document.write( "Question 1156162: Cars arrive at a CNG station at an average rate of 30 per hour. Assuming that the number of cars arriving at the CNG station follows a Poisson distribution, find the probability that
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document.write( "i) no cars arrive during a particular 5 minute interval.
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document.write( "ii) more than 3 cars arrive during a 5 minute interval.
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document.write( "iii) more than 5 cars arrive in a 15 minute interval.
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document.write( "iv) in a period of half an hour, 10 cars arrive.
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document.write( "v) less than 3 cars arrive during a 10 minute interval. \n" );
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Algebra.Com's Answer #778861 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! 30 per hour is 2.5 in every five minutes \n" ); document.write( "probability of 0 when parameter is 2.5=0.0821 \n" ); document.write( "for more than 3 do 0,1,2,3, and that is 0.7576. But want more than 3, so that is 1-0.7576=0.2424\r \n" ); document.write( "\n" ); document.write( "In a 15 minute interval the parameter is now 7.5. Do cdf for 5 and then subtract that from 1. \n" ); document.write( "It is 0.7586 probability. Can check to see if tha makes sense. Cars are coming every 2.5 minutes on average, so expect 6 cars at least in 15 minutes, so that more than 5 would be quite likely.\r \n" ); document.write( "\n" ); document.write( "In a half hour, expect 15 so 10 should be significantly less than half. The probably is 0.0486\r \n" ); document.write( "\n" ); document.write( "Less tha 3 cars arrive in a 10 minute interval. This is 0,1,2 \n" ); document.write( "cdf for 2, with parameter 1/6 of 30 or 5. \n" ); document.write( "That probability is 0.1247. Can check that with poisson pdf for parameter 5 and numbers 0,1,2 and get same result by adding the three,\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |