A random variable \(X\) has the sample space \(\{0,1\}\). The probability \(P(X = 0) = 1/4\) and \(P(X = 1) = 3/4\). What is the variance of the random variable? Hint: \(Mean\ (\mu) = \sum_{i=1}^{n} x_i p(x_i)\); \(Variance\ (\sigma^2) = \sum_{i=1}^{n} (x_i - \mu)^2 p(x_i)\)
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First find the mean using \(\mu = \sum x_i p(x_i)\), then apply the variance formula \(\sigma^2 = \sum (x_i - \mu)^2 p(x_i)\), or use the Bernoulli shortcut \(p(1-p)\).
Step 1: Recognize this as a Bernoulli random variable. $X$ takes only the values 0 and 1, with $P(X=1) = p = 3/4$. This is exactly the Bernoulli distribution with parameter $p = 3/4$. Step 2: Use the variance formula $\sigma^2 = E[X^2] - (E[X])^2$. Since $X$ only takes values 0 and 1, $X^2$ equals $X$ itself, so $E[X^2] = E[X] = p = 3/4$. Step 3: Substitute into the formula. \[ \sigma^2 = E[X^2] - (E[X])^2 = \frac{3}{4} - \left(\frac{3}{4}\right)^2 = \frac{3}{4} - \frac{9}{16} = \frac{12}{16} - \frac{9}{16} = \frac{3}{16} \] Step 4: Cross-check with the standard Bernoulli variance formula. For a Bernoulli variable, $\sigma^2 = p(1-p) = \dfrac{3}{4}\times\dfrac{1}{4} = \dfrac{3}{16}$, which matches. \[ \boxed{\sigma^2 = \frac{3}{16}} \]
