In a road-paving process, asphalt mix is delivered to the hopper of the paver by trucks that haul the material from the batching plant. The article ?"Modeling of Simultaneously Continuous and Stochastic Construction Activities for Simulation" U. of Construction Engr. and Mgmnt., 2013: 1037-1045) proposed a normal distribution with mean value 8.46 min and standard deviation .913 min for the rv X = truck haul time. a. What is the probability that haul time will be at least 10 min? Will exceed 10 min? b. What is the probability that haul time will exceed 15 min? c. What is the probability that haul time will be between 8 and 10 min? d. What value c is such that 98% of all haul times are in the interval from 8.46 — c to 8.4.6 + c? e. If four haul times are independently selected, what is the probability that at least one of them exceeds 10 min?

Answer: Step1 of 6: Given a normal distribution with a mean value 8.46 min and standard deviation 0.913 min. The random variable X = truck haul time. Step2 of 6: a). We need to find the probability that haul time will be i).at least 10 min and ii). Will exceed 10 min. That is, x 108.46 i). P(X 10) = P( 0.913 ) = P(Z1.69) = P(Z -1.69) = 0.0455 ( this value from standard normal table) Thus,the probability that haul time will be at least 10 min is 0.0455. ii). To determine the probability that haul time will exceed 10 min. That is, P(X > 10) = P(X 10) = 0.0455. Thus,the probability that haul time will exceed 10 min. Step2 of 6: b). To determine the probability that haul time will exceed 15 min. x 158.46 That is, P(X > 15) = P( > 0.913) = P(Z >7.1632) (from standard normal table) = 0. Thus,the probability that haul time will be at least 15 min is zero.