If the hospital randomly samples the additional uninsured patient costs for 7 da
ID: 3368103 • Letter: I
Question
If the hospital randomly samples the additional uninsured patient costs for 7 days, what is the probability that at least 2 of these days will have additional uninsured patient costs that are less than or equal to 125.61? Round your answer to 6 decimal places.
Greenwood Hills General Hospital has determined the following probability distribution for X, which is the daily number of patients without health insurance that apply for treatment X: number of patients without insurance 0 P(X): probability 2 .20 .19 .56 .05 In addition to the costs of treating insured patients, the hospital faces additional costs, C, of treating uninsured patients, where C $100 + 25.61 X.Explanation / Answer
here P(C)=100+25.61X
hence for C<=125.61
100+25.61X<=125.61
X<=1
probability that for an individual day additional uninsured patient costs is less than or equal to 125.61 =P(X=1)
=0.19+0.56=0.75
henec from binomial distribution probability of at least 2 of these days will have additional uninsured patient costs that are less than or equal to 125.61 =P(Y>=2)=1-P(Y<=1)=1-(P(Y=0)+P(Y=1))
=1-(7C0 (0.75)0(0.25)7+7C1 (0.75)1(0.25)6) =1-(0.0001+0.00013)=0.9987
( please revert for any clarification required)