Desperately needing homework help!

samiam619

New Member
I tried the math lab at the local college and the statistics professor was as stumped as I was!

A videotape has an average of one defect every 1000 feet. Find the probability of at least one defect in 3000 feet.

I know that the answer is 0.9502 but I have no idea how to get it. I have tried the poisson formula but cannot get it right!

statfish

New Member
http://en.wikipedia.org/wiki/Poisson_distribution

notice that λ is the expected number (on average) of occurrences that occur during the given interval.

And if X~Poisson(λ), the {X=k}={ There are k occurrences that occur during the given interval.}

Then try to explain {There are at least one occurrence during the given interval}

fernrodolfo

New Member
Not Poisson...It's Exponential..

The given random variable has a continuous domain, it may not be solved straight forward via poisson, because this variable is distribuited exponential with lambda=1/1000, so the distribution function is

f(x) = (lambda)*exp(-lambda*x) = (1/1000)*exp(-x/1000)

So the probability you're trying to find is:

p(x<3000)=Definite Integral of f(x) from o to 3000=
= 1-exp(-3) = 0.950213

I hope it helps.

JohnM

TS Contributor
The given random variable has a continuous domain, it may not be solved straight forward via poisson, because this variable is distribuited exponential with lambda=1/1000, so the distribution function is

f(x) = (lambda)*exp(-lambda*x) = (1/1000)*exp(-x/1000)

So the probability you're trying to find is:

p(x<3000)=Definite Integral of f(x) from o to 3000=
= 1-exp(-3) = 0.950213

I hope it helps.

Um..... actually the Poisson does quite nicely here, and the situation described in the problem could be a Poisson process.

P(at least 1) = 1 - P(0)

Poisson:
P(x) = mu^x * e^-mu / x!
mu = 3 (since we would expect 3 defects in 3000 ft of tape)

P(0) = 3^0 * e^-3 / 0! = e^-3 = .0498

1 - P(0) = 1 - .0498 = .9502

fernrodolfo

New Member
Nice approach John...

Never thought it that way, tnx for the lesson.