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    Continuous Probability Distributions: Normal Distribution

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    Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made a study of the maintenance costs and determined the number of miles traveled during the year followed the normal distribution. The mean of the distribution was 60,000 miles and the standard deviation 2,000 miles.

    a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more?
    b. What percent of the trucks logged more than 57,060 but less than 58,280 miles?
    c. What percent of the Fords traveled 62,000 miles or less during the year?
    d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles? Explain.

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    Solution Preview

    There is an attache file as well which may make this easier to read.

    a. What percent of the Ford Super Duty F-750's logged 65,200 miles or more?

    Mean= M = 60,000 miles
    Standard deviation = s = 2,000 miles
    x = 65,200 miles
    z = (x-M )/s = 2.6 = (65200-60000)/2000
    Cumulative Probability corresponding to z = 2.6 is = 0.9953
    Or, Probability corresponding to x<65,200 miles is Prob(Z) = 0.9953 or = 99.53%
    Therefore probability corresponding to x>65,200 miles is 1-Prob(Z) = 0.0047 = 1-0.9953 or = 0.47%

    Answer: 0.0047 or 0.47%
    ...

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