# If I have 25th, 50th and 75th percentile, can I estimate 99th percentile?

Supposing I have only:

• 25th percentile: 0.56
• 50th percentile: 0.70
• 75th percentile: 1.10

I don’t know other data (sample size, mean etc. ). Only those three value.

And assuming data is normal distributed (to make it simple), can I estimate values for other percentile?

What if they are not normally distributed?

• You could, but the arguments might be tenous. Because you know the median, you ostensibly know the mean of the distribution as well. Thus, you can estimate the standard deviation from either one of the other percentiles. That is all you need to determine the 99th percentile, however the median is not as efficient (at least, I think the property I'm thinking of is efficient) as the sample mean for estimate the mean of a normal distribution. Apr 25, 2021 at 14:19
• Since you refer to data, these percentiles are equivalent to order statistics (assuming you know the size of the dataset). This makes stats.stackexchange.com/questions/207403 and stats.stackexchange.com/questions/130156 directly relevant. The methods in both threads extend in an obvious manner to any parametric distributional family (but is useful primarily with two or fewer parameters). If you don't have the sample size, you cannot determine levels of uncertainty, making your objective nearly hopeless.
– whuber
Apr 25, 2021 at 14:27
• Sorry, accurate estimate of 99th percentile does not seem promising to me. Assuming normal "to make it simple" does not necessarily make it correct. if normal, median .70 does estimate mean as .70. However, Avg of 1st & 3rd quartiles estimates mean as .83; larger distance btw med and Q3 than btw med and Q1 may imply right-skewness, making estimation of 99th percentile even more problematic than usual. Apr 25, 2021 at 14:44
• Perhaps you can clarify the goal of this request...it seems this might be something of a precursor to a power calculation. Apr 25, 2021 at 17:02
• @DemetriPananos May you write me the formula? Apr 25, 2021 at 19:17