Let X= the cost for this type of hospital emergency room visit. Given that
X∼N(μ,σ2),μ=$328,σ=$92 Then
Z=σX−μ∼N(0,1) a. What is the probability that the cost will be more than $500?
P(X>500)=1−P(X≤500)=
=1−P(Z<92500−328)≈1−0.9692=0.0308 b. What is the probability that the cost will be less than $250?
P(X<250)=P(Z<92250−328)≈0.1983 c. What is the probability that the cost will be between $300 and $400?
P(300<X<400)=P(Z<92400−328)−P(Z<92300−328)≈
≈0.783072−0.380431=0.402641 d. If the cost to a patient is in the lower 8% of charges for this medical service, what was the cost of this patient’s emergency room visit?
p=0.08
P(Z≤z)=0.08=>z≈−1.405071
92X−328=−1.405071
X≈$198.73
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