Question

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: \(Mean\ (\mu) = \sum_{i=1}^{n} x_i p(x_i)\); \(Variance\ (\sigma^2) = \sum_{i=1}^{n} (x_i - \mu)^2 p(x_i)\)

• A. 3/16
• B. 3/4
• C. 9/16
• D. 6/16

Hint:

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 \(p(1-p)\).

Answer 1

The Correct Option is A

Step 1: Compute the mean using the given formula.
\[ \mu = \sum x_i\, p(x_i) = (0)\left(\frac{1}{4}\right) + (1)\left(\frac{3}{4}\right) = \frac{3}{4} \]
Step 2: Compute each squared deviation from the mean.
For \(x_1 = 0\): \((0 - 3/4)^2 = (3/4)^2 = 9/16\). For \(x_2 = 1\): \((1 - 3/4)^2 = (1/4)^2 = 1/16\).
Step 3: Weight each squared deviation by its probability and sum.
\[ \sigma^2 = \left(\frac{9}{16}\right)\left(\frac{1}{4}\right) + \left(\frac{1}{16}\right)\left(\frac{3}{4}\right) = \frac{9}{64} + \frac{3}{64} = \frac{12}{64} = \frac{3}{16} \] \[ \boxed{\sigma^2 = \frac{3}{16}} \]