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:
- \( \frac{581}{81} \)
- \( \frac{566}{81} \)
- \( \frac{173}{27} \)
- \( \frac{151}{27} \)
The Correct Option is B
Approach Solution - 1
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:
\[ \text{Variance} = \frac{298}{9} - \frac{2116}{81} = \frac{566}{81}. \]
Approach Solution -2
Given probability table:
X: 0, 2, 4, 6, 8
P(X): a, 2a, a+b, 2b, 3b
Total probability:
a + 2a + (a+b) + 2b + 3b = 1
4a + 6b = 1
Mean is \( \frac{46}{9} \):
0×a + 2×2a + 4×(a+b) + 6×2b + 8×3b = \(\frac{46}{9}\)
(4a + 4a + 4b + 12b + 24b) = 8a + 40b = \(\frac{46}{9}\)
Step 2: Solve for a and b
From above:
4a + 6b = 1
8a + 40b = \(\frac{46}{9}\)
Multiply first by 2: 8a + 12b = 2
Subtract:
(8a + 40b) - (8a + 12b) = \(\frac{46}{9}\) - 2
28b = \(\frac{28}{9}\)
b = \(\frac{1}{9}\)
Substitute in first equation:
4a + 6(\(\frac{1}{9}\)) = 1
4a + \(\frac{2}{3}\) = 1
4a = \(\frac{1}{3}\) ⇒ a = \(\frac{1}{12}\)
Step 3: Variance Calculation
Variance = \(E(X^2) - [E(X)]^2\)
Find \(E(X^2)\):
= 0²×a + 2²×2a + 4²×(a+b) + 6²×2b + 8²×3b
= 8a + 16a + 16b + 72b + 192b
= 24a + 280b
= 24×\(\frac{1}{12}\) + 280×\(\frac{1}{9}\)
= 2 + \(\frac{280}{9}\) = \(\frac{18}{9} + \frac{280}{9} = \frac{298}{9}\)
Variance = \(\frac{298}{9} - \left(\frac{46}{9}\right)^2\)
= \(\frac{298}{9} - \frac{2116}{81}\)
= \(\frac{2682 - 2116}{81} = \frac{566}{81}\)
Final Answer: Variance = \(\frac{566}{81}\)
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