document.write( "Question 69967: Among U.S. households, 24% have telephone answering machines. If a telemarketing campaign involves 2,500 households, find the probability that more than 650 have answering machines. \r
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document.write( "I get .2389 \n" );
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Algebra.Com's Answer #49867 by stanbon(75887) ![]() You can put this solution on YOUR website! I get 0.0096 or less than 1%. \n" ); document.write( "Converting 650 to a z-score I get: \n" ); document.write( "z(650)=[650-600]/sqrt(2500*0.24*0.76)=2.34 \n" ); document.write( "which means that 650 is 2.34 standard \n" ); document.write( "deviations to the right of the mean which \n" ); document.write( "is 600. \n" ); document.write( "The probability of the number being greater \n" ); document.write( "than 650 is very small just as the probability \n" ); document.write( "of z being greater than 2.34 is very small. \n" ); document.write( "Hope this helps. \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |