Are there any inverse distribution graph that looks like this? I want to generate random numbers according to a distribution curve like the inverse cummulative normal curve. But the curve should allow parameters to adjust the parameters as shown in the image, namely, the slopes (A, B, C), min and max values, between 0.0 and 1.0 in a meaningful manner like in terms of variance, mean etc. the usual stuff.
Is there a well known solution for this curve in terms of statistical distribution?

 A: This is one way to go, I will include some numerical calculations using R.  Suppose you have a continuous random variable $X$, which is real-valued, and have a density $f(x)$ on the interval $[x_\text{min}, x_\text{max}]$.  Then the distribution function $F(x) = \int_{x_\text{min}}^x f(u) \; du $ is continuos and increasing from zero to one on that interval, and the inverse function $Q(p) = F^{-1}(p)$ is also increasing, on the interval $[0, 1]$.
Now use implicit differention on $F(Q(p))=p$ to get $F'(Q(p)) Q'(p)=1$ or
$$
   Q'(p) = \frac1{f(x)}.
$$
This you can use to read some information from your graph: In the area "flat A" the density is high, "growth B" the density is lower, and "taper C" density increases again. This indicates a bimodal distribution!  If you take the interval $x_\text{min}, x_\text{max}$  to be $[0,1]$, you could try to fit some (asymmetric) beta distribution.  
To experiment numerically with R, we start to read off some values from your graph.  I do this rather haphazardly, you can redo with more precision.
> x
 [1] 0.085 0.090 0.095 0.110 0.150 0.180 0.220 0.250 0.500 0.900 0.980 1.000
> p
 [1] 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 1.00

Fit a monotonic spline:
> Q  <-  splinefun(p,x,method="monoH.FC")

You can plot these function in R.
Then make a more densely populated table, and use it to plot the inverse function $F(x)$:
> pp  <-  seq(0,1,by=0.01)
> xx <-  Q(pp)
> plot(xx,pp,type="l")

Gives the following plot of the cumulative distribution function:

You could continue, for example using numerical differentiation to find the density, but that I leave for you.  From this data we can calculate the expextation of $X$, $\mu=\text{E}(X)$.  We use the formula for non-negative random variables
$$
   \mu = \int_0^\infty (1-F(t))\; dt
$$
(which in our case reduces to an integral from zero to one because $1-F(t)$ is zero for $t$ larger than one.)  In R we can approximate this as

mu <- sum((1-pp))*0.01
  mu 
  1 0.505

When matching to a beta distribution with parameters $a,b$ (refer to wikipedia) we can use then the result that  $\mu = \frac{a}{a+b}$, resulting in the choice $a=1.02 b$, or approximately $a=b$.  Then you can plot beta distributions for various choices of $a$ ($b$) and see if it matches.  
