Question #97334
19. (a) In an article about the cost of health care, Money magazine reported that a visit to a hospital emergency room for something as simple as a sore throat has a mean cost of $328 (Money, January 2009). Assume that the cost for this type of hospital emergency room visit is normally distributed with a standard deviation of $92. Answer the following questions about the cost of a hospital emergency room visit for this medical service.
a. What is the probability that the cost will be more than $500?
b. What is the probability that the cost will be less than $250?
c. What is the probability that the cost will be between $300 and $400?
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?
1
Expert's answer
2019-10-25T11:32:22-0400

Let X=X= the cost for this type of hospital emergency room visit. Given that


XN(μ,σ2),μ=$328,σ=$92X\sim N(\mu, \sigma^2), \mu=\$328, \sigma=\$92

Then


Z=XμσN(0,1)Z={X- \mu\over \sigma}\sim N(0, 1)

a. What is the probability that the cost will be more than $500? 


P(X>500)=1P(X500)=P(X>500)=1-P(X\leq500)=

=1P(Z<50032892)10.9692=0.0308=1-P(Z<{500- 328\over 92})\approx1-0.9692=0.0308

b. What is the probability that the cost will be less than $250? 


P(X<250)=P(Z<25032892)0.1983P(X<250)=P(Z<{250- 328\over 92})\approx0.1983

c. What is the probability that the cost will be between $300 and $400? 


P(300<X<400)=P(Z<40032892)P(Z<30032892)P(300<X<400)=P(Z<{400- 328\over 92})-P(Z<{300- 328\over 92})\approx

0.7830720.380431=0.402641\approx0.783072-0.380431=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=0.08p=0.08

P(Zz)=0.08=>z1.405071P(Z\leq z)=0.08=>z\approx-1.405071

X32892=1.405071{X- 328\over 92}=-1.405071

X$198.73X\approx\$ 198.73


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