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I am trying to recreate the stats for 'failure frequency' - in this case the 95 percentile - shown in the attached images below.

For one set of failure data there are 27 failures over a cumulative 42889 years of operation. the Expected Value is 6.3e-4 and the shown 95% upper limit is 8.4e-4 in terms of failure frequency.

I get close if I model an inverse Poisson distribution (for the 95% CDF). But I cannot reproduce the results using what looks like the Gamma distribution PDF proposed.

See linked images:

Image from Dutch Handbook Appendix A (translated to English)1

Image from Dutch Handbook Appendix A (translated to English)2

Any feedback or suggestions appreciated.

akarich73
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1 Answers1

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If you have a Poisson process with a rate parameter of $8.411623\times 10^{-4}$ per year (so rounding to 8.4e-4)

then the probability you see at least $27$ occurrences in $42889$ years can be found using the Gamma distribution cumulative distribution function to be $95\%$

In R

> pgamma(42889, 27, 8.411623e-4)
[1] 0.95
Henry
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  • Many Thx. As a follow up question, to me the logic of using this approach to establish the 95%ile seems not quite correct as it implies a fractional number of failures for the time period. E.g. for the above rate parameter I calculate 36.08 failures. Rather I had assumed that the best way to calculate the 95%ile was to use the Poisson CDF based on a mean (or expected value) of 27, which results in 36 for the 95%ile, and estimate the failure rate from 36 over the same time period. Similar but not same result due to the difference between the discrete vs continuous distribution assumption. – akarich73 Nov 03 '21 at 00:40
  • @akarich73 You are saying $42889 \times 8.411623\times 10^{-4} \approx 36.07661$ is not an integer. But the parameter of a Poisson distribution does not need to be an integer and with that parameter the probability of getting strictly fewer than $27$ observations is $0.05$: with R `ppois(26,36.07661)` gives `0.04999997`. – Henry Nov 03 '21 at 00:41
  • If instead you used `qpois(0.95,27)` to get `36` you would hit the problem that `ppois(36,27)` gives `0.9612017` which is too high. You would still have the Poisson-Gamma equivalence: `ppois(26,36)` gives `0.0512685` while `pgamma(42889,27,36/42889)` gives `0.9487315` and these add up to $1$ but would be slightly the wrong side of $95\%$. It is only a slight change and $\frac{36}{42889} \approx 8.393761\times 10^{-4}$ which also rounds to $8.4\times 10^{-4}$. But this is the wrong way of looking at this: if you saw $0$ failures, then $0$ would also be your top figure, and that is wrong. – Henry Nov 03 '21 at 00:54