The standard deviation of a uniformly distributed random variable between 0 and 1 is

Standard Deviation of Uniformly Distributed Random Variable

The question asks for the standard deviation of a random variable that is uniformly distributed between 0 and 1. This refers to a continuous uniform distribution, often denoted as $X\sim U(a,b)$, where $a$ and $b$ are the lower and upper bounds of the distribution, respectively.

Understanding Uniform Distribution

A continuous uniform distribution means that any value within a given interval $[a,b]$ has an equal probability of occurring. The probability density function (PDF) for a continuous uniform random variable $X$ over the interval $[a,b]$ is given by:

$$f(x)=\{\begin{array}{ll}\frac{1}{b-a}& \text{for }a\le x\le b\\ 0& \text{otherwise}\end{array}$$

Formulas for Uniform Distribution

For a continuous uniform distribution $X\sim U(a,b)$, the key statistical measures are defined as follows:

- Mean (Expected Value): The mean of a uniform distribution is the average of its bounds: $$E[X]=\frac{a+b}{2}$$
- Variance: The variance measures the spread of the distribution: $$Var(X)=\frac{(b-a{)}^2}{12}$$
- Standard Deviation: The standard deviation is the square root of the variance and is a commonly used measure of dispersion: $$\sigma =\sqrt{Var(X)}=\sqrt{\frac{(b-a{)}^2}{12}}=\frac{b-a}{\sqrt{12}}$$

**Calculating Standard Deviation for $X\sim U(0,1)$ **

In this specific problem, the random variable is uniformly distributed between 0 and 1. This means:

- The lower bound $a=0$
- The upper bound $b=1$

Now, we can substitute these values into the formula for the standard deviation:

$$\sigma =\frac{b-a}{\sqrt{12}}$$

Substituting $a=0$ and $b=1$:

$$\sigma =\frac{1-0}{\sqrt{12}}$$

$$\sigma =\frac{1}{\sqrt{12}}$$

Therefore, the standard deviation of a uniformly distributed random variable between 0 and 1 is $\frac{1}{\sqrt{12}}$.