Probability Distribution Calculations and Ordering

Probability Distribution Data

The probability distribution for the variate X is given as:

Calculating Key Statistical Measures

Expected Value E(X)

The expected value, E(X), is calculated as the sum of each value multiplied by its probability:

E(X) = $\sum x \cdot \text{P}(X=x)$

E(X) = $(-3) \times \frac{1}{6} + (6) \times \frac{1}{2} + (9) \times \frac{1}{3}$

E(X) = $-\frac{1}{2} + 3 + 3 = 5.5$

So, A. E(X) = 5.5

Expected Value of X Squared E(X²)

E(X²) is calculated similarly, but with the squared values of x:

E(X²) = $\sum x^2 \cdot \text{P}(X=x)$

E(X²) = $(-3)^2 \times \frac{1}{6} + (6)^2 \times \frac{1}{2} + (9)^2 \times \frac{1}{3}$

E(X²) = $9 \times \frac{1}{6} + 36 \times \frac{1}{2} + 81 \times \frac{1}{3}$

E(X²) = $1.5 + 18 + 27 = 46.5$

So, B. E(X²) = 46.5

Variance $\sigma^2$

The variance is found using the formula $\sigma^2 = E(X^2) - [E(X)]^2$:

$\sigma^2 = 46.5 - (5.5)^2$

$\sigma^2 = 46.5 - 30.25 = 16.25$

So, D. $\sigma^2$ = 16.25

Expected Value E[(2X+1)²]

We calculate E[(2X+1)²] using the linearity of expectation:

E[(2X+1)²] = E[4X² + 4X + 1]

E[(2X+1)²] = $4 \cdot E(X^2) + 4 \cdot E(X) + 1$

E[(2X+1)²] = $4 \times 46.5 + 4 \times 5.5 + 1$

E[(2X+1)²] = $186 + 22 + 1 = 209$

So, C. E[(2X+1)²] = 209

Ordering the Measures

We have the following values:

• A. E(X) = $5.5$
• D. $\sigma^2$ = $16.25$
• B. E(X²) = $46.5$
• C. E[(2X+1)²] = $209$

Final Arrangement

Arranging these values in increasing order:

$5.5 < 16.25 < 46.5 < 209$

This corresponds to the order: A, D, B, C.