Value based on a supplied standard-deviation

Sorry if this is a very basic question, my statistics knowledge is very low!

I've got a dataset which is basically a set of greyscale values in a greyscale texture (ie. 0-255). What I want to do is take the mean value, and find out what the values would be at a specified standard-deviation to the left and right of the mean. Is this possible? I presume I'd have to presume a normal distribution?

I hope that makes sense!

• I'm missing something: you have your dataset so you compute its mean m and standard deviation s. The value at z standard deviations from the mean therefore equals m + zs for any *z you like. Where is the problem? – whuber Oct 23 '10 at 21:19
• @whuber Perhaps, the problem is that we cannot assume that the distribution is normal? It seems to me that the values are bounded from below (at 0) and from above (at 255). The normal may not make sense unless the standard deviation is very small and the mean far from the the boundary points. @Dan Could you please clarify? – user28 Oct 23 '10 at 21:51
• @Srikant That's a reasonable guess. But nothing in the current statement requires assuming any kind of distribution at all. Perhaps the question really is about how to interpret values of the form m + z*s? Regardless, although the question makes sense, it requires some amplification in order to elicit an answer that would be genuinely helpful. – whuber Oct 23 '10 at 22:26
• In most cases the values will be nearer the high range (ie. nearer 255). The SD can also vary quite a lot unfortunately. As I mentioned in my post, I'm completely new to stats, so the formula you gave whuber might be what I want. I'll need to play around with various datasets to see if it works or not. – Dan Oct 24 '10 at 9:37
• @Dan It would also help if you could tell us why you are doing this. As an aside: doing something like m + z*s may very well end with a upper bound of greater than 255 (depending on your values for m, z and s). Some sense of what you want to achieve (i.e., your goals) would help. – user28 Oct 25 '10 at 15:53