Question

Let \( X \) be the discrete random variable representing the number (\( x \)) appeared on the face of a biased die when it is rolled. The probability distribution of \( X \) is as follows: \[ X = x: \quad 1, \, 2, \, 3, \, 4, \, 5, \, 6 \] \[ P(X = x): \quad 0.1, \, 0.15, \, 0.3, \, 0.25, \, k, \, k \] The variance of \( X \) is:

• A. \( 1.64 \)
• B. \( 1.93 \)
• C. \( 2.16 \)
• D. \( 2.28 \)

Hint:

To calculate the variance, ensure you compute both \( \mathbb{E}(X^2) \) and \( (\mathbb{E}(X))^2 \). Double-check the consistency of the probability distribution.

Answer 1

The Correct Option is B

The sum of all probabilities must equal 1: \[ \sum P(X = x) = 1. \] Step 1: Solve for \( k \)
From the given probabilities, we have the equation: \[ 0.1 + 0.15 + 0.3 + 0.25 + k + k = 1. \]
Simplifying this equation: \[ 0.8 + 2k = 1. \]
Solving for \( k \): \[ k = 0.1. \] Step 2: Compute the mean \( \mu = \mathbb{E}(X) \)
The mean is defined as: \[ \mu = \sum X P(X). \]
Substitute the values for \( X \) and \( P(X) \): \[ \mu = (1)(0.1) + (2)(0.15) + (3)(0.3) + (4)(0.25) + (5)(0.1) + (6)(0.1). \]
Simplify the expression: \[ \mu = 0.1 + 0.3 + 0.9 + 1 + 0.5 + 0.6 = 3.4. \] Step 3: Compute \( \mathbb{E}(X^2) \)
The expected value of \( X^2 \) is: \[ \mathbb{E}(X^2) = \sum X^2 P(X). \]
Substitute the values: \[ \mathbb{E}(X^2) = (1^2)(0.1) + (2^2)(0.15) + (3^2)(0.3) + (4^2)(0.25) + (5^2)(0.1) + (6^2)(0.1). \]
Simplify: \[ \mathbb{E}(X^2) = (1)(0.1) + (4)(0.15) + (9)(0.3) + (16)(0.25) + (25)(0.1) + (36)(0.1). \] \[ \mathbb{E}(X^2) = 0.1 + 0.6 + 2.7 + 4 + 2.5 + 3.6 = 13.5. \] Step 4: Compute the variance \( \sigma^2 \)
The variance is calculated as: \[ \sigma^2 = \mathbb{E}(X^2) - (\mathbb{E}(X))^2. \] Substitute the computed values: \[ \sigma^2 = 13.5 - (3.4)^2. \] Simplifying the result: \[ \sigma^2 = 13.5 - 11.56 = 1.94 \, (\text{approximately } 1.93). \] Final Answer: \[ \boxed{1.93} \]