Question 1161598: The average per capita spending on health care in the 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 by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean).sd or (6000-5274)/600=1.21
probability z>1.21 is 0.1131
middle 50% occurs when z=-0.6745 to z=+0.6745 (25th and 75th percentiles
z*sd=+/- 404.69. Add to and subtract that from the mean to get the limits
($4869.31, $5678.69)
Check on a TI-83 by 2ndVARS 2 ENTER(4869.31,5678.69,5274,600) ENTER That gives you the probability of that interval.
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