Question

Let the probability distribution of random variable X be

X	-2	-1	1	2	3
P(X=x)	k	2k	2k	k	3k

Then, the value of E(X²) is

• A. \(\frac{19}{9}\)
• B. \(\frac{13}{3}\)
• C. \(\frac{35}{9}\)
• D. \(\frac{11}{3}\)
• E. \(\frac{7}{3}\)

Answer 1

The Correct Option is B

Given the probability distribution of random variable \( X \):

\( X \)	-2	-1	1	2	3
\( P(X=x) \)	\( k \)	\( 2k \)	\( 2k \)	\( k \)	\( 3k \)

Step 1: Find the value of \( k \) using the fact that probabilities must sum to 1: \[ k + 2k + 2k + k + 3k = 1 \\ 9k = 1 \\ k = \frac{1}{9} \]

Step 2: Calculate \( E(X^2) \): \[ E(X^2) = \sum x^2 P(X=x) \\\]

\[ = (-2)^2 \cdot \frac{1}{9} + (-1)^2 \cdot \frac{2}{9} + 1^2 \cdot \frac{2}{9} + 2^2 \cdot \frac{1}{9} + 3^2 \cdot \frac{3}{9} \\ \]

\[= 4 \cdot \frac{1}{9} + 1 \cdot \frac{2}{9} + 1 \cdot \frac{2}{9} + 4 \cdot \frac{1}{9} + 9 \cdot \frac{3}{9} \\\]

\[ = \frac{4}{9} + \frac{2}{9} + \frac{2}{9} + \frac{4}{9} + \frac{27}{9} \\\] \[= \frac{4 + 2 + 2 + 4 + 27}{9} \\ = \frac{39}{9} = \frac{13}{3} \]

Answer 2

First, we need to find the value of \( k \). Since the sum of probabilities for all possible values of \( X \) must equal 1, we have:

\[ k + 2k + 2k + k + 3k = 1 \] \[ 9k = 1 \] \[ k = \frac{1}{9} \]

Now we can find \( E(X^2) \):

\[ E(X^2) = \sum [x^2 \cdot P(X = x)] \quad \text{for all } x \] \[ E(X^2) = (-2)^2\left(\frac{1}{9}\right) + (-1)^2\left(\frac{2}{9}\right) + (1)^2\left(\frac{2}{9}\right) + (2)^2\left(\frac{1}{9}\right) + (3)^2\left(\frac{3}{9}\right) \] \[ = 4\left(\frac{1}{9}\right) + 1\left(\frac{2}{9}\right) + 1\left(\frac{2}{9}\right) + 4\left(\frac{1}{9}\right) + 9\left(\frac{3}{9}\right) \] \[ = \frac{4}{9} + \frac{2}{9} + \frac{2}{9} + \frac{4}{9} + \frac{27}{9} \] \[ = \frac{4 + 2 + 2 + 4 + 27}{9} \] \[ = \frac{39}{9} \] \[ = \frac{13}{3} \]

Therefore, \( E(X^2) = \frac{13}{3} \).