document.write( "Question 900743: The manager of a gym has determined that the length of time members spend training at the gym is normally distributed with a mean of 80 minutes and a standard deviation of 20 minutes. What percentage of members spend more than 2 hours training at the gym? \n" ); document.write( "
Algebra.Com's Answer #546102 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The manager of a gym has determined that the length of time members spend training at the gym is normally distributed with a mean of 80 minutes and a standard deviation of 20 minutes. What percentage of members spend more than 2 hours training at the gym? \n" ); document.write( "------------------ \n" ); document.write( "2 hrs = 120 minutes \n" ); document.write( "------------------------ \n" ); document.write( "z(120) = (120-80)/20 = 2 \n" ); document.write( "P(x > 120) = P(z > 2) = normalcdf(2,100) = 0.0228 = 2.28% \n" ); document.write( "---------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "--------------- \n" ); document.write( " |