A local hospital estimates that the number of patients admitted daily to the eme

Question

A local hospital estimates that the number of patients admitted daily to the

emergency room has a Poisson probability distribution with a mean of 4.0. What is the

probability that on a given day

a) only 2 patients will be admitted?

b) at most 6 patients will be admitted?

c) no one will be admitted?

d) What is the standard deviation of the number of patients admitted?

e) For each patient admitted, the expected daily operational expenses to the hospital

are $800. If the hospital wants to be 94.1% sure of meeting daily expenses, how

much money should it retain for operational expenses daily?

Explanation / Answer

the mean "a" gives the value of the distribution's parameter :
a = 4
and so
P( X = k) = e^-a . a^k /k!
a) P(only 2 patients will be admitted) = P(X = 2)
.... = e^-4 . 4^2 /2!
... = 8 e^-4
finally :
P(only 2 patients will be admitted) # 0,1465... = 14.65%

P(X = 0) = e^-4 . 1 ( by "convention" : 0! = 1)
P(X = 0) # 0,0183156... = 1.83%

the SD of a Poisson distribution is sqrt(a) = 2

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