Question

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 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

Answer:

(a) 0.04006

(b) 0.12924

(c) 0.29116

(d) $225.76

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($\mu=348, \sigma^{2} =87^{2}$)

The standard normal z score distribution is given by;

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

(a) Probability that the cost will be more than $500 = P(X > $500)

    P(X > 500) = P( $\frac{X-\mu}{\sigma}$ > $\frac{500-348}{87}$ ) = P(Z > 1.75) = 1 - P(Z <= 1.75)

                                                      = 1 - 0.95994 = 0.04006

(b) Probability that the cost will be less than $250 = P(X < $250)

    P(X < 250) = P( $\frac{X-\mu}{\sigma}$ < $\frac{250-348}{87}$ ) = P(Z < -1.13) = 1 - P(Z <= 1.13)

                                                      = 1 - 0.87076 = 0.12924

(c) Probability that the cost will be between $300 and $400 = P(300<X<400)

    P($300 < X < $400) = P(X < $400) - P(X <= $300)

    P(X < 400) = P( $\frac{X-\mu}{\sigma}$ < $\frac{400-348}{87}$ ) = P(Z < 0.60) = 0.72575

    P(X <= 300) = P( $\frac{X-\mu}{\sigma}$ <= $\frac{300-348}{87}$ ) = P(Z <= -0.55) = 1 - P(Z < 0.55)

                                                      = 1 - 0.70884 = 0.29116

Therefore, P($300 < X < $400) = 0.72575 - 0.29116 = 0.43459

(d) Now, it is given that the cost to a patient is in the lower 8% of charges for this medical service, i.e.;

     P(Z < x) = 0.08  ⇒ P(Z < $\frac{X-348}{87}$ ) = 0.08

From Z table it is found that the z score has a probability of 8% at a critical value of x = -1.40507

which means,  $\frac{X-348}{87}$ = -1.40507

                          X = (-1.40507 * 87) + 348 = $225.76

Therefore, the cost of this patient’s emergency room visit is $225.76 .