Sampling Distribution; Statistics. Please verify my answer Professor earns average $ \text{\$} 65,500$  per year with standard deviation of $\text{\$}3,500$.
Random sample of $64$.
a. Describe sampling distribution of sample mean $\bar{x}$  of average salary of these professors.
Since sample mean is the same as population mean, I said it is unbiased, since its mean is true to its parameter.
b. within what limit would you expect sample mean to fall with probability $0.95$
I used $1.96\text{SE} = 1.96\times 3,500/\sqrt{65}$ = $857.5$
therefore $95\text{% }$to fall within $\pm 857.5$ of the mean
c. obtain probability that $\bar{x}$ is greater than $66,000$
I used  $(66,000 - 65,500) / 3,500 = 0.1428$
and gotten $Z$ value of $0.5557$
$1-0.5557 = 0.4443$
therefore $0.4443$ chance that $\bar{x}$ is greater than $66,000$
Can anyone shed some light if I have gotten these questions correctly. 
Much appreciated!
 A: (a) I agree with your answer (plus what David said).
It is unbiased where $\mathbb E(\overline{x})=\mu$.
(b) It should be $1.96\frac{\sigma}{\sqrt{n}}$
I.e.  $1.96 \frac{3500}{\sqrt{64}} = 857.5$ ($65$ is probably a typo as you have the correct answer.)
(c) As the previous answerer has pointed out, your denominator needs to be changed from $\sigma$ to $\frac{\sigma}{\sqrt{n}}$.
$Z= \frac{\overline{x}-\mu}{\frac{σ}{\sqrt{n}}}$
$Z= \frac{66,000-65,500}{\frac{3500}{\sqrt{64}}}$
$Z≈1.143$
We now need to look for $\mathbb P(Z>1.143)$
If we want to use tables, we need to note the following:
Since we can only get $\varPhi(1.14)$ or  $\varPhi(1.15)$ from our tables, I would suggest that you interpolate the value so that you can get an approximation of  $\varPhi(1.145)$ which is closer to $1.143$.


*

*$1-\varPhi(1.14)=  1-0.87206 = 0.12794$

*$1-\varPhi(1.15)=  1-0.87493 = 0.12507$


∴ $\mathbb P(\overline{x} > 66,000)≈ \frac{0.12794+0.12507}{2}$
A: I think for part (a), a complete description of "sampling distribution of sample mean" would include the standard error (that is, the standard deviation of $\bar x$) in addition to the mean of $\bar x$. A mean is not a distribution.
(Technically neither is a mean and a standard error/deviation, but I'm supposing you are
allowed to assume a normal distribution without saying so.)
For part (b), you correctly computed the standard error 
(except for typing "$65$" when you apparently meant $64$) 
and used it to find the correct interval of salaries.
For part (c) you apparently forgot that you're dealing with the distribution of the sample mean rather than the distribution of the population, and you gave the probability that a random individual $x$ is greater than $66,000.$
This is part of the reason why I think part (a) asked for the distribution
rather than just the mean of $\bar x$; you were meant to remember that
$95$% of the time $\bar x$ will fall within $1.96$ standard errors of the mean,
not just within $1.96$ standard deviations (using the standard deviation of
the original population).
A: $a)$: True.
$b)$: It should be $\sqrt{64}$, but correct value $857.5$.
$c)$: Wrong. It should be $P\left(\bar{x} > 66,000\right) = P\left(z > \dfrac{66,000-65,500}{\dfrac{3,500}{\sqrt{64}}}\right)$ = ....
