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21 July, 17:51

Suppose abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. assume the travel times are normally distributed with a standard deviation of 10.3 min. determine the travel time x such that 22.96% of the 60 days have a travel time that is at least x.

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  1. 21 July, 20:27
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    Given that:

    mean,μ=35.6 min

    std deviation,σ=10.3 min

    we are required to find the value of x such that 22.96% of the 60 days have a travel time that is at least x.

    using z-table, the z-score that will give us 0.2296 is:-1.99

    therefore:

    z-score is given by:

    (x-μ) / σ

    hence:

    -1.99 = (x-35.6) / 10.3

    -20.497=x-35.6

    x=35.6-20.497

    x=15.103
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