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# Proportion and Probability of Household Incomes

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According to the U.S. Census Bureau, the mean household income in the United States in 2000 was \$63,091 and the median household income was \$50,054 (U.S. Census Bureau, www.census.gov, September 2011). The variability of household income is quite large, with the 85th percentile (i.e., top 15%) approximately equal to \$100,000, and an overall standard deviation of approximately \$25,000. Suppose random samples of 225 households were selected.

a) What proportion of the sample means would be below \$50,054?

b) What proportion of the sample means would be above \$100,000?

c) Why is the probability you calculated in (b) so much lower than 0.15, even though 15% of individual households have incomes above \$100,000?

d) If we select random samples of size 20, can we calculate the probabilities requested in (a) and (b)? Explain.

##### Solution Summary

This solution provides steps to determine proportion of sample means.

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