Vehicle speed on a particular bridge in China can k modeled as normally distributed ?("Fatigue Reliability Assessment for Long-Span Bridges under Combined Dynamic Loads from Winds and Vehicles," J. of Bridge Engr., 2013: 735-747). a. If 5% of all vehicles travel less than 39.12 milt and • 10% travel more than 73.24 m/h, what are the mean and standard deviation of vehicle speed? [Note: The resulting values should agree with those given in the cited article.) b. What is the probability that a randomly selected vehicle's speed is between 50 and 65 m/h? c. What is the probability that a randomly selected vehicle's speed exceeds the speed limit of 70 m/h?

Solution : Step 1 : Let X be the vehicle speed, which follows normal distribution. Here it is given that 5% of all vehicles travel less than 39.12m/h and 10% travel more than 73.24 m/h. We have to find the mean and standard deviation. It is given that P(X<39.12)= 0.05 P(X>73.24) =.1 Consider, P(X<39.12)= 0.05 P( X < 39.12 ) = 0.05 39.12 P( Z< ) =0.05 From the standard normal table we will get P(Z< -1.645) =0.05 So 39.12 = -1.645 = 39.12+1.645 ……. (1) Similarly Consider P(X>73.24) =.01 X 73.24 P( < ) = 0.1 73.24 P( Z> ) =0.1 From the standard normal table we will get P(Z> -1.28) =0.1 So 73.24 = -1.28 = 73.24+1.28 ……… (2) By solving equation (1) and (2) we can find the valueand . 39.12+1.645 = 73.24+1.28 34.12= 0.36 = 94.78 Apply this to any of the equation you will get the value of = 194.55 So Mean = 194.55 Standard deviation = 94.78 Step 2: b) We have to find the probability that a randomly selected vehicle's speed is between 50 and 65 m/h. 50 65 P(50X65) = P( Z ) 50194.55 65194.55 P( 94.78 Z 94.78 ) = P (-1.52 z -1.36) From the standard normal table we can find F(-1.52 ) =P(X 1.52) = 0.0643 F(-1.36) =P(X 1.36) = 0.0869 P(50X65) = F(-1.36) - F(-1.52 ) = 0.0226