Question

A variate \(X\) has the probability distribution:

Arrange the following in increasing order:
A. \(E(X)\)
B. \(E(X^2)\)
C. \(E[(2X+1)^2]\)
D. \(\sigma^2\)

• A. A, B, C, D
• B. B, C, D, A
• C. B, D, C, A
• D. A, D, B, C

Hint:

Always expand expressions like \((2X+1)^2\) before taking expectation.

Answer 1

The Correct Option is D

Concept: We use expectation and variance formulas: \[ E(X) = \sum xP(x), \quad E(X^2) = \sum x^2P(x) \] \[ \sigma^2 = E(X^2) - [E(X)]^2 \]

Step 1: Compute \(E(X)\).
\[ E(X) = (-3)\cdot \frac{1}{6} + 6 \cdot \frac{1}{2} + 9 \cdot \frac{1}{3} \] \[ = -\frac{3}{6} + 3 + 3 \] \[ = -\frac{1}{2} + 6 = \frac{11}{2} = 5.5 \]

Step 2: Compute \(E(X^2)\).
\[ E(X^2) = 9\cdot \frac{1}{6} + 36\cdot \frac{1}{2} + 81\cdot \frac{1}{3} \] \[ = \frac{9}{6} + 18 + 27 \] \[ = 1.5 + 18 + 27 = 46.5 \]

Step 3: Compute variance \(\sigma^2\).
\[ \sigma^2 = 46.5 - (5.5)^2 \] \[ = 46.5 - 30.25 = 16.25 \]

Step 4: Compute \(E[(2X+1)^2]\).
Expand: \[ (2X+1)^2 = 4X^2 + 4X + 1 \] Thus: \[ E[(2X+1)^2] = 4E(X^2) + 4E(X) + 1 \] \[ = 4(46.5) + 4(5.5) + 1 \] \[ = 186 + 22 + 1 = 209 \]

Step 5: Compare values.
\[ E(X) = 5.5,\quad \sigma^2 = 16.25,\quad E(X^2)=46.5,\quad E[(2X+1)^2]=209 \]

Step 6: Arrange in increasing order.
\[ 5.5 < 16.25 < 46.5 < 209 \] \[ A < D < B < C \] Final Answer: \[ \boxed{A, D, B, C} \]