Answer to Question #349150 in Statistics and Probability for nurul

Question

The average per capita spending on health care in United States is $5274. If the standard

deviation is $600 and the distribution of health care spending is approximately normal, what is the probability that a randomly selected person spends more than $6000? Find the limits of the middle 50% of individual health care expenditures.

Answer 1

a)

$$P(X>6000)=1-P(Z\le\dfrac{6000-5274}{600})$$

$$=1-P(Z\le1.21)=0.1131$$

b)

$$P(X<x_1)=P(Z<\dfrac{x_1-5274}{600})=\dfrac{1-0.5}{2}$$

$$\dfrac{x_1-5274}{600}=-0.67449$$

$$x_1=5274-600(0.67449)$$

$$x_1=4869.306$$

$$x_2=5274+600(0.67449)$$

$$x_2=5678.694$$

$$P(4869.306<X<5678.694)=0.5$$