document.write( "Question 1179066: The human resources department of a manufacturing firm found the average number of absences to be 3.4 per day. In the five days following unsuccessful labor negotiations there were a total of 39 absences, but the negotiators claimed that they were from ordinary illnesses. What is the probability of at least 39 absences in 5 days? Assume a Poisson distribution. \n" ); document.write( "
Algebra.Com's Answer #808595 by ewatrrr(24785)\"\" \"About 
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document.write( "Hi\r\n" );
document.write( "3.4/day  0r 17  per 5 days\r\n" );
document.write( "Poisson distribution:\r\n" );
document.write( "Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus\r\n" );
document.write( "P( x > 39) = 1 - P(x ≤ 38) = 1 poissoncdf( 17, 38) = 1 -0.999999 = basically 0 \r\n" );
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