Poisson distribution

Question

 A hospital director who has been studying daily emergency admission over 7 years has concluded that they are distributed according to the Poisson law, hospital records reveals that emergency admission has average 3 per day during this period. If the director is correct in assuming a Poisson distribution then find the probability that,a) exactly 1 emergency admission will occur in a given day. b) No emergency admission will occur. c) At least and at most three emergency cases will be admitted on a particular day.

Leave the answer in 5 decimals places .

Answer 1

Hello Benjamin,

A discrete random variable X has the Poisson distribution if its probability mass function is given by

p(x) = (e^-μμ^x)/x!, for x = 0,1,2,3,.....

In the above formula, the parameter μ is the mean value of X. Let X represent the number of emergency admissions to the hospital in a given day. The hospital administrator claims that X follows a Poisson distribution with mean μ = 3.

Part a) Use the formula above, with x = 1.

p(1) = (e^-33¹)/1! = 3e^-3

p(1) ≅ .14936 (rounded to 5 decimal places)

Part b) Use the Poisson distribution with x = 0.

p(0) = (e^-33⁰)/0! = e^-3

p(0) ≅ .04979

Hope that helps! (Your statement for part c appears to be incomplete, so I wasn't sure what you were supposed to find. If you need some more help with part c, just let me know.

William

Part c) Finding the probability that there are at most 3 emergency admissions is equivalent to find the probability that X≤3. X can only take on whole number values, so if X≤3, then X=0,1,2, or 3. Thus,

P(X≤3) = p(0) + p(1) + p(2) + p(3).

We already calculate p(0) and p(1). Now plug x = 2 and x = 3 into the formula above to get p(2) and p(3). (Details omitted). We then have

P(At most 3) = P(X≤3) = .04979 + .14936 + .22404 + .22404

= P(X≤3) = .64723

"At least 3 emergency admissions" means X≥3. But if X≥3, then X=3 or X=4 or X=5, etc. That is, X≥3 corresponds to infinitely many possible values of X. Instead of calculating the probability of X≥3 directly, calculate the probability of the complement of the event X≥3 and subtract this from 1. The complement of "X≥3" is "X<3", which is equivalent to X = 0,1, or 2. Therefore,

P(At least 3) = P(X≥3) = 1 - P(X<3)

= 1 - [p(0) + p(1) + p(2)]

= 1 - [.04979 + .14936 + .22404]

= .57681