document.write( "Question 321228: The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 581 MPa with a standard deviation of 13 MPa.\r
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document.write( "What is the probability that a randomly chosen sample of glass will break at, (Round your answers to 4 decimal places.)
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document.write( "(a) less than 581 MPa?
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document.write( "(b) more than 592 Mpa?
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document.write( "(c) less than 602 MPa? \n" );
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Algebra.Com's Answer #230034 by stanbon(75887) ![]() You can put this solution on YOUR website! The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 581 MPa with a standard deviation of 13 MPa. \n" ); document.write( "What is the probability that a randomly chosen sample of glass will break at, (Round your answers to 4 decimal places.)\r \n" ); document.write( "\n" ); document.write( "(a) less than 581 MPa? \n" ); document.write( "z(581) = (581-581)/13 = 0 \n" ); document.write( "P(x< 581) = P(z< 0)= 0.5000 \n" ); document.write( "--------------------------------- \n" ); document.write( " \n" ); document.write( "(b) more than 592 Mpa \n" ); document.write( "z(592) = (592-581)/13 = 11/13 \n" ); document.write( "-- \n" ); document.write( "P(x > 592) = P(z > 11/13) = normalcdf(11/13,1000) = 0.1987 \n" ); document.write( "--------------------------------- \n" ); document.write( " \n" ); document.write( "(c) less than 602 MPa? \n" ); document.write( "= P(z < z(602)) = normalcdf(0,602,581,13) = 0.9469 \n" ); document.write( "----- \n" ); document.write( "Comment: I have used several methods to find the 3 answers. \n" ); document.write( "Let me know if this is confusing. \n" ); document.write( "The normalcdf is a function on a TI-84 calculator. \n" ); document.write( "============================ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |