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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 $348. Assume that the cost for this type of hospital emergency room visit is normally distributed with a standard deviation of $87. Answer questions 3 to 6 about the cost of a hospital emergency room visit for this medical service. What is the probability that the cost will be less than $500? g

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Answer:

Probability that the cost will be less than $500 is 0.95994 .

Step-by-step explanation:

We are given that Money magazine reported that a visit to a hospital emergency room for something as simple as a sore throat has a mean cost of $348 with a standard deviation of $87.

Let X = cost of a hospital emergency room visit for this medical service

So, X ~ N([tex]\mu=348, \sigma^{2} =87^{2}[/tex])

The standard normal z score distribution is given by;

              Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

(a) Probability that the cost will be less than $500 = P(X < $500)

    P(X < 500) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{500-348}{87}[/tex] ) = P(Z < 1.75) = 0.95994 {from z table}

Therefore, the probability that the cost will be less than $500 is 0.95994 .

                                               

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