Question

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"† considers using the Poisson process for this purpose).

(a)

What is the probability that one cubic meter of discharge contains at least 5 organisms? (Round your answer to three decimal places.)

(b)

What is the probability that the number of organisms in 1.5 m³ of discharge exceeds its mean value by more than two standard deviations? (Round your answer to three decimal places.)

(c)

For what amount of discharge would the probability of containing at least 1 organism be 0.995? (Round your answer to two decimal places.)

m³

Answer 1

The number of organisms per m3 here is modelled as:

a) The probability that one cubic meter of discharge contains at least 5 organisms is computed here as:

Therefore 0.971 is the required probability here.

b) For a 1.5m3 discharge, the mean number of organisms is computed as: 10*1.5 = 15. Also variance for poisson distribution is same as mean = 15 here.

Also the standard deviation is computed as:
= Square root of variance

The probability now is computed here as:

This is computed in EXCEL here as:
=1-poisson.dist(22,15,TRUE)

The ouput came out to be 0.033

Therefore 0.033 is the required probability here.

c) Let the required amount be K, then the mean number of organisms in K m3 of discharge is computed as:
= 10k

Therefore the probability now is computed here as:
P(X >= 1) = 1 - P(X = 0) = 1 - e-10k

1 - e-10k = 0.995

e-10k = 0.005

k = -0.1Ln(0.005) = 0.53

Therefore 0.53m3 is the required amount of discharge here.