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.


  • $\begingroup$ The shape parameter I am using is 0.5 $\endgroup$
    – HorseHair
    Jun 5 '14 at 18:42
  • 3
    $\begingroup$ 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 '14 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 $\endgroup$
    – HorseHair
    Jun 5 '14 at 19:15

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/


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.