A
100+89.50×36
=3,322
b
3,322−2,900
=422
c
2,900÷89.50
=33
Answer:
a) The mean number of cases is 0.14608 cases.
b) The probability that the number of cases is exactly 0 or 1 is 0.990.
c) The probability of more than one case is 0.010
d) No, because the probability of more than one case is very small
Step-by-step explanation:
We can model this problem with a Poisson distribution, with parameter:
![\lambda=r*t=0.000011*13,280=0.14608](https://tex.z-dn.net/?f=%5Clambda%3Dr%2At%3D0.000011%2A13%2C280%3D0.14608)
a) The mean amount of cases is equal to the parameter λ=0.14608.
b) The probability of having 0 or 1 cases is:
![P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990](https://tex.z-dn.net/?f=P%28k%3D0%29%3D%5Cfrac%7B%5Clambda%5E0%20e%5E%7B-%5Clambda%7D%7D%7B0%21%7D%3D%5Cfrac%7B1%2A0.864%7D%7B1%7D%20%3D0.864%5C%5C%5C%5C%20P%28k%3D1%29%3D%5Cfrac%7B%5Clambda%5E1%20e%5E%7B-%5Clambda%7D%7D%7B0%21%7D%3D%5Cfrac%7B0.14608%2A0.864%7D%7B1%7D%20%3D0.126%5C%5C%5C%5CP%28k%5Cleq1%29%3D0.864%2B0.126%3D0.990)
c) The probability of more than one case is:
![P(k>1)=1-P(k\leq 1)=1-0.990=0.010](https://tex.z-dn.net/?f=P%28k%3E1%29%3D1-P%28k%5Cleq%201%29%3D1-0.990%3D0.010)
d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.
Answer:
i THINK that it is C
Step-by-step explanation: