1. poisson distribution

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

2. Re: poisson distribution

figure out $\lambda$ for the distribution from the wording of the problem and then evaluate the Poisson CDF for 3.

3. Re: poisson distribution

Originally Posted by romsek
figure out $\lambda$ for the distribution from the wording of the problem and then evaluate the Poisson CDF for 3.
Ignore the last post... I could have sworn you originally asked about (ii).

We can't see (iv) in your image.

4. Re: poisson distribution

Originally Posted by romsek
Ignore the last post... I could have sworn you originally asked about (ii).

We can't see (iv) in your image.
sorry. it's my fault . part iv)

6 test tube contains a total number of 5 bacteria.

5. Re: poisson distribution

Originally Posted by delso
sorry. it's my fault . part iv)

6 test tube contains a total number of 5 bacteria.
looking deeper at this ... It's not a Poisson distribution we're talking about here. It's a binomial distribution.

Your image cuts off the initial amount of water. Your work seems to indicate it's $10^5cc$ is that correct?

6. Re: poisson distribution

Originally Posted by romsek
looking deeper at this ... It's not a Poisson distribution we're talking about here. It's a binomial distribution.

Your image cuts off the initial amount of water. Your work seems to indicate it's $10^5cc$ is that correct?
yes , it's $10^5cc$ of water.. i still cant get the ans. can anyone explain further?

7. Re: poisson distribution

Originally Posted by romsek
looking deeper at this ... It's not a Poisson distribution we're talking about here. It's a binomial distribution.

Your image cuts off the initial amount of water. Your work seems to indicate it's $10^5cc$ is that correct?
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 .

8. Re: poisson distribution

How did You calculate that what You have?

9. Re: poisson distribution

Originally Posted by delso
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.
So there are an average of 1800/10000= 18/100= 9/50 "bacteria per cm".
Then 6 test tube were filled with 20cm3 of this water. Findthe probablity of
(9/50)(20)= 18/5 bacteria per test tube.

a) a test tube contain 3 bactria
b) A test tube contain at least 3 bacteria
These two problems can be done using the Poisson distribution with $\lambda= 18/5$. (a) is P(3) and (b) is 1- (P(0)+ P(1)+ P(2)).

c) exactly 4 out of 6 test tube with each contain 3 bacteria
This requires, as romsek said, the binomial distribution with p= the probability you got in (a)

d) 6 test tube conatin a total of 5 bacteria [/quote]
How many different ways can 5 bacteria be in 6 specified test tubes? They could all 5 be in one test tube, or 5 in one test tube and one in another, or 3 in one test tube and 2 in another, or 3 in one test tube and 1 in each of two other test tubes, etc.

i got stucked at part d .[/QUOTE]

10. Re: poisson distribution

i got stucked at part d .[/QUOTE][/QUOTE]
d) 6 test tube conatin a total of 5 bacteria [/quote]
How many different ways can 5 bacteria be in 6 specified test tubes? They could all 5 be in one test tube, or 5 in one test tube and one in another, or 3 in one test tube and 2 in another, or 3 in one test tube and 1 in each of two other test tubes, etc.

for part d , can you show me all the possibility of having 5 bacteria in 6 test tube? any other easy method to get the ans for this? it seems quite complicated if i do in this way..

double post~