document.write( "Question 1021583: A computer repair person is ‘beeped’ each time there is a call for service. The number of beeps per hour has a Poisson distribution with mean 2 beeps per hour. Calculate the probability that there will be: \r
\n" ); document.write( "\n" ); document.write( "i. At least two beeps in an hour.
\n" ); document.write( "ii. One beep in next three hours.
\n" ); document.write( "iii. No beep in 45 minutes.
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Algebra.Com's Answer #637314 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
We use the pmf of Poisson r.v.\r
\n" ); document.write( "\n" ); document.write( "i. \"p%28x%29+=+%282%5Ex%2Fx%21%29e%5E%28-2%29\"
\n" ); document.write( "The answer is p(2)+p(3)+p(4) +... = 1 - p(0) - p(1) = \"1-e%5E-2-2e%5E-2\" = 0.594, to 3 decimal places. (Here \"mu+=+2\". \r
\n" ); document.write( "\n" ); document.write( "ii. Here \"mu+=+3%2A2+=+6\"
\n" ); document.write( "==> \"p%281%29+=+%286%5E1%2F1%21%29e%5E%28-6%29+=+6e%5E%28-6%29+=+0.01487\" to five decimal places.\r
\n" ); document.write( "\n" ); document.write( "iii. Here \"mu+=+%283%2F4%29%2A2+=+3%2F2\"
\n" ); document.write( "==>\"p%280%29+=+%281.5%5E0%2F0%21%29e%5E%28-1.5%29+=+e%5E%28-1.5%29+=+0.22313\" to five decimal places.
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