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Let μ be the mean annual salary of Major League Baseball players for 2002. Assume that the standard deviation of the salaries of these players is $107,000.

What is the probability that the 2002 mean salary of a random sample of 50 baseball players was within $20,000 of the population mean, μ . Assume that n/N ≤ 0.05 .

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Is this a homework problem? What have you done so far? What is $n$? What is $N$? Do you know anything else about the distribution? –  Calvin Lin Jan 13 '13 at 6:45
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Not making assumptions about the distribution of salaries, I'd say the question is impossible to answer. You do know that your sample will have the same expectation value as the population mean, and that the variance of the mean of the sample can be calculated, taking advantage of the smallness of $n/N$, as

$$\sqrt{\frac{\sigma^2}{n}(1-\frac{n-1}{N-1})} \approx \sqrt{\frac{\sigma^2}{n}} = \$15,132$$

If we take the distribution to be normal, then answering your question is a walk in the park from here. But that's not stated, so I don't think we can do better than Chebyshev's inequality. $\$20,000$ is $k=1.322$ of the value we got for the standard deviation of the sample mean. So the probability of that the mean of a sample is beyond $\$20,000$ cannot be more than $1/k^2=0.572$. So the sought probability is at least $42.8\%$.

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