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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.

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up vote 2 down vote accepted

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 $$X=\mu-\sqrt{3}\,\sigma + (2\sqrt{3}\,\sigma) U=\mu+(\sqrt{3}\,\sigma)(2U-1).$$

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Thanks for the great explanation. So given my mean of $3.5$ and sdt of $1,3$ and $U$ of $0,2$, $a$ would be $1,2483$, $b$ would be $5,7516$ and $X=2,14896$? – Favolas May 3 '12 at 8:58
@Favolas: I checked your calculation above. Agree. – André Nicolas May 3 '12 at 13:00
Thanks. Just one more question. Could this be done $X = mean - 0.5 + sdt * sqrt(12) * U$ ? – Favolas May 3 '12 at 14:27
I will add a formula of this type (but different) to the post. – André Nicolas May 3 '12 at 14:32
Thanks. Don't quite understand the $Subtract$ part but thanks for the last formula. Comparing yours with mine give quite different results. Many many thanks – Favolas May 3 '12 at 15:21

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 $X$, $$ \begin{eqnarray} \mathbb{P}( x_1 < X \leqslant x_2) &=& \mathbb{P}\left( x_1 < \mu + 2 \sqrt{3} \sigma \left( U - \frac{1}{2} \right) \leqslant x_2\right) \\ &=& \mathbb{P}\left( \frac{1}{2}+\frac{x_1 - \mu}{2 \sqrt{3} \sigma} < U \leqslant \frac{1}{2} + \frac{x_2 - \mu}{2 \sqrt{3} \sigma} \right) \\ &=& x_2-x_1 \end{eqnarray} $$

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