Respuesta :
Answer:
There is a 93.28% probability that one cubic meter of discharge contains at least 6 organisms.
Step-by-step explanation:
The number of organisms in a cubic meter of discharge is a Poisson process. So, we use the following definition:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
[tex]e = 2.71828[/tex] is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
Organisms are present in ballast water discharged from a ship according to a Poisson process with a concentration of 10 organisms/m3. This means that [tex]\mu = 10[/tex]
What is the probability that one cubic meter of discharge contains at least 6 organisms?
This is [tex]P(X \geq 6)[/tex]. We know that either we have less than 6 organisms, or we have at least 6 organism. The sum of the probabilities is decimal 1. So
[tex]P(X < 6) + P(X \geq 6) = 1[/tex]
[tex]P(X \geq 6) = 1 - P(X < 6)[/tex]
In which
[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex].
Each one of these probabilities can be found by the poisson formula.
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-10}*10^{0}}{(0)!} = 0.00004[/tex]
[tex]P(X = 1) = \frac{e^{-10}*10^{1}}{(1)!} = 0.0004[/tex]
[tex]P(X = 2) = \frac{e^{-10}*10^{2}}{(2)!} = 0.0023[/tex]
[tex]P(X = 3) = \frac{e^{-10}*10^{3}}{(3)!} = 0.0078[/tex]
[tex]P(X = 4) = \frac{e^{-10}*10^{0}}{(0)!} = 0.0189[/tex]
[tex]P(X = 5) = \frac{e^{-10}*10^{1}}{(1)!} = 0.0378[/tex]
So
[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
[tex]P(X < 6) = 0.00004 + 0.0004 + 0.0023 + 0.0078 + 0.0189 + 0.0378 = 0.06724[/tex]
Finally
[tex]P(X \geq 6) = 1 - P(X < 6) = 1 - 0.06724 = 0.9328[/tex]
There is a 93.28% probability that one cubic meter of discharge contains at least 6 organisms.
