{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

A question asks to find which manufacturer produces

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online