document.write( "Question 285981: in Rivendell memorial hospital the time to complete surgery in a routine tubal ligation without complication is normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes. The next procedure has been scheduled in the same operating room 60 minutes after the beginning of a tubal ligation procedure.\r
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document.write( "Allowing 20 minutes to vacate and prepare the operating room between procedures, what is the probability that the next procedure will have to be delayed? \n" );
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Algebra.Com's Answer #207386 by stanbon(75887) ![]() You can put this solution on YOUR website! in Rivendell memorial hospital the time to complete surgery in a routine tubal ligation without complication is normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes. The next procedure has been scheduled in the same operating room 60 minutes after the beginning of a tubal ligation procedure. \r \n" ); document.write( "\n" ); document.write( "Allowing 20 minutes to vacate and prepare the operating room between procedures, what is the probability that the next procedure will have to be delayed? \n" ); document.write( "-------------------- \n" ); document.write( " If tubal ligation takes > 40 min (procedure delayed) \n" ); document.write( "------------------------------------ \n" ); document.write( "z(40) = (40-30)/8 = 5/4 \n" ); document.write( "------ \n" ); document.write( "P(procedure delayed) = P(x>40) = P(z> 5/4) = normalcdf(5/4,100) = 0.1056 \n" ); document.write( "================================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |