Question

The variance of the following probability distribution is:

\[ \begin{array}{|c|c|} \hline x & P(X) \\ \hline 0 & \frac{9}{16} \\ 1 & \frac{3}{8} \\ 2 & \frac{1}{16} \\ \hline \end{array} \]

• A. \(\frac{1}{8}\)
• B. \(\frac{5}{8}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{3}{8}\)

Hint:

The variance of a probability distribution is calculated by subtracting the square of the expected value from the expected value of \(X^2\).

Answer 1

The Correct Option is D

The variance of a probability distribution is computed using the formula: \[ \text{Variance} = E(X^2) - (E(X))^2 \] Where: - \(E(X)\) represents the expected value (mean). - \(E(X^2)\) represents the expected value of \(X^2\). Calculation Steps: Step 1: Determine \(E(X)\), the expected value: \[ E(X) = \sum x \cdot P(X) \] \[ E(X) = (0 \times \frac{9}{16}) + (1 \times \frac{3}{8}) + (2 \times \frac{1}{16}) \] \[ E(X) = 0 + \frac{3}{8} + \frac{2}{16} = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \] Step 2: Determine \(E(X^2)\), the expected value of \(X^2\): \[ E(X^2) = \sum x^2 \cdot P(X) \] \[ E(X^2) = (0^2 \times \frac{9}{16}) + (1^2 \times \frac{3}{8}) + (2^2 \times \frac{1}{16}) \] \[ E(X^2) = 0 + \frac{3}{8} + \frac{4}{16} = \frac{3}{8} + \frac{1}{4} = \frac{3}{8} + \frac{2}{8} = \frac{5}{8} \] Step 3: Compute the variance: \[ \text{Variance} = E(X^2) - (E(X))^2 \] \[ \text{Variance} = \frac{5}{8} - \left(\frac{1}{2}\right)^2 = \frac{5}{8} - \frac{2}{8} = \frac{3}{8} \] Result: The variance for the provided probability distribution is \(\frac{3}{8}\).