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)\"\" \"About 
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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
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