Generating uniformly distributed random variables with given mean and deviation I don't understand maths very well so sorry for my newbie question.
Imagine I have a bank and clients of this bank. Clients arrive following a uniform distribution with $3.5$ mean and $1.3$ standard deviation.
How can I calculate a random number with this data?
I know that $f(x) = \frac{1}{(b-a)}$ on the interval $[a,b]$. So 
$$P(c\leq X \leq d) = \frac{(d-c)}{(b-a)}$$
In my example, $a$ and $b$ will be generated using a random generator but I don't understand how I can generate a random number following a uniform distribution with my mean and standard deviation.
 A: First will come the pretty complete theory. Then we look at your particular situation.
Theory: We need an expression for the variance of $X$. The variance  is $E(X^2)-(E(X))^2$. For $E(X^2)$, we need to calculate
$$\int_a^b \frac{x^2}{b-a}\,dx,$$
which is $\frac{b^3-a^3}{3(b-a)}$. This simplifies to $\frac{b^2+ab+a^2}{3}$.  
I imagine that you know that  $E(X)=\frac{b+a}{2}$. One can do this by integration, but it is clear by symmetry that the mean is halfway between $a$ and $b$. 
So we know that the variance is $\frac{b^2+ab+a^2}{3}-\frac{(b+a)^2}{4}$. Bring to a common denominator, simplify. We get that 
$$\text{Var}(X)=\frac{(b-a)^2}{12} \tag{$\ast$}.$$
More simply, you can search under uniform distribution, say on Wikipedia. They will have the expression $(\ast)$ for the variance of $X$.
Your problem:  We know that  $\frac{b+a}{2}=3.5$. We also know that the standard deviation of $X$ is $1.3$, so the variance is $(1.3)^2=1.69$.
So, by $(\ast)$,  $\frac{(b-a)^2}{12}=1.69$, and therefore $b-a=\sqrt{(12)(1.69)}\approx 4.5033$. We also know that $b+a=(2)(3.5)=7$. Now that we know $b-a$ and $b+a$, it is easy to find $a$ and $b$.
For your simulation, presumably you are starting from a random number generator that generates numbers that are more or less uniformly distributed on $[0,1)$. If $U$ represents the output of such a generator, we simulate $X$ by using $a+(b-a)U$. And we now know $a$ and $b$.
Added If you want a general formula instead of a procedure, let $\mu=\frac{a+b}{2}$ be the mean, and $\sigma=\frac{b-a}{\sqrt{12}}=\frac{b-a}{2\sqrt{3}}$ be the standard deviation. Then $\frac{b-a}{2}=\sqrt{3}\,\sigma$. 
We get $a=\frac{b+a}{2}-\frac{b-a}{2}=\mu-\sqrt{3}\,\sigma$. It follows that we can take
$$X=\mu-\sqrt{3}\,\sigma + (2\sqrt{3}\,\sigma) U=\mu+(\sqrt{3}\,\sigma)(2U-1).$$
A: Let $X$ be a uniformly distributed continuous random variable with mean $\mu$ and standard deviation $\sigma$. Then 
$$
    X = \mu + 2 \sqrt{3} \sigma \left( U - \frac{1}{2} \right)
$$
where $U$ is a random variables uniformly distributed on a unit interval. 
Indeed:
$$
  \mathbb{E}(X) = \mu + \sqrt{3} \sigma \mathbb{E}(2U-1) = \mu
$$
$$
  \mathbb{Var}(X) = (2 \sqrt{3} \sigma)^2 \mathbb{Var}(U-1/2) = 12 \sigma^2 \mathbb{Var}(U) = \sigma^2
$$
and, assuming both $x_1$ and $x_2$ in the domain of $X$,
$$ \begin{eqnarray}
   \mathbb{P}( x_1 < X \leqslant x_2) &=& 
   \mathbb{P}\left( x_1 < \mu + 2 \sqrt{3} \sigma \left( U - \frac{1}{2} \right) \leqslant x_2\right) \\
  &=& \mathbb{P}\left( \frac{1}{2}+\frac{x_1 - \mu}{2 \sqrt{3} \sigma} < U \leqslant \frac{1}{2} + \frac{x_2 - \mu}{2 \sqrt{3} \sigma}  \right) \\
   &=& x_2-x_1
\end{eqnarray}
$$
