A random variable X has the following probability distribution:
|
X |
0 |
1 |
2 |
3 |
|
P(X) |
0.2 |
0.1 |
0.3 |
0.4 |
The variance of X will be
Answer & explanation
Correct answer: option 2
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: $\sigma^2 = E(X^2) - (E(X))^2$
$E(X^2) = \sum X^2 \cdot P(X) = 0^2 \cdot 0.2 + 1^2 \cdot 0.1 + 2^2 \cdot 0.3 + 3^2 \cdot 0.4 = 0 + 0.1 + 1.2 + 3.6 = 4.9$
$\sigma^2 = 4.9 - (1.9)^2 = 4.9 - 3.61 = 1.29$