Is there a plateau-shaped distribution? I am looking for a distribution where the probability density decreases quickly after some point away from the mean, or in my own words a "plateau-shaped distribution".  
Something in between the Gaussian and the uniform.
 A: See my "Devil's tower" distribution in here [1]: 
$f(x) = 0.3334$, for $|x| < 0.9399$;
$f(x) = 0.2945/x^2$, for $0.9399 \leq |x| < 2.3242$; and
$f(x) = 0$, for $2.3242 \leq |x|$.

The "slip-dress"distribution is even more interesting.
It is easy to construct distributions having whatever shape you want.
[1]: Westfall, P.H. (2014)
   "Kurtosis as Peakedness, 1905 – 2014. R.I.P."
   Am. Stat. 68(3): 191–195. doi:10.1080/00031305.2014.917055
   public access pdf: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4321753/pdf/nihms-599845.pdf
A: You may be looking for distribution known under the names of generalized normal (version 1), Subbotin distribution, or exponential power distribution. It is parametrized by location $\mu$, scale $\sigma$ and shape $\beta$ with pdf
$$ \frac{\beta}{2\sigma\Gamma(1/\beta)} \exp\left[-\left(\frac{|x-\mu|}{\sigma}\right)^{\beta}\right] $$
as you can notice, for $\beta=1$ it resembles and converges to Laplace distribution, with $\beta=2$ it converges to normal, and when $\beta = \infty$ to uniform distribution.

If you are looking for software that has it implemented, you can check normalp library for R (Mineo and Ruggieri, 2005). What is nice about this package is that, among other things, it implements regression with generalized normally distributed errors, i.e. minimizing $L_p$ norm.

Mineo, A. M., & Ruggieri, M. (2005). A software tool for the exponential power distribution: The normalp package. Journal of Statistical Software, 12(4), 1-24.
A: Lots of nice answers. The solution proffered here has 2 features: (i) that it has a particularly simple functional form, and (ii) that the resulting distribution necessarily produces a plateau-shaped pdf (not just as a special case). I'm not sure if this already has a name in the literature, but absent same, let us call it a Plateau distribution with pdf $f(x)$:
$$f(x) = k \frac{1}{1 + x^{2 a}}  \quad \quad \text{for } x \in \mathbb{R}$$
where:


*

*parameter $a$ is a positive integer, and 

*$k$ is a constant of integration:  $k = \frac{a}{\pi} \sin \left(\frac{\pi}{2 a}\right)$


Here is a plot of the pdf, for different values of parameter $a$:

.
As parameter $a$ becomes large, the density tends towards a Uniform(-1,1) distribution. The following plot also compares to a standard Normal (gray dashed):

A: Another one (EDIT: I simplified it now. EDIT2: I simplified it even further, though now the picture doesn't really reflect this exact equation):
$$f(x) = \frac{1}{3 \cdot \alpha} \cdot
\log{\left(
  \frac{\cosh{\left(\alpha \cdot a\right)}+
    \cosh{\left(\alpha \cdot x\right)}}
    {\cosh{\left(\alpha \cdot b\right)}+
    \cosh{\left(\alpha \cdot x\right)}}
  \right)}
$$
Clunky, I know, but here I took advantage of the fact that $\log(\cosh(x))$ approaches a line as $x$ increases.
Basically you have control over how smooth is the transition ($alpha$). If $a = 2$ and $b = 1$ I guarantee it's a valid probability density (sums to 1). If you choose other values then you'll have to renormalize it.

Here is some sample code in R:
f = function(x, a, b, alpha){
  y = log((cosh(2*alpha*pi*a)+cosh(2*alpha*pi*x))/(cosh(2*alpha*pi*b)+cosh(2*alpha*pi*x)))
  y = y/pi/alpha/6
  return(y)
}

f is our distribution. Let's plot it for a sequence of x
plot(0, type = "n", xlim = c(-5,5), ylim = c(0,0.4))
x = seq(-100,100,length.out = 10001L)

for(i in 1:10){
  y = f(x = x, a = 2, b = 1, alpha = seq(0.1,2, length.out = 10L)[i]); print(paste("integral =", round(sum(0.02*y), 3L)))
  lines(x, y, type = "l", col = rainbow(10, alpha = 0.5)[i], lwd = 4)
}
legend("topright", paste("alpha =", round(seq(0.1,2, length.out = 10L), 3L)), col = rainbow(10), lwd = 4)

Console output:
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = 1"
#[1] "integral = NaN" #I suspect underflow, inspecting the plots don't show divergence at all
#[1] "integral = NaN"
#[1] "integral = NaN"

And plot:

You could change a and b, approximately the start and end of the slope respectively, but then further normalization would be needed, and I didn't calculate it (that's why I'm using a = 2 and b = 1 in the plot).
A: @StrongBad's comment is a really good suggestion. The sum of a uniform RV and gaussian RV can give you exactly what you're looking for if you pick the parameters right. And it actually has a reasonably nice closed form solution.
The pdf of this variable is given by the expression:
$$\dfrac{1}{4a}\left[\mathrm{erf}\left(\dfrac{x+a}{\sigma\sqrt{2}}\right)-\mathrm{erf}\left(\dfrac{x-a}{\sigma\sqrt{2}}\right)
\right]$$
$a$ is the "radius" of the zero-mean uniform RV.
$\sigma$ is the standard deviation of the zero-mean gaussian RV.

A: If you are looking for something very simple, with a central plateau and the sides of a triangle distribution, you can for instance combine N triangle distributions, N depending on the desired ratio between the plateau and the descent. Why triangles, because their sampling functions already exist in most languages. You randomly sort from one of them. 
In R that would give:
library(triangle)
rplateau = function(n=1){
  replicate(n, switch(sample(1:3, 1), rtriangle(1, 0, 2), rtriangle(1, 1, 3), rtriangle(1, 2, 4)))
}
hist(rplateau(1E5), breaks=200)



A: Here's a pretty one: the product of two logistic functions.
(1/B) * 1/(1+exp(A*(x-B))) * 1/(1+exp(-A*(x+B)))

This has the benefit of not being piecewise.
B adjusts the width and A adjusts the steepness of the drop off. Shown below are B=1:6 with A=2. Note: I haven't taken the time to figure out how to properly normalize this.

A: There's an infinite number of "plateau-shaped" distributions. 
Were you after something more specific than "in between the Gaussian and the uniform"? That's somewhat vague.
Here's one easy one: you could always stick a half-normal at each end of a uniform:

You can control the "width" of the uniform relative to the scale of the normal so you can have wider or narrower plateaus, giving a whole class of distributions, which include the Gaussian and the uniform as limiting cases.
The density is:
$\frac{h}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2\sigma^2}(x-\mu+w/2)^2} 
\mathbb{I}_{x\leq \mu-w/2}  \\
+ \:\frac{h}{\sqrt{2\pi}\sigma}\quad\mathbb{I}_{\mu-w/2< x\leq \mu+w/2} \\
+ \frac{h}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2\sigma^2}(x-\mu-w/2)^2} 
\mathbb{I}_{x > \mu+w/2}   $
where $h = \frac{1}{1 + w/(\sqrt{2\pi}\sigma)}$
As $\sigma \to 0$ for fixed $w$, we approach the uniform on $(\mu-w/2,\mu+w/2)$ and as $w \to 0$ for fixed $\sigma$ we approach $N(\mu,\sigma^2)$.
Here are some examples (with $\mu=0$ in each case):

We might perhaps call this density a "Gaussian-tailed uniform".
