Variance Calculation for Random Variable X

The problem asks for the variance of a random variable $X$ with a given sample space and probabilities.

Step 1: Calculate Mean ($\mu$)

The mean (or expected value) $\mu$ is calculated using the formula $\mu = \sum x_i p(x_i)$.

The random variable $X$ takes values $0$ and $1$ with probabilities $P(X=0) = \frac{1}{4}$ and $P(X=1) = \frac{3}{4}$.

$ \mu = (0 \times P(X = 0)) + (1 \times P(X = 1)) $

Substitute the probabilities:

$ \mu = (0 \times \frac{1}{4}) + (1 \times \frac{3}{4}) $

$ \mu = 0 + \frac{3}{4} = \frac{3}{4} $

Step 2: Calculate Variance ($\sigma^2$)

The variance $\sigma^2$ is calculated using the formula $\sigma^2 = \sum (x_i - \mu)^2 p(x_i)$.

$ \sigma^2 = (0 - \mu)^2 P(X=0) + (1 - \mu)^2 P(X=1) $

Substitute the values of $\mu$, $P(X=0)$, and $P(X=1)$:

$ \sigma^2 = (0 - \frac{3}{4})^2 \times \frac{1}{4} + (1 - \frac{3}{4})^2 \times \frac{3}{4} $

Calculate the terms:

$ \sigma^2 = (-\frac{3}{4})^2 \times \frac{1}{4} + (\frac{1}{4})^2 \times \frac{3}{4} $

$ \sigma^2 = (\frac{9}{16} \times \frac{1}{4}) + (\frac{1}{16} \times \frac{3}{4}) $

$ \sigma^2 = \frac{9}{64} + \frac{3}{64} $

$ \sigma^2 = \frac{12}{64} $

Step 3: Simplify Variance

Simplify the resulting fraction:

$ \sigma^2 = \frac{12}{64} = \frac{3 \times 4}{16 \times 4} = \frac{3}{16} $

The variance of the random variable $X$ is $\frac{3}{16}$.