**Re: Confidence and Reliability - Reference to a 90/95 confidence and reliability leve**

[FONT=arial,helvetica,sans-serif][SIZE=-1]I could not locate the table to which you refer, but I did locate the formula that the table would have been based upon.

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Bayes Success-Run Theorem (based on the binomial distribution).

R = (1-C) ^ (1/n) ;

R = Reliability (or probability of success) ,

C = confidence level ,

n = sample size. For "0" failures allowed on test

Given R=.95, C = .90 , solve for n

Ln(R)=(1/n)*Ln(1-C)

-.05129=(1/n)*(-2.30259)

n= 44.89 or 45.

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