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Unformatted text preview: Statistics 403 Problem Set 5 Due in lab on Friday, October 15th 1. The following questions are about a population of people, some of whom are home owners, and some of whom are renters. (a) Suppose 65% of the population are homeowners, the homeowners have a mean annual income of $52,000, and the mean annual income of the population is $45,000. What is the mean annual income of the renters? Solution: Apply the double expectation theorem: 45000 = E (Income) = E Own E (Income  Own) = 0 . 65 · E (Income  Own = 1) + 0 . 35 · E (Income  Own = 0) = 0 . 65 · 52000 + 0 . 35 · E (Income  Own = 0) , where Own = 1 means the person owns their home, and Own = 0 means that the person rents their home. Solving we get E (Income  Own = 0) = 32000 . (b) Suppose the mean annual incomes of the homeowners and renters are $43,000 and $29,000, respectively, and the mean annual income in the population is $33,000. What proportion of the population owns their home? Solution: 33000 = E Own E (Income  Own) = p · E (Income  Own = 1) + (1 p ) · E (Income  Own = 0) = p · 43000 + (1 p ) · 29000 = 14000 p + 29000 . Thus p = 4000 / 14000 ≈ . 29. (c) Suppose the home owners have a mean annual income of $47,000 and the standard 1 deviation of home owners’ incomes is $7,000. Then suppose that the renters have a mean annual income of $31,000 and the standard deviation of renters’ incomes is $4,500. If 72% of the population are renters, what is the standard deviation of incomes in the population? Solution: Apply the law of total variation to get the variance of incomes in the population: var(Income) = E Own var(Income  Own) + var Own E (Income  Own) The first term is E Own var(Income  Own) = 0 . 28 · 7000 2 + 0 . 72 · 4500 2 = 28300000 . The overall mean is . 28 · 47000 + 0 . 72 · 31000 = 35480 ....
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 Winter '12
 KerbyShedden
 Statistics, Standard Deviation, Variance, $7,000, $8,500, $4,500, $55,000

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