Week_6_Poisson

Week_6_Poisson - The mean number of business failues per...

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Unformatted text preview: The mean number of business failues per hour in the United States in a recent year was about 8. Find the probability that: a. exactly four businesses will fail in any given hour. b. at least four businesses will fail in any given hour. c. more than four businesses will fail in any given hour. Solution: Define "success" as "business will fail within a given hour". x = number of businesses that fail in any given hour; u = mean number of successes in any given hour x P(x) Percentage 0 0.000335 0.03% 1 0.002684 0.27% 2 0.010735 1.07% 3 0.023626 2.86% 4 0.057252 5.73% 5 0.091604 9.16% 6 0.122138 12.21% 7 0.139587 13.95% 8 0.139587 13.96% 9 0.124077 12.41% 10 0.099262 9.93% 11 0.072190 7.22% 12 0.043127 431% _ , _ _ 13 0329616 296% Important Note: When it = 20, the sum of the probabllitles IS 14 0015924 159% "close" to 100%. Technically, x goes on forever. We only need 15 0009025 090% to continue until we see the sum of the probabilities getting close 16 0004513 045% to 100%. In this case, 20 works; in other cases, the distribution 17 0.002124 0.21% 18 0. 000944 0.09% 19 0.000397 0.04% 20 0.000159 0.02% Total 0.999906 99.99% Answers: Let x = the number of businesses that fail in any given hour. Part A. From the distribution above, P(exaclly 4 will faii in any given hour ) = P(x=4) = 0.057252 = 5.7% Part B. P(x is at least 4) = P( x = 4 or 5 or 6 or )= P(x=4) + P(x=5) + P(x=6) + = l- P( x = 3 or2 or 1 or 0)=1-(P(x=0)+ P(x=1) + P(x=2) + P(x=3) ) = l - ( 0.028626 + 0.010735 + 0.002684 + 0.000335 ) = 1 - 0.042380 = 0.957620 = 95.8% Part C. P(x is more than 4) = P( x = 5 or 6 or 7" or ) = P(x=5) + P(x=6) + P(x=?) + = Answer from Part b minus P(x=4) = 0.957620 - 0.057252 = 0.900368 = 90.0% See the Next Page for a Formula View of the above table. x Ptx) Percentage o =POISSON(A61.8,FALSE) =BB1 1 =POISSON(A62,8,FALSE) =B62 . 2 =POISSON(A53,8.FALSE) =BB3 '3 3 =POISSON(A64r8.FALSE) =Be4 “ ,, 4 =POISSON(AB5,8,FALSE) =365 Percemage 5 =POISSON(A66,8.FALSE) :35?) 6 =POISSON(A67,8.FALSE) 2B6? 7 =PO|SSON(A68,8.FALSE) =368 8 =PO|SSON(A69,8.FALSE) =BSQ 9 =PO|SSON(A70,8,FALSE) =B7o 10 =PO|SSON(A71,8,FALSE) =B7‘1 11 =PO|SSON(A72,8,FALSE) =B72 12 =PO|SSON(A73,8,FALSE) =373 13 =POISSON(A74,8,FALSE) =B74 14 =POISSON(A75,8,FALSE) =375 15 =POISSON(A?6,8,FALSE) =B76 16 =POISSON(A77,8,FALSE) =37? 17 =POISSON(A78,8,FALSE) =373 13 =POISSON(A79,8,FALSE) =379 19 =POISSON(A80,8,FALSE) =BBO 20 =POISSON(A81,8,FALSE) =BB1 Total =SUM(Be1:BB1) =332 Important Note: When x = 20, the sum of the probabilities is "close" to 100%. Technically, x goes on forever. We only need to continue until we see the sum ofthe probabilities getting close to l00%. [n this case, 20 works; in other cases, the distribution may need to go to a larger value. ...
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This note was uploaded on 05/17/2011 for the course BIS 115 taught by Professor Wright during the Spring '10 term at DeVry Chicago.

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Week_6_Poisson - The mean number of business failues per...

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