Question

A random variable $X$ has the following probability distribution: \[ \begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{25}{36} & k & \frac{1}{36} \\ \hline \end{array} \] If the mean of the random variable $X$ is $\frac{1}{3}$, then the variance is:

• A. $1$
• B. $\frac{5}{18}$
• C. $\frac{7}{18}$
• D. $\frac{11}{18}$

Answer 1

The Correct Option is B

The mean $E(X)$ is given as $\frac{1}{3}$. First, solve for $k$: \[ E(X) = \sum X \cdot P(X) = 0 \cdot \frac{25}{36} + 1 \cdot k + 2 \cdot \frac{1}{36} = k + \frac{2}{36}. \] Equating $E(X) = \frac{1}{3}$: \[ k + \frac{2}{36} = \frac{1}{3} \implies k = \frac{1}{3} - \frac{1}{18} = \frac{6}{18} - \frac{1}{18} = \frac{5}{18}. \] Now compute $E(X^2)$: \[ E(X^2) = \sum X^2 \cdot P(X) = 0 \cdot \frac{25}{36} + 1^2 \cdot \frac{5}{18} + 2^2 \cdot \frac{1}{36}. \] Simplify: \[ E(X^2) = \frac{5}{18} + \frac{4}{36} = \frac{5}{18} + \frac{2}{18} = \frac{7}{18}. \] The variance is: \[ \text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{7}{18} - \left(\frac{1}{3}\right)^2 = \frac{7}{18} - \frac{1}{9} = \frac{7}{18} - \frac{2}{18} = \frac{5}{18}. \]