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The average middle-distance runner at a local high school runs the mile in 4.5 minutes, with a standard deviation of 0.3 minute. The percentage of a runners that will run the mile in less than 4 minutes is %.

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  1. .5/.3 = 1.67 standard deviations below mean
    Look at table for normal distribution
    My quick table is very rough
    for z = -1.6 F(z) = .055
    for z = -1.7 F(z) = .045
    so about F(1.67) = about .05 or 5 %

    By the way, I do not believe it. Suspect sigma much lower than 0.3 min or possibly not a normal distribution at all.

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