Statistics refresher help

[Eirgorn]

Hey guys,

It seems it's been too many years since I took Probability and Statistics and I was wondering if anyone that has either taken it more recently or uses it often could help me figure out what method I need to use to solve a problem Im working on.

Im trying to figure out what the average time would be for four events to take place given a known rate and time interval between opportunities for the events to take place. I've made some guesstimates for these figures, but they are close enough for what Im looking for.

The rate is appx 2% (0.02) for each of the four events and there is a chance one or more of these will occur every 2.5 minutes. Each event is independant of the others.

So the probability of all four happening at the first interval would be 0.02^4 = 1.6e-7, but how would I find the average time for all 4 events to take place?

I think I would use an exponential distribution since it's memoryless - but Ive pretty much forgotten how I need to set it up.

Let me know if I need to post more info, and thanks for any help!

 

[ArJuN]

Ha, I'm sitting in AP Stat right now. But that doesn't mean I know anything...I guess I can ask my teacher lol.

 

[Paperdoc]

Seems to me, no matter which time interval you pick, the probability of having all four independent events occur during that interval is 0.02^4. Think of that number as the Quotient: "number of observations that produce the desired outcome (4 simultaneous events) / total number of observations". Now, 2.5 minutes is "time per one observation". So in terms of unit analysis, 2.5 minutes / (0.02^4) has units of:
"Minutes per observation " x "number of observations" / "number of successful observations", which will resolve to "minutes per successful observation". That's the average time it should take to ahcieve "success" as defined by "four simultaneous events".

 

[HomeBrewerDude]

I was going to suggest the poisson distn but I'm not sure since you're interested in the intersection of 4 events. I suppose you could set p(event)=.02

p(y) = ((lambda^y)(e^y))/y! where y = 0,1,2,3....

 

[Eirgorn]

Thanks for the replies all - I think my OP was misleading though. When I said all 4 events could occur at the same time I didnt mean that was what I was looking for, just that it could happen.

So:
t=2.5 Event 1 happened
t=5 no event
t=7.5 no event
t=10 no event
t=12.5 Event 2 and 3 happen
t=15 Event 4 happens - Finished

Could be an example (even thought that would be very lucky for them to all happen so soon)

So im trying to find the Mean time it would be expected to take all 4 events to happen, and then perhaps even the probability that all 4 events occur before time T (insert any time)

For example P(T=2.5) would be the 0.02^4 stated earlier

That make any more sense?

 

[HomeBrewerDude]

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

 

[FleshLight]

Originally posted by: HomeBrewerDude
I was going to suggest the poisson distn but I'm not sure since you're interested in the intersection of 4 events. I suppose you could set p(event)=.02

p(y) = ((lambda^y)(e^-y))/y! where y = 0,1,2,3....

Fixed.

Remember that lambda = alpha * t, so alpha would 0.02 and t would be 2.5 minutes. Since you want the probability of all 4 happening, Y = 4 and just plug and chug.

For the poisson distribution, the expected value E is equal to lambda.