# Poisson distribution

• Nov 6th 2006, 11:34 AM
JackStolerman
Poisson distribution
Calls for dial-in connections to a computer centre arrive at an average rate of four per minute. The calls follow a Poisson distribution. If a call arrives at the beginning of a 1-minute interval, what is the probability that a second call will not arrive in the next 20 seconds?

I'm having trouble putting this into the Poisson probability distribution formula. It seems different from some of the other problems I've done. How do you solve this?

Jack
• Nov 6th 2006, 12:26 PM
JakeD
Quote:

Originally Posted by JackStolerman
Calls for dial-in connections to a computer centre arrive at an average rate of four per minute. The calls follow a Poisson distribution. If a call arrives at the beginning of a 1-minute interval, what is the probability that a second call will not arrive in the next 20 seconds?

I'm having trouble putting this into the Poisson probability distribution formula. It seems different from some of the other problems I've done. How do you solve this?

Jack

For a Poisson process, if the arrivals per unit of time have a Poisson distribution with mean $\lambda,$ the waiting time between arrivals has an exponential distribution with mean $\lambda^{-1}$ and cumulative distribution function $F(x) = 1 - e^{-\lambda x}.$ Can you take it from here?
• Nov 6th 2006, 04:03 PM
JackStolerman
Well I'm pretty sure that I have to integrate.

Also is the formula the same as:
f(y)=(1/B)*e^(-y/B) where y is greater than or equal to zero and less than infinity.
I have to integrate this, but what are the bounds and everything?

Jack
• Nov 6th 2006, 04:50 PM
JackStolerman
OK. I think I got it, but could someone confirm this:

I let Y be the time between the arrival of two dial-in connections in minutes. We have P(Y>1/3)
(lambda)t=4 and (1/lambda)=(1/4)
Thus, f(y)=(1/0.25)e^(-y/0.25)=4e^(-4y)
P(Y>1/3)=integration of 4e^(-4y) from (1/3) to infinity. This equals e^(-4/3).

Is this the probability?
• Nov 6th 2006, 08:41 PM
JakeD
Quote:

Originally Posted by JakeD
For a Poisson process, if the arrivals per unit of time have a Poisson distribution with mean $\lambda,$ the waiting time between arrivals has an exponential distribution with mean $\lambda^{-1}$ and cumulative distribution function $F(x) = 1 - e^{-\lambda x}.$ Can you take it from here?

Quote:

Originally Posted by JackStolerman
OK. I think I got it, but could someone confirm this:

I let Y be the time between the arrival of two dial-in connections in minutes. We have P(Y>1/3)
(lambda)t=4 and (1/lambda)=(1/4)
Thus, f(y)=(1/0.25)e^(-y/0.25)=4e^(-4y)
P(Y>1/3)=integration of 4e^(-4y) from (1/3) to infinity. This equals e^(-4/3).

Is this the probability?

Yes. Note the cumulative distribution function is $F(x) = P(Y \le x)$ so the probability $P(Y > 1/3) = 1-F(1/3) = e^{-\lambda /3} = e^{-4/3}.$