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Poisson Distribution

  1. Oct 30, 2008 #1
    [SOLVED] Poisson Distribution

    1. The problem statement, all variables and given/known data

    Let X be the number of people entering the ICU in a hospital. From Historical data, we know the average number of people entering ICU on any given day is 5

    a) What is the probability that the number of people entering the ICU on any given day is less than 2. Do you think this is a rare event?

    b) What is the probability of people entering the ICU on any 2 consecutive day is less than 2. Do you think this is a rare event?

    2. Relevant equations

    [tex]P(X = K) = \frac{\mu^k e^{-\mu}}{k!}[/tex]

    3. The attempt at a solution

    a) [tex]P (X < 2) = P(X=0) + P(X=1) = \frac{5^0 e^{-5}}{0!} + \frac{5^{1} e^{-5}}{1!}
    = e^{-5} + 5e^{-5} = 6e^{-5} = 0.0404[/tex]

    b) Since it's 2 consecutive days =>P (X < 2)*P (X < 2) = P (X < 2)^2 = 0.0404^2 = 0.00163

    How would I determine if it's a rare event? Would I just compare it to 5 people entering the ICU for both cases?

    Thank You
     
    Last edited: Oct 31, 2008
  2. jcsd
  3. Oct 31, 2008 #2

    Mark44

    Staff: Mentor

    Your work looks fine (although a couple of signs are switched the fractions in part a -- corrected in following work).
    a) The probability is only about .04, so about 1 chance in 25. I'd say that's a fairly rare occurrence.
    b) The probability is much less, hence a much rarer event.
     
  4. Oct 31, 2008 #3
    Thank You.
     
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