A die is tossed once. If the random variable X is defined as \(X = \begin{cases}1, & \text{if the die results in an odd number}\\-1, & \text{if the die results in an even number}\end{cases}\), then the variance of X is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
Three of the six faces are odd and three are even, so
\(\displaystyle P(X = 1) = \frac{3}{6} = \frac{1}{2},\quad P(X = -1) = \frac{1}{2}\)
The mean:
\(\displaystyle E(X) = 1 \cdot \frac{1}{2} + (-1) \cdot \frac{1}{2} = 0\)
Since \(X^2 = 1\) in both cases,
\(E(X^2) = 1\)
The variance:
\(\mathrm{Var}(X) = E(X^2) - [E(X)]^2\)
\(= 1 - 0 = 1\)