I have a database of 10-star movie ratings (similar to IMDB). For each movie the raw data is a distribution of votes from 0-star all the way to 10-star, and I have also computed the mean and standard deviation of the votes for each movie.

Now consider the set of all the computed standard deviations. Is there a known statistical distribution that this set is expected to follow in the general case? And if so, how do I compute its parameters?

Addendum: We cannot assume anything (namely normality) about the distribution of the movie scores themselves.

  • $\begingroup$ I don't think that there is a general analytical form for the distribution of the sd of an unknown distribution, but I could be wrong. Depending on what you need this for, you could try resampling (with replacement), calculating the sd's of your resamples, and plotting that distribution. $\endgroup$ – generic_user Apr 25 '15 at 16:44

Are the movie scores themselves normally distributed?

Given a set of samples with a normal distribution, the distribution of the standard deviations would be a scaled chi-squared distribution. See the Wolfram Math page on the standard deviation distribution.

Now if the original samples (movie scores) do not follow a normal distribution. it would make more sense to talk about the distribution of variances. And that would depend on what distribution the scores best follow.

  • $\begingroup$ Movie ratings can't really be normal. I addition, the chi-squared distribution is more directly connected to the variance than the SD. $\endgroup$ – gung Apr 25 '15 at 14:59
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    $\begingroup$ Agreed on both points. Seems like the original question is whether the standard deviation follows a known (same) distribution no matter the distribution of the original samples. Clearly it does not. I'm not sure how to answer the question better without it being more detailed. $\endgroup$ – paisanco Apr 25 '15 at 15:04
  • $\begingroup$ I've edited the question to clarify: we cannot assume anything about the distribution of the movie scores. $\endgroup$ – Jon Smark Apr 25 '15 at 16:01

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