, C Secure I https://webwork.latec Fall201, 18-Math242-Adatorwovor/La lech Stat

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, C Secure I https://webwork.latec Fall201, 18-Math242-Adatorwovor/La lech Stat Ch4/4/?user-dpr010&effectiveUser-dpr0108; key=gEN4VegtSLUgsH4ytG6a3 WeBWorK LLOUSIANA TECH Logged in as dpr10 Lug Cul webwork fall201718-math242-adatorwovor latech stat ch4 4 MAIN MENU Courses Homework Sels LaTech Stat Ch4: Problem 4 Lalech Stat Ch4 Problem 4 User Settings Previous Problem Problem List Next Problem Crades (1 point) A soft drink botler purchases glass boitles from a vendor. The bottles are required to have an internal pressure of at least 150 pounds per square inch (ps). A prospective borte vendor claims that its production process yields bottles with a mean internal pressure of 157 psi and a standard deviation of psi. The bottier stikes an agreement with the vendor that penmits the bottler to sample trom the production process to verlty the clalm. the botter randomly selects 60 botles from the last 10000 produced, measures the internal pressure of each, and finds the mean pressure for the sarrmple to be 0.2 psi below the process mean cited by the vendor Problems Problerm 1 Problem 2 Problem 3 Problem Prablem5 a) Assuming that the vendar is correct in his claim, what is the probability of obtaining a sample mean this far or farther below the process mean? (b) If the standard deviation were 3 psi as claimed, but the mean was 155 psi, what is the probability of obtaining a sample mean of 156.8 psi or below? Problern 7 (c) It the process mean were 157 psl as clalmed, but the standard devlatlon was 1.8 psl, what is the probabllity ot obtalning a sample mean of 156.8 psl or below? Note: You can earn partial credit on this probiem. Preview My Answers Subrnit You have attempted this problem times. You have unlmited attempts remaining. :02 PM O Type here to search rP _20, 11/13/2017

Explanation / Answer

a) std error of mean =std deviation/(n)1/2 =3/(60)1/2 =0.3873

therefore P(X<156.8)=P(Z<(156.8-157)/0.3873)=P(Z<-0.5164)=0.3028

b)  P(X<156.8)=P(Z<(156.8-155)/0.3873)=P(Z<4.6476)=0.999998~ 1.0000

c)  std error of mean =std deviation/(n)1/2 =1.8/(60)1/2 =0.0.2224

P(X<156.8)=P(Z<(156.8-157)/0.3873)=P(Z<-0.8607)=0.1947

please revert

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