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August 2

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Get Mean/Median of Standard Distribution from Percent

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Suppose that I have a normal distribtion of data. It is numbers from 1 to some huge number. The distribution is a frequency. So, to make this simple, assume it goes from 1 to 100 and know the mean/median is 50 (it is normal, mean=median). With this, I know what percent of values are between 1 and 30. But, what if I don't know the mean/median and I don't know the maximum number. Instead, I know the percent of values that go from 1 to 30. So, I know 10% of the values are from 1 to 30. What is the mean/median? It seems to me that it is easy to calculate, but I don't know how standard deviations work well enough to convert this to a formula where I enter the percent of values from 1 to 30 and I get a median value. 97.82.165.112 (talk) 15:20, 2 August 2022 (UTC)[reply]

A normal distribution is determined by two parameters: the mean and the variance To estimate these two parameters given data derived from a sample, the estimation procedure needs at least two values as input. If all you know is that the fraction of values in a given range amounts to 10%, you have only one value, which is not enough. So the problem is underdetermined. For example, about 10% of the values drawn from the normal distribution with and lie in the range from 1 to 30. But this is also the case when and  --Lambiam 16:09, 2 August 2022 (UTC)[reply]
Correct. If I know the standard deviation, I would know how to caculate it. I do know the total number of values, which is how I can make the claim that the number of values that range from 1 to 30 is 10% of the total values. I simply had no idea how to use that knowledge to figure out the mean (or median) without somehow also know the standard deviation. I assumed that there was some method because it is used for calculating half-lifes. I know 100 events will occur. I know that it took 3 days for 20 events to occur. The half life (when 50 events occur) is X. Somehow, scientists calculate X. 97.82.165.112 (talk) 16:25, 2 August 2022 (UTC)[reply]
An exponential distribution is determined by a single parameter, so if you know that the events follow an exponential distribution, one value suffices to estimate it. For the example, After 6 days about 20% of the remaining 80 events will have occurred too, bringing the total to 20 + 16 = 36 events. Three days later the expected number of events equals 36 + 0.20 × 64 = 48.8 events. The best estimate for event 50 is after 9.3 days. Without knowing more about the process that gives rise to the events, though, it is impossible to know the family of distributions. If the events are the hatching of eggs of some unknown species, all bets are off.  --Lambiam 17:21, 2 August 2022 (UTC)[reply]
I see. I happen to know that what I'm working with is very close to a normal distribution, not an exponential distribution. So, I'm confusing one process for one distribution with a different distribution. The good thing is that there is a reason why I couldn't figure out how to calculate it. 97.82.165.112 (talk) 18:15, 2 August 2022 (UTC)[reply]
It sounds like the actual distribution you want to use is a binomial distribution at p = 1/2, where the number n of boxes is the unknown parameter. 66.44.49.56 (talk) 14:29, 8 August 2022 (UTC)[reply]