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Chrome File Edit View History Bookmarks People Window Help Student P x Homewor xDo Home M Inbox -m XSection 5 x Michae . C ià Secure https://www.mathd.com/Student playerHomework.aspx7homeworkide 46.. BUS 352 TR700 Spring 2018 Homework: Chapter 6 Homework Save Score: 0 of 1 pt 4 of 5 (4 complete) Hw Score: 48.33%, 2.42 of 5 6.2.13-T EQuestion Help A particular manufacturing design requires a shaft with a diameter between 18.89 mm and 19.012 mm. The manufacturing process yields shafts with diameters normally distributed, with a mean of 19.005 mm and a standard deviation of 0.004 mm. Complete parts (a) through (c) a. For this process what is the proportion of shafts with a diameter between 18.89 mm and 19.00 mm? The proportion of shafts with diameter between 18.89 mm and 19.00 mm is 1056 Round to four decimal places as needed.) b. For this process what is the probability that a shaft is acceptable? The probability that a shaft is acceptable is (Round to four decimal places as needed.) Enter your answer in the answer box and then click Check Answer. Check Anwer 18 MacBoolk

Explanation / Answer

Ans:

Given that

mean=19.005

standard deviation=0.004

a)

z(18.89)=(18.89-19.005)/0.004=-28.75

z(19)=(19-19.005)/0.004=-1.25

P(18.89<x<19)=P(-28.75<z<-1.25)

=P(z<-1.25)-P(z<-28.75)

=0.1056-0

=0.1056

b)

z(19.012)=(19.012-19.005)/0.004=1.75

P(acceptable)=P(18.89<x<19.012)=P(-28.75<z<1.75)

=P(z<1.75)-P(z<-28.75)

=0.9599-0

=0.9599

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