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Unformatted text preview: 1 − F T (30) = 31 − 30 60 60 − 30 60 = 1 30 , and P [ A  B ] = P [ AB ] P [ B ] = F T (31) − F T (30) F T (31) = 31 − 30 60 31 60 = 1 31 . Correction: Please substitute λ for μ in this problem (2 places), as λ is conventionally used for the rate parameter . Also assume λ is measured in hours. Ans: The pdf of the failure time X is given from Section 2.7 in general as f X ( t ) = α ( t ) exp( − Z t α ( τ ) dτ ) . In this case the rate is constant and α ( t ) = λ, thus f X ( t ) = λ exp( − λt ) . If A = { failure in 100 hours or less } , then P [ A ] = P [ X ≤ 100] = Z 100 λe − λt dt = 1 − e − 100 λ . In order for this probability to be less than or equal to . 05 , we need e − 100 λ ≥ . 95 or λ ≤ 1 100 ln 1 . 95 We get 5 . 13 × 10 − 4 2...
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This note was uploaded on 04/15/2010 for the course ECSE 4500 taught by Professor Woods during the Spring '08 term at Rensselaer Polytechnic Institute.
 Spring '08
 WOODS

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