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[SOLVED] Poisson Distribution
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?
[tex]P(X = K) = \frac{\mu^k e^{-\mu}}{k!}[/tex]
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
Homework Statement
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?
Homework Equations
[tex]P(X = K) = \frac{\mu^k e^{-\mu}}{k!}[/tex]
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
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