Answer to Question #97334 in Statistics and Probability for mani

Question

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?

Answer 1

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

$$X\sim N(\mu, \sigma^2), \mu=\$328, \sigma=\$92$$

Then

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

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

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

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

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

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

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

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

$$\approx0.783072-0.380431=0.402641$$

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?

$$p=0.08$$

$$P(Z\leq z)=0.08=>z\approx-1.405071$$

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

$$X\approx\$ 198.73$$