# Confidence Interval within min and max values

I have set of numbers and know it's size, mean, standard deviation, minimum and maximum. When I calculate regular confidence interval for mean I get something like this (-20;50). But the source value can only be positive, so it would look bad on a graph. I'd like to find confidence interval within given minimum and maximum values. It's just max values have random peaks and I want to smooth it with CI. Is there some way to do it?

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There is almost certainly a better way to do what you're doing, but I would want to know more about what you're doing before trying to say what that is. In particular, how do you know that the values can only be positive? Could it be that you're finding a confidence interval appropriate to a sample from a normal distribution? Observations from a normal distribution will not always be positive, so that's not a good way to proceed unless you have a standard deviation that's so small compared to the sizes of the observations that a normal approximation is appropriate. – Michael Hardy Oct 7 '11 at 22:15
Values are pings. They can't be negative. Standard deviation isn't small at all. Basically I need a good way to show statistics of pings on a graph for a period of time. So what's a better way you're thinking about? – Nickname Oct 7 '11 at 22:41
What do you mean by "pings"? (That the standard deviation is large could be deduced from what you told us initially.) – Michael Hardy Oct 7 '11 at 23:21
Ping is the round-trip time for the signal sent from one computer to another. – Nickname Oct 7 '11 at 23:46
One might want to see the data before being sure how to proceed. You might consider whether the logarithms of the times look like a sample from a normal distribution. If so, find a confidence interval by the usual methods you'd use for samples from a normal distribution, then take antilogarithms to get a confidence interval for what you started with. This would be for the median of the population, not for the mean, since lognormal distributions are like that. – Michael Hardy Oct 8 '11 at 0:21