Fast Service Truck Lines uses the Ford Super Duty F-750exclusively. Management m

Question

Fast Service Truck Lines uses the Ford Super Duty F-750exclusively. Management made a study of the maintenance costsand determined the number of miles traveled during the yearfollowed the normal distribution.  The mean of thedistribution was 60,000 miles and the standard deviation 2,000miles. A. What percent of the Ford Super Duty F-750s logged65,200 miles or more? B. What percent of the trucks logged more than 57, 060but less than 58,280 miles? C. What percent of the Fords traveled 62,000 miles orless during the year? D. Is it reasonable to conclude that any of the truckswere driven more than 70,000 miles? Explain. Fast Service Truck Lines uses the Ford Super Duty F-750exclusively. Management made a study of the maintenance costsand determined the number of miles traveled during the yearfollowed the normal distribution.  The mean of thedistribution was 60,000 miles and the standard deviation 2,000miles. A. What percent of the Ford Super Duty F-750s logged65,200 miles or more? B. What percent of the trucks logged more than 57, 060but less than 58,280 miles? C. What percent of the Fords traveled 62,000 miles orless during the year? D. Is it reasonable to conclude that any of the truckswere driven more than 70,000 miles? Explain.

Explanation / Answer

. 1.   To convert from the normal distribution to thestandard normal distribution: z = (x-)/ 2.   z = (x-60000)/(2000) . 3.   Part A: 4.   z = (65200-60000)/(2000) = 2.6 5.   from the standard normal tables:   2.6corresponds to 0.4953 6.   % with 65,200 miles or more = (100%)*P[z 2.6] = (100%)*(0.5 - 0.4953) = 0.47% . 7. Part B: 8.   z = (57060-60000)/(2000) = -1.47 9.   z = (58280-60000)/(2000) = -0.86 10.   from the standard normal tables:   1.47corresponds to 0.4292 11.   from the standard normal tables:   0.86corresponds to 0.3051 12.   % between 57,060 and 58,280 miles = (100%)*P[-1.47 z -0.86] = (100%)*(0.4292-0.3051) =12.41% . 13.   Part C: 14.   z = (62000-60000)/(2000) = 1.00 15.   from the standard normal tables:   1.00corresponds to 0.3413 16.   % with 62,000 miles or less = (100%)*P[z 1.0] = (100%)*(0.5 + 0.3413) = 84.13% . 17.   Part D: 18.   It is unreasonable to conclude that any of thetrucks were driven more than 70,000 miles. 19.   z = (70000-60000)/2000 = 5.00 20.   from the standard normal tables:   5.00corresponds to 0.4999997 21. % with more than 70,000 miles = (100%)*P[ z 5.00] = (100%)*(0.500 - 0.4999997) = 0.00003% .

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