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Mathematics 12 Online
OpenStudy (anonymous):

Poisson Distribution. How do you go about finding the distribution between a range using Poisson distribution? Photo is attached.

OpenStudy (anonymous):

OpenStudy (anonymous):

for part a) i know X=3, and lambda = (2.5*2) = 5 then i used the formula and crunched for P(x=3) \[P(x=3) = \frac{ 5^{3} e ^{-5} }{ 3! } = 0.1404\] part b) P(25<=x<=50) = ? I was thinking about integration but I am not sure if that is the right way.

OpenStudy (anonymous):

Or since the question asked for between, find the midpoint?

OpenStudy (anonymous):

i tried adding 25 and 50... but i don't feel it is correct

OpenStudy (perl):

i think you have to use exponential distribution

OpenStudy (perl):

poisson is a discrete random variable, the probability of P(n) for integer values

OpenStudy (anonymous):

lets see

OpenStudy (perl):

$$ \Large { P(a \leq x \leq b ) = e^{-\lambda a} - e^{-\lambda b } } $$

OpenStudy (anonymous):

i have lambda to be 0.9375, do you agree?

OpenStudy (perl):

how did you get this lambda

OpenStudy (anonymous):

sorry, i made a mistake, it's .4, inverse of the mean.

OpenStudy (perl):

right, that sounds better

OpenStudy (perl):

2.5 eruptions per 100 years = .025 eruptions per year

OpenStudy (anonymous):

my answer is really small.= .000045

OpenStudy (perl):

$$ \Large { P(a \leq x \leq b ) = e^{-\lambda a} - e^{-\lambda b } \\ \therefore \\ P(25 \leq x \leq 50 ) = e^{-\frac{2.5}{100}\cdot 25} - e^{-\frac{2.5}{100}\cdot 50 } = 0.2487566 } $$

OpenStudy (anonymous):

that is a different approach to obtain the lambda. I didn't think of it but it looks correct since the probability can be at 25th and it is 0.25

OpenStudy (perl):

is this a multiple choice question

OpenStudy (anonymous):

It is a practice exam, showing the workings gives the points.

OpenStudy (perl):

is .2487 correct?

OpenStudy (anonymous):

I do not know. The professor will be posting the answers later.

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