X (The expect cracks in 20km) and the Variance of X. I 20km (b) [3 points] Find the number of cracks in 2him which have the largest probability. (c) [4 points] What's the probability that at least one crack requires repair in 5km of the highway? (d) [4 points] Let Y be the number of cracks in 5km, sketch the (CDF) Cumulative Distribution Function and graph up to y = 4.5. (e) [4 points] If we should order the material to fix the cracks beforehand, how many packages of the material (One package for one crack) shall we order to ensure that all the cracks in 5km can be fixed with at least 95% chance? - End of Assignment - 212 IN tv A 14