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The population mean annual salary for environmental complicance specialists is about \$60,000. A random sample of 35 specialists is drawn from this population. How would I find the probability that the mean salary of the sample is less than \$57,000? Assume standard deviation = \$6,500. Thank you!

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If you assume that population distribution is a normal distribution with mean $\mu = \$ 60,000$, and standard deviation $\sigma = \$6,500$. Let $\{x_1, \ldots, x_m \}$ be sample of size $m$, where each $x_i$ is i.i.d. from $\mathcal{N}(\mu, \sigma)$. You are asking to compute $$ \mathbb{P}\left( \frac{1}{m} \left( x_1 + \cdots + x_m \right) \le z \right) $$ But the sum of normal variables has normal distribution. To determine it, one needs to compute its mean and variance: $$ \mu_m = \mathbb{E}\left( \frac{1}{m} \left( x_1 + \cdots + x_m \right) \right) = \frac{1}{m} \left( \underbrace{\mu + \cdots + m}_{\text{m times}} \right) = \mu $$ $$ \sigma_m^2 = \mathbb{Var} \left( \frac{1}{m} \left( x_1 + \cdots + x_m \right) \right) = \frac{1}{m^2} \left( \underbrace{\sigma^2 + \cdots + \sigma^2}_{\text{m times}} \right) = \frac{\sigma^2}{m} $$ Thus $$ \begin{eqnarray} \mathbb{P}\left( \frac{1}{m} \left( x_1 + \cdots + x_m \right) \le z \right) &=& \Phi\left( \frac{z - \mu_m }{\sigma_m} \right) = \Phi\left( \frac{z-\mu}{\sigma} \sqrt{m} \right) \\ &=& \Phi\left( \frac{57,000 -60,000}{6,500} \sqrt{35} \right) = \Phi(-2.73) = 0.0032 \end{eqnarray} $$

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This is very helpful, thank you, but it is not the correct way of doing this problem, and is not the right answer. –  math Dec 15 '11 at 6:34
    
@math I was using the wrong numbers, so the final probability was coming out way off. –  Sasha Dec 15 '11 at 7:06
    
+1 for the calculation which gives $0.011455964$ when you change $57000$ to $57500$ as @math says is the correct figure to be used in comments on InterestedGuest's answer (though the main question remains unedited). I don't know why math says that this is not the correct way of doing the problem. In my opinion, this answer is not deserving of the down-vote that someone has given it. –  Dilip Sarwate Dec 15 '11 at 13:36
    
Another answer (initially incorrect but corrected later) to this question has been deleted, presumably by its author InterestedGuest. In the comments (now also deleted) on that answer, the OP admitted that what was desired was the probability that the sample mean was no larger than $57500$, not $57000$, and that the numerical value of this probability was known to him (from an answer sheet or back of the book?) as $0.114\ldots$. The OP's allegation that "it is not the correct way of doing this problem, and is not the right answer." is sheer nonsense. –  Dilip Sarwate Dec 16 '11 at 14:24
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