Question

A random variable \( X \) is defined by \[ X = \begin{cases} -2 & \text{with probability} \, \frac{1}{3}, \\ 3 & \text{with probability} \, \frac{1}{2}, \\ 1 & \text{with probability} \, \frac{1}{6}. \end{cases} \] The value of \( E(X^2) \) is \(\underline{\hspace{2cm}}\) (round off to one decimal place).

Hint:

To calculate \( E(X^2) \), square the values of the random variable, multiply by their respective probabilities, and sum the results.

Answer 1

Correct Answer: 6

We are given the probability distribution for the random variable \( X \). To calculate \( E(X^2) \), we use the formula: \[ E(X^2) = \sum x^2 \cdot P(X = x) \] Substituting the values: \[ E(X^2) = (-2)^2 \cdot \frac{1}{3} + 3^2 \cdot \frac{1}{2} + 1^2 \cdot \frac{1}{6} \] Simplifying: \[ E(X^2) = 4 \cdot \frac{1}{3} + 9 \cdot \frac{1}{2} + 1 \cdot \frac{1}{6} \] \[ E(X^2) = \frac{4}{3} + \frac{9}{2} + \frac{1}{6} \] Finding a common denominator: \[ E(X^2) = \frac{8}{6} + \frac{27}{6} + \frac{1}{6} = \frac{36}{6} = 6.0 \] Thus, the value of \( E(X^2) \) is: \[ \boxed{6.0} \]