Question

If the mean of the following probability distribution of a random variable $X$: \begin{array}{|c|c|c|c|c|c|} \hline X & 0 & 2 & 4 & 6 & 8 \\\ \hline P(X) & a & 2a & a+b & 2b & 3b \\\ \hline \end{array} is $\frac{46}{9}$, then the variance of the distribution is:

• A. $\frac{581}{81}$
• B. $\frac{566}{81}$
• C. $\frac{173}{27}$
• D. $\frac{151}{27}$

Answer 1

Answer: (B)

$\frac{566}{81}$

Answer 2

1. From the given probability distribution, the total probability must equal 1:

$\sum P_i = 1 \implies a + 2a + a + b + 2b + 3b = 1.$

Simplify:

$4a + 6b = 1 \quad \cdots \text{(I)}$

2. The mean is given by:

$E(X) = \sum P_i X_i = \frac{46}{9}.$

Substitute the probabilities:

$E(X) = 0 \times a + 2 \times 2a + 4 \times (a + b) + 6 \times 2b + 8 \times 3b.$

Simplify:

$4a + 4b + 12b + 24b = \frac{46}{9},$

$8a + 40b = \frac{46}{9} \quad \cdots \text{(II)}.$

3. Solve equations (I) and (II) simultaneously: From (I): $b = \frac{1}{9} - \frac{2a}{3}$. Substitute $b = \frac{1}{9}$ and solve to find:

$a = \frac{1}{12}, \quad b = \frac{1}{9}.$

4. Variance is calculated as:

$\text{Variance} = E(X^2) - (E(X))^2.$

Step 1: Find $E(X^2)$:

$E(X^2) = \sum P_i X_i^2 = 0^2 \times a + 2^2 \times 2a + 4^2 \times (a + b) + 6^2 \times 2b + 8^2 \times 3b.$

Simplify:

$E(X^2) = 4a + 16(a + b) + 72b + 192b.$

Substitute $a = \frac{1}{12}, b = \frac{1}{9}$:

$E(X^2) = \frac{298}{9}.$

Step 2: Find $(E(X))^2$:

$(E(X))^2 = \left(\frac{46}{9}\right)^2 = \frac{2116}{81}.$

Step 3: Calculate variance:

$\text{Variance} = \frac{298}{9} - \frac{2116}{81} = \frac{566}{81}.$