Unformatted text preview: manufacturer with the lower fraction of defective rivets, what is the probability that a rivet in the combined batch will be defective? Total probability P(D) = P(D  RR) P(RR) + P(D  RT) P(RT) = 1.38*10
4 (0.20) + 4.23*10
8 (0.80) = 2.77*10
5 c) 9 points Suppose the manufacturer with a higher fraction of defective rivets can improve the process to adjust the standard deviation, but not the mean breaking strength. What new standard deviation should this manufacturer set to have the same probability of defective rivets as the other manufacturer (and therefore capture more of the market)? In order for RR t o have same probability of defective rivets with RR, ZRR=ZRT
5.357=(5000
7000)/σnew σnew=373.33psi Comments on Grading This problem is a normal distribution problem. When you know that the distribution is normal you can find the cumulative...
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 Spring '08
 Clark
 the00, Professor00, λnew, λnewt

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