Central limit theorem when the mean is not constant I've heard that there is a soft version or interpretation of the CLT that says —in summary— that it can be applied also to a sequence of independent real-valued random variables that share the same distribution shape but not the mean (for instance, a sequence of $n$ normal variables $X_i \sim N(\mu_i, \sigma)$ where $\mu_i$ can be different for each $i$).
Is that right? Where can I find it (textbook or article)?
And, a last one: In this case (where $X_i$ are independently and equally distributed except for their means), what would the distribution of $Y=\frac{1}{n}\sum_{i=1}^n{X_i}$ approximately be, according to CLT?

EDIT:
I want to stress that $X_i\sim N(\mu_i,\sigma)$ was just an example.
What I have is $X_i \sim F_i$, where $F_i$ stands for the CDF of $X_i$, and where all $F_i$'s are equal to the same distribution (with a known constant variance $\sigma^2$) except for the fact that they have different averages $\mu_i$.

EDIT:
Just to make it clearer, what I would like to know is whether or not it is correct to state this:
$Y=\frac{1}{n}\sum_{i=1}^n{X_i}$ is approximately distributed as $N(\mu_Y,\sigma_Y)$, where


*

*$\mu_Y$ can be estimated as $\bar{x} = \frac{x_1+ \cdots + x_n}{n}$ and

*$\sigma_Y$ can be estimated as $\frac{s}{\sqrt{n}}$, $s$ being the sample (quasi)standard deviation,


using a random sample $(x_1, \dots, x_n)$ of $(X_1, \dots, X_n)$.
 A: A precise reference for the Lindeberg-Feller CLT is Theorem 4.12 in Foundations of Modern Probability by Olav Kallenberg.  I paraphrase it below:

Let $(\xi_{n,j}; j = 1,\ldots,n)_{n=1}^\infty$ be a triangular array
  of row-wise independent random variables with mean $0$ and
  $\sum_{j=1}^n \mathbb{E}[\xi_{n,j}^2] \to 1$ as $n\to\infty$.
Suppose that for any $\epsilon > 0$ we have 
  $\sum_{j=1}^n \mathbb{E}[\xi_{n,j}^2; |\xi_{n,j}|>\epsilon] \to 0$.  (*)
Then $\sum_{j=1}^n \xi_{n,j} \overset{d}{\to} N(0,1)$.

Suppose the $X_i$ are arbitrary independent random variables. Let $\mu_j = \mathbb{E}[X_j]$ and $\sigma_j^2 = Var[X_j]$.
Take $\xi_{n,j} = \frac{X_j - \mu_j}{\sqrt{\sum_{i = 1}^n \sigma_i^2}}$.  Then the theorem says
$$
\frac{1}{\sqrt{\sum_{i = 1}^n \sigma_i^2}} \sum_{j = 1}^n (X_j - \mu_j) \overset{d}{\to} N(0,1),
$$
and thus
$$
\frac{1}{\sqrt{n}} \sum_{j=1}^n (X_j - \mu_j) \overset{d}{\to} N(0,\sigma^2)
$$
if the limit $\sigma^2 = \lim_{n\to\infty}\sum_{i=1}^n \sigma_i^2 / n$ exists.  To put in another way, the sample mean $Y_n = \frac{1}{n} \sum_{i=1}^n X_i$ satisfies
$$
\sqrt{n}(Y_n - \frac{1}{n}\sum_{i=1}^n \mu_i)  \overset{d}{\to} N(0,\sigma^2),
$$
i.e. $Y_n$ has approximate distribution $N(\frac{1}{n}\sum_{i=1}^n \mu_i, \sigma^2/n)$.  However, be warned that the theorem does not say exactly what the rate of convergence is!
I leave it as an exercise for you to see what the Lindeberg condition (*) is.
