document.write( "Question 1128023: About 12.5% of restaurant bills are incorrect. If 200 bills are selected at random, find the probability that at least 22 will contain an error. \n" ); document.write( "
Algebra.Com's Answer #744537 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! The mean of this is 200*.125=25 \n" ); document.write( "the variance is np(1-p)=25*.875=21.875 \n" ); document.write( "sd is sqrt (variance)=4.68 \n" ); document.write( "use normal approximation with mean 25 and sd 4.68 \n" ); document.write( "continuity correction, use 21.5 to cover at least 22. \n" ); document.write( "z>=(21.5-25)/4.68 \n" ); document.write( "z>-0.75 for probability 0.7734 \n" ); document.write( " \n" ); document.write( " |