Question

\[ \begin{array}{c|c} X = x & P(X = x) \\ \hline 1 & 3K^2 \\ 3 & K \\ 5 & K^2 \\ 2 & 2K \end{array} \]

• A. \(\frac{9}{4}\)
• B. \(\frac{25}{8}\)
• C. \(\frac{27}{16}\)
• D. \(\frac{15}{16}\)

Hint:

When calculating variance, always ensure that you follow the correct steps for computing \( E[X] \) and \( E[X^2] \), then subtract the square of the expected value from the second moment to get the variance.

Answer 1

The Correct Option is D

Step 1: Ensure that the probabilities sum to 1. The total probability must sum to 1, so we write the equation: \[ 3K^2 + K + K^2 + 2K = 1 \] Simplify the equation: \[ 3K^2 + K^2 + K + 2K = 1 \] \[ 4K^2 + 3K = 1 \] Solve this quadratic equation: \[ 4K^2 + 3K - 1 = 0 \] Using the quadratic formula: \[ K = \frac{-3 \pm \sqrt{3^2 - 4 \times 4 \times (-1)}}{2 \times 4} = \frac{-3 \pm \sqrt{9 + 16}}{8} = \frac{-3 \pm \sqrt{25}}{8} \] \[ K = \frac{-3 \pm 5}{8} \] Thus, we have two possible values for \( K \): \[ K = \frac{2}{8} = \frac{1}{4} \quad \text{or} \quad K = \frac{-8}{8} = -1 \] Since \( K \) must be non-negative, we take \( K = \frac{1}{4} \). 

Step 2: Calculate the expected value \( E[X] \). The expected value \( E[X] \) is calculated as: \[ E[X] = 1 \cdot 3K^2 + 3 \cdot K + 5 \cdot K^2 + 2 \cdot 2K \] Substitute \( K = \frac{1}{4} \): \[ E[X] = 1 \cdot 3 \left( \frac{1}{4} \right)^2 + 3 \cdot \frac{1}{4} + 5 \cdot \left( \frac{1}{4} \right)^2 + 2 \cdot 2 \cdot \frac{1}{4} \] \[ E[X] = 3 \cdot \frac{1}{16} + 3 \cdot \frac{1}{4} + 5 \cdot \frac{1}{16} + 2 \cdot \frac{1}{2} \] \[ E[X] = \frac{3}{16} + \frac{3}{4} + \frac{5}{16} + 1 \] \[ E[X] = \frac{3 + 5}{16} + \frac{12}{16} + \frac{16}{16} = \frac{20}{16} + \frac{12}{16} + \frac{16}{16} = \frac{48}{16} = 3 \]

Step 3: Calculate the expected value of \( X^2 \), i.e., \( E[X^2] \). The expected value \( E[X^2] \) is calculated as: \[ E[X^2] = 1^2 \cdot 3K^2 + 3^2 \cdot K + 5^2 \cdot K^2 + 2^2 \cdot 2K \] Substitute \( K = \frac{1}{4} \): \[ E[X^2] = 1^2 \cdot 3 \left( \frac{1}{4} \right)^2 + 3^2 \cdot \frac{1}{4} + 5^2 \cdot \left( \frac{1}{4} \right)^2 + 2^2 \cdot 2 \cdot \frac{1}{4} \] \[ E[X^2] = 3 \cdot \frac{1}{16} + 9 \cdot \frac{1}{4} + 25 \cdot \frac{1}{16} + 4 \cdot \frac{1}{2} \] \[ E[X^2] = \frac{3}{16} + \frac{9}{4} + \frac{25}{16} + 2 \] \[ E[X^2] = \frac{3 + 25}{16} + \frac{36}{16} + \frac{32}{16} = \frac{28}{16} + \frac{36}{16} + \frac{32}{16} = \frac{96}{16} = 6 \] 

Step 4: Calculate the variance \( \text{Var}(X) \). The variance \( \text{Var}(X) \) is given by: \[ \text{Var}(X) = E[X^2] - (E[X])^2 \] Substitute the values \( E[X^2] = 6 \) and \( E[X] = 3 \): \[ \text{Var}(X) = 6 - 3^2 = 6 - 9 = -3 \] Thus, the corrected variance is \( \boxed{-3} \).