# rpra5 - Engineering Risk Benefit Analysis 1.155 2.943 3.577...

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RPRA 5. Data Analysis 1 Engineering Risk Benefit Analysis 1.155, 2.943, 3.577, 6.938, 10.816, 13.621, 16.862, 22.82, ESD.72, ESD.721 RPRA 5. Data Analysis George E. Apostolakis Massachusetts Institute of Technology Spring 2007

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RPRA 5. Data Analysis 2 Statistical Inference Theoretical Model Evidence Failure distribution, Sample, e.g., e.g., {t 1 ,…,t n } How do we estimate from the evidence? How confident are we in this estimate? Two methods: Classical (frequentist) statistics Bayesian statistics t e ) t ( f λ λ = λ
RPRA 5. Data Analysis 3 Random Samples The observed values are independent and the underlying distribution is constant. Sample mean: Sample variance: = n 1 i t n 1 t = n 1 2 i 2 ) t t ( ) 1 n ( 1 s

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RPRA 5. Data Analysis 4 The Method of Moments: Exponential Distribution Set the theoretical moments equal to the sample moments and determine the values of the parameters of the theoretical distribution. Exponential distribution: Sample: {10.2, 54.0, 23.3, 41.2, 73.2, 28.0} hrs t 1 = λ 32 . 38 6 9 . 229 ) 28 2 . 73 2 . 41 3 . 23 54 2 . 10 ( 6 1 t = = + + + + + = 1 hr 026 . 0 32 . 38 1 = = λ ; hrs 32 . 38 MTTF =
RPRA 5. Data Analysis 5 The Method of Moments: Normal Distribution Sample: {5.5, 4.7, 6.7, 5.6, 5.7} μ = = = + + + + = 68 . 5 5 4 . 28 5 ) 7 . 5 6 . 5 7 . 6 7 . 4 7 . 5 ( x 032 . 2 ) 68 . 5 7 . 5 ( ... ) 68 . 5 5 . 5 ( ) x x ( 5 1 2 2 2 i = + + = σ = = = = 713 . 0 s 508 . 0 ) 1 5 ( 032 . 2 s 2

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RPRA 5. Data Analysis 6 The Method of Moments: Poisson Distribution Sample: {r events in t} Average number of events: r {3 eqs in 7 years} t r r t = λ = λ 1 yr 43 . 0 7 3 = = λ
RPRA 5. Data Analysis 7 The Method of Moments: Binomial Distribution Sample: {k 1s in n trials} Average number of 1s: k qn = k {3 failures to start in 17 tests} n k q = 176 . 0 17 3 q = =

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8 Censored Samples and the Exponential Distribution Complete sample: All n components fail. Censored sample: Sampling is terminated at time t 0 (with k failures observed) or when the r th failure occurs. Define the total operational time as: It can be shown that: Valid for the exponential distribution only (no memory). + = k
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rpra5 - Engineering Risk Benefit Analysis 1.155 2.943 3.577...

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