What is an intuitive meaning of $E(\overline { X } )$ and $Var(\overline { X } )$? Let $X$ be a random variable distributed over, for example say, the Binomial Distribution. Then $P(X)$ is the probability of getting $x$ successful trials in $n$ total trials. 
So I saw a notation that represents the mean of random variables that made me I feel sceptical about my understanding of all the notations I have known. So here's my understanding of the notations:

When it says the expectation of $X$, $E(X)$, does it mean over a long
  run, $E(X)$ is the likely number of successful trials? In other words,
  the expected value of $X$ is the expected number of successful trials
  we would expect in a long run?
When it says the variance of $X$, $Var(X)$, does it mean how spread
  out the probability of successful trials are? Like how far apart the
  probability between the successful trials are?

Now, here's the confusing part. I see a notation like this: $\overline { X } =\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ { X }_{ i } }  $ and this is called the mean of all the random variables. But it doesn't seem to make sense to me. $X$ is the random variable and carries the value that is the number of successful trials. The average of $X$ is like the average number of successful trials?
Does it then mean $\overline { X } =E(X)$?
Then, there is also the expectation of the mean of all the random variables, $E(\overline { X } )$. So does this represent the average of the average of all the random variables, which means $E(\overline { X } )=E(E(X))$? But at this point, I couldn't understand what it means intuitively. What does it mean here to say the average of the average of all random variables? 
Similarly, $Var(\overline { X } )$ is also a confusing term to me. Since $\overline { X } $ is just the average value, what spread does it have?
What is the intuitive meaning of this $\overline { X } $ mean of all random variables $X$ and what does this add on to the meaning of $E(\overline { X } )$ and $Var(\overline { X } )$?
 A: You have a vast population of people who have different heights, and you choose one at random.  That person's height is $X$.  $E(X)$ is the average height of everyone in the population.  Then you pick $20$ people at random.  Their heights are $X_1,\ldots,X_{20}$.  Their average height is $\bar{X} = (X_1+\cdots+X_{20})/20$.  That is a random variable because if you pick another set of $20$, it has a different value---thus it varies randomly.  The expected value $E(\bar{X})$ is the same as the expected value $E(X)$.  But the variance $\operatorname{var}(\bar{X})$ is smaller than the variance $\operatorname{var}(X)$, because on average one sample of $20$ differs less from another sample of $20$ than one individual differs from another individual.
A: You should think of $E(X)$ and $Var(X)$ (when they exist) as parameters which describe the distribution of the random variable $X$.  They are numbers.  They are not random variables.  So, although you can write $E(E(X))$, it's not very useful since the expected value of a constant is just the constant.  So you have $E(E(X)) = E(X)$.
Sometimes it is helpful to call $E(X)$ the "population average."
Now, you should think of $\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$ as simply a function of $n$ random variables, albeit a very useful one.  You shouldn't equate it in your mind with the population average $E(X)$ (a number).  Rather $\bar{X}_n$ is a random variable. 
Given some outcome $\omega$, you observe the sample values $x_1 = X_1(\omega), \dots, x_n = X_n(\omega)$ and you can compute the so called "sample average," which is a particular realization of $\bar{X}_n$, namely 
$$\bar{X}_n(\omega) = \frac{1}{n}\sum_{i=1}^n X_i(\omega) = \frac{1}{n}\sum_{i=1}^n x_i$$
Now, this sample average is a number, but it could be quite different from the population average, the number $E(X)$.
However, the Law of Large Numbers says that if $n$ is large enough then (any realization of) $\bar{X}_n$ will be close to $E(X)$.
And, yes, you can take the expected value of the random variable $\bar{X}_n$, as you suggest, and you will simply have
$$E(\bar{X}_n) = \frac{1}{n}\sum_{i=1}^n E(X_i) = E(X)$$
(The last equality assumes the $X_i$ all come from a distribution with mean $E(X)$.)
As for the variance, no, $Var(X)$ does not measure the spread between successes in repeated trials.  Rather, it is a measure of the spread of the possible values of the function $X$ about its mean $E(X)$.  If the variance is larger (smaller) that means there is a larger (smaller) chance $X$ will take on a value far from $E(X)$.  In your example, the possible values are 0 and 1, with probabilities $(1-p)$ and $p$, say.  The variance would be much bigger if you chose, say, 0 and 100 as the possible values for $X$.
Finally, note that the variance of $\bar{X}_n$ gets smaller as $n$ gets larger.  This is telling you that the larger $n$ is, the more likely the observed value of $\bar{X}_n$ will be close to $E(X)$.
A: The distinction among $\bar X$, $E(\bar X)$ and $E(X)$ is indeed not obvious and in my personal oppinion, often not well clarified in introductory statistics courses. The same is valid for $V(X)$ and $S^2$ (variance from a sample).
Let's use your suggestion and consider a $X$ r.v. Binomial(n,p). For any discrete r.v. $X$, if $E(X)$ and $V(X)$ exist, $$E(X)=\sum_{x\in \Omega_x} x P(X=x)\ \text{and}\ \ V(X)=E[(X-E(X))^2]$$
in which $\Omega_x$ is the set of possible values for $X$. You can understand $E(X)$ and $V(X)$ as theoretical constants associated with the PMF. For the Binomial$(n,p)$, $E(X)=n p$ and $V(X)=np(1-p)$, constants that depend on the parameters $n$ and $p$ from the Binomial. You can regard them as the theoretical mean and theoretical variance associated with the Binomial(n,p).
Now consider an independent sample size $m$ from this variable $X$ Binomial$(n,p)$ given by $X_1,X_2,\ldots,X_m$ and compute the average and the sample variance using:
$$\bar X_m=\dfrac{\sum_{i=1}^{m} X_i}{m}\ \ \text{and}\ \ S^2=\dfrac{\sum_{i=1} (X_i-\bar X_m)^2}{m-1}$$
Because of an important property of the operator $E(\cdot)$ it is true for any r.v.s $X$ and $Y$ and constants $a$ and $b$ that $$E(aX+bY)=aE(X)+bE(Y)$$.
Using this property we can compute $E(\bar X_m)$:
$$E(\bar X_m)=\dfrac{\sum_{i=1}^{m} E(X_i)}{m}=\dfrac{m E(X)}{m}=E(X),$$
as for any $X_i$ from the sample, $E(X_i)=E(X)$. With some additional effort we can also show that $E(S^2)=V(X).$
This is the reason why $E(\bar X_m)=E(X)$. And by the Law of Large Numbers (LLN) it is true that (under regularity conditions) $\bar X_m$ converges to $E(X)$ as $m\rightarrow \infty$ by statistical notions. By the same reason $S^2$ converges to $V(X)$ as $m\rightarrow \infty$.
Let's return to our Binomial$(n,p)$ with $E(X)=n p$ and $V(X)=np(1-p)$ and do some experimentation. If, for instance, $n=10$ and $p=0.3$, we have $E(X)=10\times 0.3=3$ and $V(X)=10\times 0.3\times 0.7=2.1$
Using a statistical software, let us obtain 2 samples size m=5 from this Binomial$(n=10,p=0.3)$ (using for instance rbinom$(5,10,0.3)$ in software R):

*

*Sample 1: 1,2,2,7,4  with $\bar X_5=3.2$ and $S^2=5.7$

*Sample 2: 4,4,1,1,1  with $\bar X_5=2.2$ and $S^2=2.7$
Notice that, in both samples, $\bar X_5$ and $S^2$ are not equal $E(X)=3$ and $V(X)=2.1$. The values will tend to be closer as the sample size increases, as theoreticaly expected by the LLN.
