A random variable X has the following probability distribution:

The variance of X will be

1.29

The correct answer is Option (2) → 1.29

Given probability distribution:

$X$: 0, 1, 2, 3

$P(X)$: 0.2, 0.1, 0.3, 0.4

Mean (expected value) $\mu = E(X) = \sum X \cdot P(X)$

$\mu = 0 \cdot 0.2 + 1 \cdot 0.1 + 2 \cdot 0.3 + 3 \cdot 0.4 = 0 + 0.1 + 0.6 + 1.2 = 1.9$

Variance: $\sigma^2 = E(X^2) - (E(X))^2$

$E(X^2) = \sum X^2 \cdot P(X) = 0^2 \cdot 0.2 + 1^2 \cdot 0.1 + 2^2 \cdot 0.3 + 3^2 \cdot 0.4 = 0 + 0.1 + 1.2 + 3.6 = 4.9$

$\sigma^2 = 4.9 - (1.9)^2 = 4.9 - 3.61 = 1.29$