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I've have a script that creates Weibull distributions with a static shape parameter and a variable scale parameter, the scale parameter changing to yield a distribution that meets a specified mean while maintaining the shape parameter.

I have calculated that the standard deviation for my distributions always equals 2*sqrt(5). This is really small for what I am doing: I am using the distribution to model webpage read-time based on a known average (calculated from the number of words on the page). Times, including standard deviation, are in seconds. 2*sqrt(5) doesn't seem like a whole lot of wiggle room for read-time variation when trying to emulate an "average" web browsing user.

How can I approximate how many standard deviations "wiggle room" is appropriate for my application? I was surprised, since I thought that "standard" deviation implied that a majority of samples would be within one standard deviation of the mean. I can't believe it's so tight around my average. It doesn't seem like a "standard" deviation is "standard" at all, since in other distributions I've looked at, a large percentage of samples fall within one of them. This assumes I didn't make a calculation error -- I'm not finding an error yet at least.

Thanks

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  • $\begingroup$ The shape parameter I am using is 0.5 $\endgroup$
    – Ken - Enough about Monica
    Jun 5, 2014 at 18:42
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    $\begingroup$ It is possible for all the data to be one or more SDs from their mean. However, at least three-quarters of the data will lie within two SDs of the mean. If that's not the case, then you have made a computational error. $\endgroup$
    – whuber ♦
    Jun 5, 2014 at 19:10
  • $\begingroup$ @whuber - thank you for that bit of information. I don't have a data set myself, I'm trying to very roughly emulate someone else's results to just (again very roughly) approximate a process. I guess I will try to play with the distributions until 3/4 of the output tends to lie within two standard deviations $\endgroup$
    – Ken - Enough about Monica
    Jun 5, 2014 at 19:15

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Perhaps this paper,"A Simple Normal Approximation for Weibull Distribution with Application to Estimation of Upper Prediction Limit" by H. V. Kulkarni and S. K., will assist in answering your questions with respect to the UPL.

Link: http://www.hindawi.com/journals/jps/2011/863274/

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