Question

X is the random variable that can take any one of the values 0, 1, 7, 11 and 12. The probability mass function for \( X \) is \[ P(X = 0) = 0.4, \quad P(X = 1) = 0.3, \quad P(X = 7) = 0.1, \quad P(X = 11) = 0.1, \quad P(X = 12) = 0.1. \] Then, the variance of \( X \) is

• A. 20.81
• B. 28.40
• C. 31.70
• D. 10.89

Hint:

To calculate the variance, first find the expected value of \( X \), then the expected value of \( X^2 \), and subtract the square of the expected value from the expected value of \( X^2 \).

Answer 1

The Correct Option is A

The variance of a random variable \( X \) is given by:

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

Where:

• \( E(X) \) is the expected value (mean) of \( X \).
• \( E(X^2) \) is the expected value of \( X^2 \).

We will first calculate \( E(X) \) and \( E(X^2) \).

Step 1: Calculate \( E(X) \)

The expected value \( E(X) \) is given by:

\[ E(X) = \sum_{x} x \cdot P(x). \]

Substituting the values from the given probability mass function:

\[ E(X) = 0 \times 0.4 + 1 \times 0.3 + 7 \times 0.1 + 11 \times 0.1 + 12 \times 0.1. \] \[ E(X) = 0 + 0.3 + 0.7 + 1.1 + 1.2 = 3.3. \]

Step 2: Calculate \( E(X^2) \)

The expected value of \( X^2 \) is given by:

\[ E(X^2) = \sum_{x} x^2 \cdot P(x). \]

Substituting the values from the probability mass function:

\[ E(X^2) = 0^2 \times 0.4 + 1^2 \times 0.3 + 7^2 \times 0.1 + 11^2 \times 0.1 + 12^2 \times 0.1. \] \[ E(X^2) = 0 + 0.3 + 49 \times 0.1 + 121 \times 0.1 + 144 \times 0.1. \] \[ E(X^2) = 0 + 0.3 + 4.9 + 12.1 + 14.4 = 31.7. \]

Step 3: Calculate the Variance

Now that we have both \( E(X) \) and \( E(X^2) \), we can calculate the variance:

\[ \text{Var}(X) = E(X^2) - [E(X)]^2 = 31.7 - (3.3)^2. \] \[ \text{Var}(X) = 31.7 - 10.89 = 20.81. \]

Thus, the variance of \( X \) is \( 20.81 \).

The correct answer is option (A).