Organisms are present in ballast water discharged from a ship according to a Poisson process with a concentration of 10 organisms/m3 [the article ?"Counting at Low Concentrations: • The Statistical Challenges of Verifying Ballast Water Discharge Standards" (Ecological Applications, 2013: 339-351) ?considers using the Poisson process for this purpose]. a. What is the probability that one cubic meter of dis-charge contains at least 8 organisms? b. What is the probability that the number of organisms in 1.5 m3 of discharge exceeds its mean value by more than one standard deviation? c. For what amount of discharge would the probability of containing at least 1 organism be .999?

Answer : Step 1 of 3 : Given, Organisms are present in ballast water discharged from a ship according to a Poisson process with a concentration of 10 organisms/m3 Let X be the organisms in cubic meter. The probability density function of Poisson distribution is e x P(X = x) = x! , where, x = 0, 1, 2, 3, ….. a) The claim is to find the probability that one cubic meter of discharge contains at least 8 organisms Therefore, P( x 8 ) = 1 - P(x 7 ) = 1 - [ P(x = 0) +P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) ] e10100 e10101 e10102 e10103 e10104 e10105 = 1 - [ 0! + 1! + 2! + 3! + 4! + 5! + e10106 e10107 6! + 7! ] 10 10 x = 1 - [ e x! ] = 1 - [0.0000454(1 + 10 + 50 + 166.67 + 416.667 + 833.33 + 1388.89 + 1987.13] = 1 - 0.2203 P( x 8 ) = 0.7797 Therefore, the probability value is 0.7797. Step 2 of 3 : b) 3 The claim is to find the probability that the number of organisms in 1.5 m of discharge exceeds its mean value by more than one standard deviation the number of organisms in 1.5 m 3 we have 10 organisms/m3 So 10×1.5 = 15 Using the cumulative distribution table for poisson distribution table P(x > + ) = 1 - P(x + ) = 1 - F( 15 + 15) = 1 - F(15 + 3.837) = 1 - F(18.837) = 1 - F(19) = 1 - P(x 19) = 1 - (0.8752) = 0.1248