document.write( "Question 1164283: The human resources division in a large government department has shown that illness is given
\n" ); document.write( "as the reason for absenteeism from work in 80% of absenteeism cases. Suppose a group of 100
\n" ); document.write( "employees who had been absent are randomly selected. Find the approximate probability that
\n" ); document.write( "at most 86 of them gave illness as their reason for absence.
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Algebra.Com's Answer #788636 by Boreal(15235)\"\" \"About 
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use normal approximation
\n" ); document.write( "mean is np=100*0.80=80
\n" ); document.write( "variance is np(1-p)=80*0.2=16
\n" ); document.write( "sd is sqrt(16)=4
\n" ); document.write( "the approximate probability would be z<=(x+0.5-mean)/sd=(86.5-80)/4=1.625. The +0.5 is the continuity correction factor.
\n" ); document.write( "that probability is 0.9479\r
\n" ); document.write( "\n" ); document.write( "The exact probability is 0.9531.
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