Question

The average per capita spending on health care in the United Statesis $5274. If the standard deviation is $600 and the distribution of health care spending is approximatelynormal, what is the probability that a randomly selected person spends more than $6000? Find the limits of the middle 50% of individual health careexpenditures.

Answer 1

Find Z:
Z = (x - μ) / σ
Z = (6000 - 5274) / 600
Z = 1.21

Now look at a Z table for normal distributions.
Z = 1.21 corresponds to a probability of 0.8869.
This means that P(X ≤ 6000) = 0.8869.
So P(X > 6000) = 1 - 0.8869 = ______

The middle 50% is bordered by probabilities of 25% (0.25) and 75% (0.75).
From the Z table, these values most closely match Z values of -0.67 and +0.67.

Lower limit:
Z = (x - μ) / σ
-0.67 = (x - 5274) / 600
Solve for x.

Repeat with +0.67 to find x for the upper limit.