Respuesta :
Answer:
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
[tex]\lambda=10\,org/m^3[/tex]
a) The probability of having at least 8 organisms is equal to the sum of the probabilities of having from 0 to 8 organisms:
[tex]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[/tex]
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:
[tex]\lambda V=10\,org/m^3\cdot1.5m^3=15org[/tex]
The standard deviation of this distribution is
[tex]\sigma=\sqrt{\lambda}=\sqrt{15}\approx 3.873[/tex]
We have to calculate the probability of exceeding 19 organisms in 1.5m3:
[tex]k=15+3.873 \approx19[/tex]
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:
[tex]P(k>19)=1-P(k\leq19)\\\\P(k>19)=1-\sum_{i=0}^{19}P(k=i)=1-\sum_{i=0}^{19}\frac{\lambda V^i e^{-\lambda V}}{i!} \\\\P(k>19)=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(k>19)=1-0.875=0.125[/tex]
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.
[tex]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[/tex]
