document.write( "Question 1202116: A grocery store counts the number of customers who arrive during an hour. The average over a year is 19 customers per hour. Assume the arrival of customers follows a Poisson distribution. (It usually does.)
\n" ); document.write( "Find the probability that at least one customer arrives in a particular one minute period. Round your answer to 3 decimals.
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\n" ); document.write( "Find the probability that at least two customers arrive in a particular 5 minute period. Round your answer to four decimals.
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Algebra.Com's Answer #837656 by Boreal(15235) $\"\"$ $\"About$

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19 per hour is 19/12 per 5 minute period, since it is proportional to time.
\n" ); document.write( "it is 19/60 per 1 minute period.
\n" ); document.write( "probability that at least 1 customer arrives in a 1 minute period is 1-P(0) arrive in that time.
\n" ); document.write( "P(0)=e^-19/60=0.7286, so the answer is 1-0.729=0.271.
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\n" ); document.write( "at least 2 means 1-P(0)-P(1)
\n" ); document.write( "P(0)=e^-19/12=0.2053
\n" ); document.write( "P(1)=e-19/12*19/12^1/1!=0.3250
\n" ); document.write( "so at least two customers would be 1-P(0)-P(1)=1-0.2053-0.3250=0.4697\r
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