Sampling distribution of the mean confusion Say we have some population and decide to randomly take a sample of size $n$ from this population.  What does it then mean to talk about the distribution of the sample mean?
In other words, what do we mean by the distribution of the sample mean here?

I’m quite confused about this as from what I understood from my textbook, it doesn’t make sense to talk about a distribution of a sample mean in this scenario (since we are looking at one sample).  Obviously, I’m wrong, but I’m not seeing why. 
 A: With the help of @Stephen, it seems you have a good intuitive understanding
of the idea of 'standard error.' Here is an elementary statement of some
basic facts so that you will know the precise terminology involved.
Estimation of $\mu$ when $\sigma$ is known. If you are starting with a normal population with unknown mean $\mu$ and known SD $\sigma,$ then the sample mean $\bar X$ is the estimator of $\mu.$ In this
context $SD(\bar X) = \sigma/\sqrt{n}$ is called the standard error of the mean.
Perhaps the most common application is that a 95% confidence interval for $\mu$ is of the form $\bar X \pm 1.96\sigma/\sqrt{n}.$
Estimation of $\mu$ when $\sigma$ is unknown. If you are starting with a normal population with unknown mean $\mu$ and variance $\sigma^2,$ then $\bar X$ is estimator of $\mu$ and the sample variance
$S^2$ is the estimator of $\sigma^2;$ with $E(\bar X) = \mu,\,$ $E(S^2) = \sigma^2,\,$ $Var(\bar X) = \sigma^2/n,\,$ and $SD(\bar X) = \sigma/\sqrt{n}.$
Then the standard deviation of the mean $SD(\bar X) = \sigma \sqrt{n}$ is again
called the standard error of the mean. However, because $\sigma$ is unknown, the
the estimated standard error of the mean is $S/\sqrt{n}.$ By abbreviation or sloppiness, the word "estimated" is sometimes dropped and one refers to $S/\sqrt{n}$ as the "standard error of the mean."
In this case, a 95% CI for $\mu$ is of the form $\bar X \pm t^*S/\sqrt{n},$
where $t^*$ (cutting probability 2.5% from the upper tail of Student's t
distribution with $n - 1$ degrees of freedom) can be found from printed
tables or using software.
A: Suppose $n=1$ and $X_1,X_2,X_3$ are independent and identically distributed and $$X_1 = \begin{cases} 1 & \text{with probability } 1/2, \\ 2 & \text{with probability } 1/2. \end{cases}$$
The sample mean is $(X_1+X_2 + X_3)/3.$
You have
$$
(X_1,X_2,X_3) = \begin{cases} (1,1,1) \\ (1,1,2) \\ (1,2,1) \\ (1,2,2) \\ (2,1,1) \\ (2,1,2) \\ (2,2,1) \\ (2,2,2) \end{cases}
$$
each with probability $1/8.$ Therefore
$$
\frac{X_1+X_2 + X_3} 3 = \begin{cases} 1 & \text{with probability } 1/8, \\ 4/3 & \text{with probability } 3/8, \\ 5/3 & \text{with probability } 3/8, \\ 2 & \text{with probability } 1/8. \end{cases}
$$
That is the probability distribution of the sample mean.
A: As a relative newcomer to statistics, this confused me also. Similar wording is often used when comparing a single sample to the sampling distribution. When wording like this is used, they do not mean the sampling distribution was generated from that one sample, they are referring to the sampling distribution generated from all possible samples of the same size as the single sample.
