Does the central limit theorem apply to these probability density functions? Let's say you have n uniform random variables from 0 to 1. The distribution of the average of these variables approaches normal with increasing n according to the central limit theorem. What if however, instead of all the variables being random, one of them was guaranteed to be 0, and one of them was guaranteed to be 1. This would arise in the following case: Let's say you have n=7 randomly generated numbers from 0 to 1 and they are, from smallest to largest, [.1419 .1576 .4854 .8003 .9572 .9649 .9706]. If you were to subtract the smallest number from all of the numbers and then divide all the numbers by the new maximum you would end up with [0 .0189 .4145 .7945 .9838 .9931 1]. In this way you have a set of n numbers where n-2 of them are random and the other two are guaranteed to be 0 and 1. I would like to know whether the central limit theorem still applies to numbers generated in this way. By visual inspection using MATLAB, it actually appears to approach normal quicker than when the numbers are all random, but I would like a mathematical reason as to why, especially considering that the central limit theorem states that all the numbers must be random. 
 A: Denote $X_i, i=1,...,n$ the $U(0,1)$ independent RVs. The transformation described by the OP is (using the usual notation for order statistics),
$$Z_i = \frac {X_i-X_{n,(1)}}{X_{n,(n)}-X_{n,(1)}} = R_n^{-1}\cdot (X_i-X_{n,(1)})$$
where the double index in the minimum and maximum order statistic serve to remind us that they are functions of $n$. $R_n$ is the range of the untransformed sample.  
We want to consider
$$\frac 1n \sum_{i=1}^nZ_i \equiv \bar Z_n = R_n^{-1}\frac 1n \sum_{i=1}^nX_i - R_n^{-1}X_{n,(1)}$$
We have that
$$R_n^{-1} \xrightarrow{p} 1,\;\;\; \frac 1n \sum_{i=1}^nX_i \xrightarrow{p} \frac 12,\;\; X_{n,(1)}\xrightarrow{p} 0$$
So in all, applying Slutsky's lemma,
$$\bar Z_n \xrightarrow{p} \frac 12 = \text{plim} \frac 1n \sum_{i=1}^nX_i \equiv \text{plim}\bar X_n$$
So the sample average of the transformed sample is also a consistent estimator of the common expected value of the $X$'s.  Note that $\text{Var}(\bar X_n) = \frac 1{12n}$
Then, consider the manipulation
$$\sqrt{12n}\left(\bar Z_n - \frac 12\right) = \\R_n^{-1}\cdot \sqrt{12n}\left(\bar X_n - \frac 12 \right)  +\sqrt{12n}\left(\frac 12R_n^{-1} -\frac 12\right) -\sqrt{12n}R_n^{-1}X_{n,(1)}$$
We examine each of the three components in turn:  
A) By the CLT we have that  $\sqrt{12n}\left(\bar X_n - \frac 12 \right) \xrightarrow{d}\mathcal N (0,1)$. Since also $R_n^{-1} \xrightarrow{p} 1$, then by Slutsky the first term converges in distribution to $\mathcal N (0,1)$.
B) We can write
$$\sqrt{12n}\left(\frac 12R_n^{-1} -\frac 12\right) = \sqrt{3}\left(\frac {n(1-R_n)}{\sqrt nR_n}\right)$$
In Dasgupta 2008 ch. 8 p. 108 Example 8.12, one can find for the sample range from an i.i.d. sample of $U(0,1)$ uniforms that $n(1-R_n) \xrightarrow{d} \frac 12 \mathcal \chi^2(4)$). So the numerator above converges while the denominator goes to infinity. So the whole term goes to zero.  
C) We know that the minimum order statistic from a sample of non-negative random variables, needs to be scaled by $n$ in order to converge in distribution (see this post). In other words convergence is "fast", and scaling the third term only by $\sqrt n$ doesn't cut it. Therefore we have that  $\sqrt{12n}R_n^{-1}X_{n,(1)} \rightarrow0$.
So, we conclude that
$$\sqrt{12n}\left(\bar Z_n - \frac 12\right) \xrightarrow{d} \mathcal N(0,1)$$
as does $\bar X_n$, for the same shifting and scaling.
A: Say your sample is $(x_1, \ldots, x_n)$, you are looking at the random variable 
$$z = \frac{\bar{x} - \min_i x_i}{\max_i x_i - \min_i x_i}$$
In the case of the uniform distribution, $z$ converges almost surely to $\bar{x}$, and since $\bar{x}$ converges in distribution to a normal distribution, so does $z$.
