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Probability - Random Samples

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 10th percentile approximately equal to \$111,600, 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 \$55,000?

b) If random samples of size 20 were selected, can you use the methods discussed in this chapter to calculate the probabilities requested in (a)-(c)? Explain.
Yes. These probabilities are independent of sample size.

Solution Preview

(a) x = 55000, mu = 57045, sigma = 25000

z = (x - mu)/(sigma/sqrt n)

z = (55000 - ...

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