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Math Help - finding percent of people that fall bellow a certain value in a list of data

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    [RESOLVED]finding percent of people that fall bellow a certain value in a list of dat

    I have a file containing the times taken to do something. The data is fits a normal distribution. I am trying to find the percent of people that take longer than 1800 seconds. So I find the z-score z=\frac{X-\mu}{\sigma}=\frac{1800-1369.53}{257.504}\approx 1.67 By symmetry I take -1.67 and from a table I look up -1.67 and find 0.0475 which gives me 4.75%
    Am I good up to this point?

    If I haven't made a mistake yet, what's confusing me is that my statistical analysis program (StatsGraphics Centurion) makes me think it should be about 5.26%. This is because the program created a class with the interval (1815,1908] which has the relative frequency of 0.0526. Shouldn't this number be smaller than 0.0475 because the interval (1815,1908] is smaller than [1800,1900]. Important: the largest value is 1869 so that's why I took the one class interval (1815, 1908]
    Last edited by superdude; June 7th 2010 at 11:22 PM. Reason: resolved
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