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I have a problem related to a formula and I need help from all of you. It may use some statistics formula to solve the problem. Your help would be appreciated.

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We have following values (Percentile and respected values)

3% :——————: 7.7999999999999998

15% :——————: 8.5999999999999996

50% :——————: 9.5999999999999996

85% :——————: 10.800000000000001

97% :——————: 11.800000000000001

Now we need to find the percentile of given value with respect to above values

Suppose Given Value -> 8.2 Then what will be the percentile of 8.5 (It should be less than 15% and greater than 3%)?

Then what will be the percentile of 10.0 (It should be less than 85% and greater than 50%)?

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  • $\begingroup$ As long as you do not know anything more about the distribution of the data, you may interpolate linearly to get estimates for other pecerntiles. $\endgroup$ – trancelocation Sep 27 '18 at 6:17
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The intervals you suggest are reasonable. Unless you have additional information or want to make assumptions, they are the best you can do.

I suppose the information you have is from a sample and you're asking for approximate sample percentiles. Are those supposed to serve as estimates of population percentiles? I assume you don't have access to the entire sample.

You could use the information given in your question to make a plot of the estimated CDF (as below). If it looks like the CDF of a known distribution, perhaps try to fit that. Otherwise, use linear interpolation.

enter image description here

In this case, the population may be that of a normal distribution. If so, you can estimate the mean as about 9.65 (median is 9.6, but other quantiles suggest a little higher) and from the quantiles you can deduce that the SD is about 1.03 or 1.04. Superimposing the CDF of $\mathsf{Norm}(\mu=9.65,\,\sigma=1.035)$ [heavy blue curve] on the estimated CDF shows a pretty good match.

diff(qnorm(c(.15,.85)))
[1] 2.072867            # spread btw 15th & 85th pctl for NORM(0,1)
2.027/(10.7-8.7)
[1] 1.0135              # rough est of pop SD based on data

enter image description here

If you believe data may have come from this normal distribution, then the required percentile estimates are 8th percentile for 8.2 and 63rd percentile for 10. [Computations in R below. In R, pnorm is a normal CDF.]

pnorm(8.2,9.65,1.035)
[1] 0.08061209          # About 8% of sample below 8.2
pnorm(10,9.65,1.035)
[1] 0.6323803           # About 63% of sample below 10

These estimates are in the intervals you suggested. Perhaps they are more useful than what you would have gotten by linear interpolation using the estimated (broken line) CDF. I will leave the linear interpolation to you. Results should be about the same.

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