Question

If the probability distribution of a random variable \( X \) is as follows, then the variance of \( X \) is: \[ X = x \quad 2 \quad 3 \quad 5 \quad 9 \] \[ P(X = x) = k \quad 2k \quad 3k^2 \quad k^2 \]

• A. \( \frac{61}{4} \)
• B. \( \frac{7}{2} \)
• C. 12
• D. 3

Hint:

To calculate the variance of a discrete random variable, first determine \( E(X) \) and \( E(X^2) \) using the probability distribution, then apply the formula \( \text{Var}(X) = E(X^2) - (E(X))^2 \).

Answer 1

The Correct Option is D

We are provided with the probability distribution for the random variable \( X \) and need to determine its variance. 

Step 1: Determine the value of \( k \) Since the sum of the probabilities must equal 1, we have: \[ P(X = 2) + P(X = 3) + P(X = 5) + P(X = 9) = 1. \] Substituting the given probabilities, we get: \[ k + 2k + 3k^2 + k^2 = 1. \] Simplify the expression: \[ 3k + 4k^2 = 1. \] This can be rearranged as a quadratic in \( k \): \[ 4k^2 + 3k - 1 = 0. \] Using the quadratic formula: \[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] with \( a = 4 \), \( b = 3 \), and \( c = -1 \), we have: \[ k = \frac{-3 \pm \sqrt{9 - 4(4)(-1)}}{8} = \frac{-3 \pm \sqrt{9 + 16}}{8} = \frac{-3 \pm \sqrt{25}}{8}. \] This gives: \[ k = \frac{-3 + 5}{8} = \frac{2}{8} = \frac{1}{4} \quad \text{or} \quad k = \frac{-3 - 5}{8} = \frac{-8}{8} = -1. \] Since probabilities cannot be negative, we select \( k = \frac{1}{4} \). 

Step 2: Compute the Expected Value \( E(X) \) The expected value is given by: \[ E(X) = \sum_{i} x_i \, P(X = x_i). \] Substitute the values: \[ E(X) = 2\left(\frac{1}{4}\right) + 3\left(2 \times \frac{1}{4}\right) + 5\left(3 \times \left(\frac{1}{4}\right)^2\right) + 9\left(\left(\frac{1}{4}\right)^2\right). \] Simplify: \[ E(X) = \frac{2}{4} + \frac{6}{4} + \frac{15}{16} + \frac{9}{16} = \frac{8}{4} + \frac{24}{16}. \] \[ E(X) = 2 + \frac{3}{2} = \frac{7}{2}. \] 

Step 3: Compute \( E(X^2) \) The second moment is: \[ E(X^2) = \sum_{i} x_i^2 \, P(X = x_i). \] Substitute the values: \[ E(X^2) = 2^2\left(\frac{1}{4}\right) + 3^2\left(2 \times \frac{1}{4}\right) + 5^2\left(3 \times \left(\frac{1}{4}\right)^2\right) + 9^2\left(\left(\frac{1}{4}\right)^2\right). \] Simplify: \[ E(X^2) = \frac{4}{4} + \frac{18}{4} + \frac{75}{16} + \frac{81}{16} = 1 + \frac{9}{2} + \frac{156}{16}. \] Express all terms with a common denominator: \[ 1 = \frac{4}{4}, \quad \frac{9}{2} = \frac{18}{4}, \quad \frac{156}{16} = \frac{39}{4}, \] so that: \[ E(X^2) = \frac{4 + 18 + 39}{4} = \frac{61}{4}. \] 

Step 4: Compute the Variance \( \text{Var}(X) \) Variance is given by: \[ \text{Var}(X) = E(X^2) - \left(E(X)\right)^2. \] Substitute the values: \[ \text{Var}(X) = \frac{61}{4} - \left(\frac{7}{2}\right)^2 = \frac{61}{4} - \frac{49}{4} = \frac{12}{4} = 3. \] Thus, the variance of \( X \) is: \[ \boxed{3}. \]