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RELIABILITY DEMONSTRATION TESTING Example 2 - Time Terminated Test without Replacement: Twenty items are placed on test for 100 hours with seven failures occuring at the 10, 16,17, 25, 31, 46 and 65 hour points. Determine the one-sided lower 90% confidence interval. Solution: The total number of part hours accumulated is: 10 + 16+17 +25+ 31 +46+ 65+ (13 non-failed items) (100 hours) = 1510 hrs. The MTBF is 1510 hours/7 failures = 216 hrs. From Table 5-4, Note 3 applies, d = 2(7+1) = 16. The factor from the table is .0848 for the 90% one-sided lower limit. Therefore, we are 90% confident that the MTBF is greater than (,0848)(1510 hours) = 128 hours. 4.0 Poisson Distribution: The Poisson distribution is useful in calculating the probability that a certain number of failures will occur over a certain length of time for systems exhibiting exponential failure distributions (e.g., non-redundant electronic systems). The Poisson model can be stated as follows: PM-^ where P(r) = probability of exactly r failures occurring X = the true failure rate per hour (i.e., the failure rate which would be exhibited over an infinite period) t = the test time r = the number of failure occurrences e = 2.71828 I = factorial symbol (e.g., 4! = 4x3x2x 1 = 24, 0! = 1,11 = 1) The probability of exactly 0 failures results in the exponential form of this distribution which is used to calculate the probability of success for a given period of time (i.e., P(0) = e"^). The probability of more than one failure occurring is the sum of the probabilities of individual failures occurring. For example, the probability of two or less failures occurring is P(0) + P(1) + P(2). Table 5-5 is a tabulation of exact probabilities used to find the probability of an exact number of failures occurring. Table 5-6 is a tabulation of cumulative probabilities used to find the probability of a specific number of failures, or less, occurring. A-44 ROME LABORATORY RELIABILITY ENGINEER'S TOOLKIT

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