Question

Given a discrete random variable X with probability distribution:

X	-2	-1	0	1	2
P(X)	$ \frac{k^2}{3} $	$ k^2 $	$ \frac{2k^2}{3} $	$ \frac{k}{2} $	$ \frac{k}{2} $

Find the mean (expected value) of X.

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{5} \)
• C. \( \frac{11}{2} \)
• D. \( \frac{13}{2} \)

Hint:

Always first use the normalization condition \( \sum P(x_i) = 1 \) to find unknown constants in the distribution.

Answer 1

The Correct Option is A

Step 1: Use total probability = 1 to find \( k \): \[ \frac{k^2}{3} + k^2 + \frac{2k^2}{3} + \frac{k}{2} + \frac{k}{2} = 1 \Rightarrow 2k^2 + k = 1 \Rightarrow 2k^2 + k - 1 = 0 \Rightarrow k = \frac{1}{2},\quad k^2 = \frac{1}{4} \] Step 2: Compute mean: \[ E(X) = \sum x_i P(x_i) = (-2)\cdot \frac{1}{12} + (-1)\cdot \frac{1}{4} + 0 + 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{4} = -\frac{2}{12} - \frac{3}{12} + \frac{3}{12} + \frac{6}{12} = \frac{4}{12} = \frac{1}{3} \]