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:

\[f(x) = \begin{cases} \frac{1}{b-a} & \text{for } a \le x \le b \\ 0 & \text{otherwise} \end{cases}\]

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 \(\frac{1}{\sqrt{12}}\).