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I don't have raw data/numbers; instead I have the 10th percentile and the 90th percentile. From that, I need to calculate the 60th and 100th percentiles.
I'm struggling with how to accomplish that; does anyone have a suggestion? Thanks-
I don't have raw data/numbers; instead I have the 10th percentile and the 90th percentile. From that, I need to calculate the 60th and 100th percentiles.
I'm struggling with how to accomplish that; does anyone have a suggestion? Thanks-
Perhaps..
60th percentile
=(90th percentile less 10th percentile)/(0.9-0.1)*0.6
100th percentile
=(90th percentile less 10th percentile)/(0.9-0.1)*1
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Thank you for the reply. Doesn't your calculation use percentages rather than percentiles, though?
Its the same logic..
The percentile function is based on position of values in a range simalr to what a percentage is. Does the formula give you expected results?
Thanks again for the reply. I'm starting to wonder if it's even possible to get a true percentile from other percentiles . . . perhaps I can get 'close' with your suggested formula, but I'm doubting if it will produce a true percentile. Instead, it appears it may produce a percentage increase from the provided percentiles. What are your thoughts?
just fyi, baseing a percentage/percentile on other percentages/percentiles is mathematically incorrect. as you say, it will get you close, but it will not me (100%) accurate
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Two diverse examples suggest that you might get close (even spot on!) but it all depends on how scattered your underlying data is.
Keep refreshing the file (press F9) to see how the second set of data changes
in your example, you've divided by .8. Can you explain?
Thats the difference between 90 percentile and 10 percentile i.e. 0.9 less 0.1
that would be represented by =.8 rather than /.8
How is the data shaped? If it is a bell curve, and 10 is the 10th percentile, the 20th percentile will be much less than 20 (depending on the std. deviation)
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If you observe the formula its /.8 * 0.6Originally Posted by kelib
/0.8 gives you 100 percentile based on value of 80th percentile. And the *0.6 brings that back to 60th percentile.
Again these are approximates, as it all depends on how the original data is shaped (which is unknown!)
Unfortunately, I don't have raw data; I have only other percentiles.
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