Question: According to the U.S. Census Bureau, the mean household income in the United States in 2000 was $57,045 and the median household income was $42,148 (U.S. Census Bureau, "Money Income in the United States: 2000," www.census.gov, September 2001). The variability of household income is quite large, with the 90th percentile (i.e., top 10%) approximately equal to $111,600, and an overall standard deviation of approximately $25,000. Suppose random samples of 225 households were selected.

1. What proportion of the sample means would be below $55,000?
2. What proportion of the sample means would be above $111,600?
3. Why is the probability you calculated in (b) so much lower than 0.10, even though 10% of individual households have incomes above $111,600?
4. If random samples of size 20 were selected, can you calculate the probabilities requested in (a) and (b)? Explain.

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