# Thread: Probability distribution for the number of arrivals in a second

1. ## Probability distribution for the number of arrivals in a second

Here's the question I'm having a little trouble with:

Packets arrive at a Network channel at a mean arrival rate of 72 packets per minute. Assuming that the traffic system is Markovian, find the probability distribution for the number of arrivals in a second.

I am assuming from the above that the mean arrival rate in seconds is 1.2 packets per second

I am also assuming the I have to use:

Poisson Probability Distribution: Prob(x arrivals in a unit time)

λxe-λ
p(x) = -----
x!

Lambda = 1.2 , but what do i need to use for x ( 1 to 60???)
Am i on thr right track with this question??

2. Originally Posted by mikebyrne
Here's the question I'm having a little trouble with:

Packets arrive at a Network channel at a mean arrival rate of 72 packets per minute. Assuming that the traffic system is Markovian, find the probability distribution for the number of arrivals in a second.

I am assuming from the above that the mean arrival rate in seconds is 1.2 packets per second

I am also assuming the I have to use:

Poisson Probability Distribution: Prob(x arrivals in a unit time)

λxe-λ
p(x) = -----
x!

Lambda = 1.2 , but what do i need to use for x ( 1 to 60???)
Am i on thr right track with this question??
The support of the Poisson distribution can be found here: Poisson distribution - Wikipedia, the free encyclopedia

3. I'm just looking for confirmation that im correct in saying

Lambda = 1.2

X=1

Any confirmation would be great

4. Originally Posted by mikebyrne
I'm just looking for confirmation that im correct in saying

Lambda = 1.2

X=1

Any confirmation would be great
My first reply clearly implied agreement that X (the number of packets per second) has a Poisson distribution with a mean of 1.2. It then gave a link to where to get the support for a Poisson distribution.

I have no idea what X=1 is meant to mean (it certainly is not the possble values of X).

5. X = the exact number of packets to arrive in a second

1 = 36.1% chance of 1 packet arriving in a second
2 = 21.7%
3 = 8.7%
4 = 2.6%
5 = 0.6%

Am i right in assuming the above??

6. Originally Posted by mikebyrne
X = the exact number of packets to arrive in a second

1 = 36.1% chance of 1 packet arriving in a second
2 = 21.7%
3 = 8.7%
4 = 2.6%
5 = 0.6%

Am i right in assuming the above??
All you have to do is substitute the values of X into the Poisson pmf and press some buttons on a calculator. Why do you need someone to check what is just basic arithmetic?

7. It not the math i need confirmation on more the english

Its the assumption that:

X = the exact number of packets to arrive in a second

1 = 36.1% = The % chance of 1 packet arriving in a second

8. Originally Posted by mikebyrne
It not the math i need confirmation on more the english

Its the assumption that:

X = the exact number of packets to arrive in a second

1 = 36.1% = The % chance of 1 packet arriving in a second
I have defined X in post #4. I have agreed with you on the distribution of X. I have told you where to find the possible values of X (the support).

The above collectively defines the probability distribution the question asks you to find. I don't know why you are now calcuating probabilities for different values of X. The posted question does not ask you to do that.