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Math Help - Probability distribution for the number of arrivals in a second

  1. #1
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    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??
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  2. #2
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    Quote Originally Posted by mikebyrne View Post
    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
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  3. #3
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    I'm just looking for confirmation that im correct in saying

    Lambda = 1.2

    X=1


    Any confirmation would be great
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  4. #4
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    Quote Originally Posted by mikebyrne View Post
    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).
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  5. #5
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    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??
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  6. #6
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    Quote Originally Posted by mikebyrne View Post
    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?
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  7. #7
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    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
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  8. #8
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    Quote Originally Posted by mikebyrne View Post
    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.
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