document.write( "Question 1206804: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 110.3 cm and a standard deviation of 1.7 cm. For shipment, 7 steel rods are bundled together.\r
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document.write( "Find the probability that the average length of a randomly selected bundle of steel rods is more than 112 cm. \n" );
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Algebra.Com's Answer #844471 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "mu = 110.3 = population mean \n" ); document.write( "sigma = 1.7 = population standard deviation\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "n = 7 = sample size \n" ); document.write( "xbar = sample mean\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Compute the z score when xbar = 112 \n" ); document.write( "z = (xbar - mu)/( sigma/sqrt(n) ) \n" ); document.write( "z = (112 - 110.3)/( 1.7/sqrt(7) ) \n" ); document.write( "z = 2.65 approximately when rounding to 2 decimal places\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then use a Z table to determine that \n" ); document.write( "P(Z < 2.65) = 0.99598 \n" ); document.write( "which leads to \n" ); document.write( "P(Z > 2.65) = 1-P(Z < 2.65) \n" ); document.write( "P(Z > 2.65) = 1-0.99598 \n" ); document.write( "P(Z > 2.65) = 0.00402 \n" ); document.write( "This value is approximate.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "To get more accuracy you can use a TI84 or similar stats calculator. \n" ); document.write( " \n" ); document.write( " |