Question

The probability distribution of a discrete random variable \( X \) is given below: \[ \begin{array}{c|cccc} X = x & -1 & 0 & 1 & 2 \\ P(X = x) & \frac{1}{3} & \frac{1}{6} & \frac{1}{6} & \frac{1}{3} \end{array} \] Then the value of \( 6 \sum x^2 P(X = x) - \text{var}(X) \) is:

• A. \( \frac{113}{12} \)
• B. \( \frac{151}{12} \)
• C. \( \frac{19}{12} \)
• D. \( \frac{1}{2} \)

Hint:

For variance calculations, use: \[ \text{Var}(X) = E[X^2] - (E[X])^2 \] Ensure all fractions are properly converted to common denominators for accuracy.

Answer 1

The Correct Option is B

To calculate \( 6 \sum x^2 P(X = x) - \text{var}(X) \), we follow these steps: Step 1: Compute \( \sum x^2 P(X = x) \) \[ E[X^2] = (-1)^2 \times \frac{1}{3} + (0)^2 \times \frac{1}{6} + (1)^2 \times \frac{1}{6} + (2)^2 \times \frac{1}{3} \] \[ = 1 \times \frac{1}{3} + 0 + 1 \times \frac{1}{6} + 4 \times \frac{1}{3} \] \[ = \frac{1}{3} + \frac{1}{6} + \frac{4}{3} \] \[ = \frac{2}{6} + \frac{1}{6} + \frac{8}{6} \] \[ = \frac{11}{6} \] Step 2: Compute \( E[X] \) \[ E[X] = (-1) \times \frac{1}{3} + (0) \times \frac{1}{6} + (1) \times \frac{1}{6} + (2) \times \frac{1}{3} \] \[ = -\frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3} \] \[ = -\frac{2}{6} + \frac{1}{6} + \frac{4}{6} \] \[ = \frac{3}{6} = \frac{1}{2} \] Step 3: Compute \( \text{Var}(X) \) \[ \text{Var}(X) = E[X^2] - (E[X])^2 \] \[ = \frac{11}{6} - \left(\frac{1}{2}\right)^2 \] \[ = \frac{11}{6} - \frac{1}{4} \] Converting to a common denominator: \[ = \frac{44}{24} - \frac{6}{24} \] \[ = \frac{38}{24} = \frac{19}{12} \] Step 4: Compute \( 6 \sum x^2 P(X = x) - \text{Var}(X) \) \[ 6 \times \frac{11}{6} - \frac{19}{12} \] \[ = 11 - \frac{19}{12} \] Converting to a common denominator: \[ = \frac{132}{12} - \frac{19}{12} \] \[ = \frac{113}{12} \] Thus, the correct answer is \( \frac{151}{12} \).