Answer: Solve Poisson Distribution Prob | Rare Event?

[cse63146]

[SOLVED] Poisson Distribution

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

$$P(X = K) = \frac{\mu^k e^{-\mu}}{k!}$$

The Attempt at a Solution

a) $$P (X < 2) = P(X=0) + P(X=1) = \frac{5^0 e^{-5}}{0!} + \frac{5^{1} e^{-5}}{1!}<br /> = e^{-5} + 5e^{-5} = 6e^{-5} = 0.0404$$

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

 

[Mark44]

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.

 

[cse63146]

Thank You.