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

The Correct Option is B

1. The sum of probabilities in a distribution must equal 1:

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

Simplifying this equation yields:

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

2. The expected value E(X) is calculated as:

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

Substituting the given probabilities:

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

Simplifying this expression:

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

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

3. Solving equations (I) and (II) simultaneously. From (I), we can express b as \(b = \frac{1}{9} - \frac{2a}{3}\). Substituting this into the equations and solving for a and b gives:

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

4. The variance is computed using the formula:

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

Step 1: Calculate \(E(X^2)\):

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

Simplifying the expression for \(E(X^2)\):

\[ E(X^2) = 4a + 16(a + b) + 72b + 192b. \]

Substituting the values \(a = \frac{1}{12}\) and \(b = \frac{1}{9}\):

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

Step 2: Calculate \((E(X))^2\):

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

Step 3: Compute the variance:

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