Probability Question Finding Salary 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!
 A: 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
$$ \begin{eqnarray}
  \mathbb{P}\left( \frac{1}{m} \left( x_1 + \cdots + x_m \right) \le z \right) &=& \Phi\left( \frac{z - \mu_m }{\sigma_m} \right) = \Phi\left( \frac{z-\mu}{\sigma} \sqrt{m} \right) \\ &=& 
 \Phi\left( \frac{57,000 -60,000}{6,500} \sqrt{35} \right) = \Phi(-2.73) = 0.0032
\end{eqnarray}
$$
