Does a uniform random distribution become a normal distribution? I have $N$ numbers from uniform random distribution. Now I have done a transformation on these $N$ numbers as: each $x$ is converted to $\frac{(x-\mu_N)}{\sigma_N}$. Where $\mu_N$ is the mean of $N$ numbers and $\sigma_N$ is the standard deviation of those $N$ numbers. Does the converted distribution follow a normal distribution?
 A: Try it:
> X <- runif(500000)
> hist(X, 20)
> hist((X-mean(X))/sd(X), 20)


The plot above is slightly misleading, as there is a plotting artifact that makes it look as if it was not exactly uniform, but it is. To convince yourself, draw a plot for uniform distribution with bounds equal to the lowest and highest value from the sample above, the plot would look the same. Since R has a problem with drawing histograms for data that does not have "round" bounds, I've drawn the probability distribution function of this distribution.
> hist(runif(500000, min((X-mean(X))/sd(X)), max((X-mean(X))/sd(X))), 20, freq=F)
> curve(dunif(x, min((X-mean(X))/sd(X)), max((X-mean(X))/sd(X))), min((X-mean(X))/sd(X)), max((X-mean(X))/sd(X)), col="red", add=T, lw=3)


It's flat and it's exactly the same distribution you got by scaling the data. What the scaling did is it shifted each of the values by mean and multiplied it by the standard deviation. If you transform a random variable in such a way, it will change the mean, standard deviation, and bounds of the distribution, but not the shape of the distribution. It creates a location-scale variant of uniform distribution. If $f(x)$ is the probability distribution function of uniform distribution, then after subtracting mean $a$ and dividing by standard deviation $b$, you get
$$
g(x) = \tfrac{1}{b} f\big(\tfrac{x-a}{b}\big)
$$
so it cannot have a different shape than $f$. It's just wider and shorter, so its area under the curve remains equal to $1$, as required by a probability distribution.
The result is not surprising, as subtracting the mean and dividing by standard deviation doesn't transform any distribution to Gaussian.
A: In order for your sample to follow a normal distribution, you may want to feed it to a normal distribution's quantile function, which is:
$\mu+\sigma\sqrt2 erf^{-1}(2x-1)$ where $erf^{-1}$ is the inverse error function. Or you could simply feed you sample into the function qnorm() in R:
N <- 1000
sample <- runif(1000)
normal <- qnorm(sample, mean=10, sd=4)
hist(normal)

In both cases, you choose the mean ($\mu$) and standard distribution ($\sigma$) yourself.

