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somecelxis
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Homework Statement
i am having problem with part iv ) . the ans is 0.04519 . can anyone tell me how to do this ? i have solved part i , ii and iii ..p/s line 1:
A tank contain 10^5 cm3 of water
somecelxis said:Homework Statement
i am having problem with part iv ) . the ans is 0.04519 . can anyone tell me how to do this ? i have solved part i , ii and iii ..p/s line 1:
A tank contain 10^5 cm3 of water
Homework Equations
The Attempt at a Solution
Ray Vickson said:You seem to make a habit of violating PF standards: you keep posting thumbnails, when--officially--you are supposed to type things out. I cannot read your thumbnails on some media, and on the computer I am using now the part you want to know about (part (iv)) does not even show up in the attached picture!
somecelxis said:A water tank contains 10^5 cm3 of water that is free from bacteria. A number of 1800 bacteria were added to water and water is stirred to disrtibute the bactria randomly. Then 6 test tube were filled with 20cm3 of this water. Findthe probablity of
a) a test tube contain 3 bactria
b) A test tube contain at least 3 bacteria
c) exactly 4 out of 6 test tube with each contain 3 bacteria
d) 6 test tube conatin a total of 5 bacteria
i got stucked at part d .
Ray Vickson said:The bacteria amount in 6 test tubes is the sum of 6 independent, identically-distributed Poisson random variables. Do you know what that distribution is? (Of course, in reality the 6 test-tube bacteria contents are not really independent, but since the total volume is small compared with the overall tank volume, we can assume independence without making much of an error.)
That wasn't quite complete. There are two facts which, together, mean you can treat the samples as independent:somecelxis said:why the total volume is small compared with the overall tank volume, we can assume independence without making much of an error.??
What are you calculating there, and how? Please show all steps.so my working would be (e^-0.36)x 0.36 = 0.2511
haruspex said:That wasn't quite complete. There are two facts which, together, mean you can treat the samples as independent:
1. The sample volumes are tiny compared to the tank volume
2. The number of bacteria sampled, in total, is small compared with the total in the tank.
As a result, taking one sample does not much change the circumstances for the next sample.
What are you calculating there, and how? Please show all steps.
somecelxis said:6c5 ( (0.2511)^5 )x ( (1-0.2511)^1 )= 4.49x10^-3
i assume each tube contain only 1 bacteria.
Orodruin said:By making this assumption you are effectively saying that the only way of catching 5 bacteria is to catch 1 in each of 5 samples and getting the last sample empty. What about the possibility of catching all bacteria in one sample? What about that of catching 3 in one, 2 in one, 1 in one, and none in 3?
Does it matter which of the samples the bacteria are caught in?
somecelxis said:this is based on what the homework helper form physiscs forum have said . If i consider catching 3 in one, 2 in one, 1 in one, and none in 3?
then i should also consider catching 3 in two, catching 4 in two , catching 5 in two , there would be too many combinations . any other simpler way look at tis question?
If you are referring to my question:somecelxis said:this is based on what the homework helper form physiscs forum have said .
please ignore that and follow Orodruin's hint.For (d), what are all the possible distributions of the 5 between the 6 test tubes?
Poisson distribution is a statistical distribution that represents the probability of a certain number of events occurring in a fixed time or space, given the average rate of occurrence and assuming independence between events.
In the context of a tank of water, Poisson distribution can be used to calculate the probability of a certain number of water molecules in a given volume of the tank, given the average density of water molecules and assuming random distribution within the tank.
The formula for calculating Poisson distribution is P(x;μ) = (e^-μ) * (μ^x) / x!, where x is the number of events, μ is the average rate of occurrence, e is the mathematical constant e, and x! is the factorial of x.
Poisson distribution can be applied in various practical situations, such as predicting the number of customer arrivals at a store, the number of defects in a manufacturing process, or the number of accidents on a highway within a certain time period.
Poisson distribution assumes that events occur independently at a constant rate, which may not always be the case in real-life situations. It also assumes that the average rate of occurrence is known, which may not always be accurately estimated. Additionally, Poisson distribution is not applicable for events that have a continuous range of values.