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I know that a standard normal distribution has mean 0, sd=1, skewness=0, and kurtosis=0. I can then transform a value and calculate which percentile it sits. 1.96 is in the top 97.5%

Can I perform a similar calculation when the skewness and kurtosis imply the distribution is not normal? ie. skewness=-2.2, kurtosis=3.1

If I were to deal with this data set, should I perform a transformation to try and make it normal, as I would like to predict where a value would fall in an almost normal distribution (ie with mean=0, sd=1, skewness=-2.2, and kurtosis = 3.1).

In short, I have a value of 1.96, will this fall in top 97.5% of the data? Given a value, can I predict where it would sit in the distribution?

Everything I read online explains how to calculate skewness and kurtosis but not find a way of predicting where a value would sit given these factors to consider as well

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    $\begingroup$ There is no rule that a normal distribution has mean 0 and SD 1. Nor does calculating percentiles depend in any sense whatsoever on a distribution being normal. You can do it regardless of the distribution, normal or otherwise. The question also seems to end in mid-air. Perhaps you should back up and tell us your overall goal. It may sometimes be a good idea to transform non-normal distributions but it is hard to advise on a good transformation given skewness and kurtosis alone and especially without knowing why you think a transformation is needed for your goals. $\endgroup$
    – Nick Cox
    Sep 6, 2017 at 11:32
  • $\begingroup$ 1. "1.96 is in the top 95%" ... for a standard normal 1.96 is at the 97.5 percentile. Not the 95th. 2. Your question seems to just stop mid-sentence. Please check $\endgroup$
    – Glen_b
    Sep 6, 2017 at 12:13
  • $\begingroup$ thanks fr pointing out the error of my question. It has been updated and corrected :) $\endgroup$
    – frank
    Sep 6, 2017 at 12:36
  • $\begingroup$ Beside the indicated duplicate also see stats.stackexchange.com/questions/56844/… ... a number of other posts touch on this. $\endgroup$
    – Glen_b
    Sep 7, 2017 at 4:45

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