A discrete random variable X has the Poisson distribution if its probability mass function is given by
p(x) = (e-μμx)/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.
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]