Question:

A random variable X has the following probability distribution:
X-2-1012
P(X)0.20.10.30.20.2

The variance of X will be:

Updated On: Nov 15, 2024
  • 0.1
  • 1.42
  • 1.89
  • 2.54
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Variance is calculated as:

\[ \text{Var}(X) = E(X^2) - [E(X)]^2. \]

First, calculate \( E(X) \):

\[ E(X) = \sum X \cdot P(X) = (-2)(0.2) + (-1)(0.1) + (0)(0.3) + (1)(0.2) + (2)(0.2). \]

\[ E(X) = -0.4 - 0.1 + 0 + 0.2 + 0.4 = 0.1. \]

Next, calculate \( E(X^2) \):

\[ E(X^2) = \sum X^2 \cdot P(X) = (-2)^2(0.2) + (-1)^2(0.1) + (0)^2(0.3) + (1)^2(0.2) + (2)^2(0.2). \]

\[ E(X^2) = 4(0.2) + 1(0.1) + 0(0.3) + 1(0.2) + 4(0.2) = 0.8 + 0.1 + 0 + 0.2 + 0.8 = 1.9. \]

Finally, calculate the variance:

\[ \text{Var}(X) = E(X^2) - [E(X)]^2 = 1.9 - (0.1)^2 = 1.9 - 0.01 = 1.89. \]

Was this answer helpful?
0
0