Calculating Variance for Uniform Distribution

The problem provides a probability density function (PDF) for a continuous random variable $X$:

$f(x) = \begin{cases} 0.5, & 0 < x < 2 \\ 0.0, & \text{Otherwise} \end{cases}$

This describes a uniform distribution on the interval $(0, 2)$. We are given that the mean, $E[X] = 1$, and we need to find the variance, $\text{Var}(X)$.

Variance Formula

The variance is calculated using the formula:

$ \text{Var}(X) = E[X^2] - (E[X])^2 $

Calculating Expected Value of $X^2$

First, we calculate $E[X^2]$:

$ E[X^2] = \int_{-\infty}^{\infty} x^2 f(x) dx $

Substitute the given PDF:

$ E[X^2] = \int_{0}^{2} x^2 (0.5) dx $

$ E[X^2] = 0.5 \int_{0}^{2} x^2 dx $

$ E[X^2] = 0.5 \left[ \frac{x^3}{3} \right]_{0}^{2} $

$ E[X^2] = 0.5 \left( \frac{2^3}{3} - \frac{0^3}{3} \right) $

$ E[X^2] = 0.5 \left( \frac{8}{3} \right) = \frac{4}{3} $

Calculating Variance

Now, use the variance formula with $E[X^2] = \frac{4}{3}$ and the given $E[X] = 1$:

$ \text{Var}(X) = E[X^2] - (E[X])^2 $

$ \text{Var}(X) = \frac{4}{3} - (1)^2 $

$ \text{Var}(X) = \frac{4}{3} - 1 $

$ \text{Var}(X) = \frac{1}{3} $

Final Result

Convert the fraction to a decimal:

$ \text{Var}(X) = \frac{1}{3} \approx 0.3333... $

Rounding to two decimal places, the variance is 0.33.

This value falls between 0.32 and 0.34.