I came across this derivation which I don't understand: If $X_1, X_2, ..., X_n$ are random samples of size n taken from a population of mean $\mu$ and variance $\sigma^2$, then
$\bar{X} = (X_1 + X_2 + ... + X_n)/n$
$E(\bar{X}) = E(X_1 + X_2 + ... + X_n)/n = (1/n)(E(X_1) + E(X_2) + ... + E(X_n))$
$E(\bar{X}) = (1/n)(\mu + \mu + ...n ~\text{times}) = \mu$
This is where I'm lost. The argument used is $E(X_i) = \mu$ because they are identically distributed. In reality this isn't true. Suppose I have a sample, $S=\{1,2,3,4,5,6\}$ and then if randomly select 2 numbers with replacement and repeat this procedure 10 times, then I get 10 samples: (5, 4) (2, 5) (1, 2) (4, 1) (4, 6) (2, 4) (6, 1) (2, 4) (3, 1) (5, 1). This is how it looks like for 2 random variables $X_1, X_2$. Now if I take the expectation value of $X_1$ I get,
$E(X_1) = 1.(1/10) + 2.(3/10) + 3.(1/10) + 4.(2/10) + 5.(2/10) + 6.(1/10) = 34/10 = 3.4$
But the expected value of the population is 3.5. What is actually wrong in my reasoning?