document.write( "Question 438819: The manager of a local gym has determined that the length of time members spend at the gym is a normally distributed variable with a mean of 80 minutes and a standard deviation of 20 minutes.\r
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document.write( "What proportion of members spend between 80 and 100 minutes at the gym? Use the standard normal distribution tables on p564 of your Wegner textbook. \n" );
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Algebra.Com's Answer #303356 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The manager of a local gym has determined that the length of time members spend at the gym is a normally distributed variable with a mean of 80 minutes and a standard deviation of 20 minutes. \n" ); document.write( "What proportion of members spend between 80 and 100 minutes at the gym? \n" ); document.write( "--------------- \n" ); document.write( "z(80) = 0 \n" ); document.write( "z(100) = (100-80)/20 = 1 \n" ); document.write( "--- \n" ); document.write( "P(80<= x <= 100) = P(0<= z <=1) = 0.3413 \n" ); document.write( "============================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |