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
\n" ); document.write( "\n" ); 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.
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Algebra.Com's Answer #303356 by stanbon(75887)\"\" \"About 
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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?
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\n" ); document.write( "z(80) = 0
\n" ); document.write( "z(100) = (100-80)/20 = 1
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\n" ); document.write( "P(80<= x <= 100) = P(0<= z <=1) = 0.3413
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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