Mathematics, 13.12.2019 05:31 drelisech6467

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 ) what is the probability that one cubic meter of discharge contains at least 8 organisms? (round your answer to three decimal ) 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? (round your answer to three decimal ) for what amount of discharge would the probability of containing at least 1 organism be 0.994? (round your answer to two decimal places.) m3

Answers: 3

Show answers

Answers

Answer from: addisynshepherd

a) The probaility is 0.333.

b) The probability is 0.125.

c) The volume is 0.512 m3.

Step-by-step explanation:

Organisms are present in ballast water discharged with a concentration of 10 organisms/m3.

That is our rate of the Poisson process

$\lambda=10\,org/m^3$

a) The probability of having at least 8 organisms is equal to the sum of the probabilities of having from 0 to 8 organisms:

$P(k\leq8)=\sum_{i=0}^{8}P(k=i)=\sum_{i=0}^{8}\frac{\lambda V^i e^{-\lambda V}}{i!} \\\\P(k\leq8)=0.000+0.000+0.002+0.008+0.019+0.038+0.063+0.090+0.113\\\\P(k\leq8)=0.333$

Note: In this case, the volume V is 1 m3.

b) In this case, the volume is 1.5m3 so we have to multiply the rate by 1.5. Then it becomes:

$\lambda V=10\,org/m^3\cdot1.5m^3=15org$

The standard deviation of this distribution is

$\sigma=\sqrt{\lambda}=\sqrt{15}\approx 3.873$

We have to calculate the probability of exceeding 19 organisms in 1.5m3:

$k=15+3.873 \approx19$

We have that the probability of having more than 19 org. is equal to one substracting the probability of having equal or less than 19 org:

$P(k19)=1-P(k\leq19)\\\\P(k19)=1-\sum_{i=0}^{19}P(k=i)=1-\sum_{i=0}^{19}\frac{\lambda V^i e^{-\lambda V}}{i!} \\\\P(k19)=1-(0.000+0.000+0.000+0.000+0.001+0.002+0.005+0.010+0.019+0.032+0.049+0.066+0.083+0.096+0.102+0.102+0.096+0.085+0.071+0.056)\\\\P(k19)=1-0.875=0.125$

c) We have to calculate the volume such that there is a probability P=0.994 of having at least one organism in the water. This can be calculated as one less the probability of having zero organisms.

$P(x\leq1)=1-P(0)=1-\frac{\lambda V^0 e^{-\lambda V}}{0!} \\\\P((x\leq1)=1-\frac{1\cdot e^{-10 V}}{1}=1-e^{-10V}=0.994\\\\e^{-10V}=1-0.994=0.006\\\\-10V=ln(0.006)=-5.116 \\\\V=5.116/10=0.512$