09 304996 7991 30095012 013 03 043 part

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Unformatted text preview: lerance will this provide? Assembly tolerance: (80.09 – (30+49.96)) – (79.91-(30.09+50.12)) = 0.13 – (-0.3) = 0.43 PART A b. (1.5 pts) What type of assembly fit is this assembly designed for? PART B Transition fit c. (5 pts) Based on the statistical tolerancing method, what is the probability that some of the assemblies will experience interference? Show the necessary steps for full credit. µA = 80; µB = 30.045; µC = 50.04; µAssembly Clearance = - 0.085; σA = 0.18/6 = 0.03; σB = 0.09/6 = 0.015; σC = 0.16/6 = 0.0267; σAssembly Clearance = 0.043 P(Assembly Clearance < 0) = P(z < (0.085/0.043 = 1.977)) = 0.976 or 97.6% 2 PART C...
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