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
\n" ); document.write( "
\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( "

Algebra.Com's Answer #207386 by stanbon(75887)\"\" \"About 
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( "
\n" );