Question

Let X₁ and X₂ be two independent and identically distributed random variables having \(X^2_2\) distribution and W = X₁ + X₂ . Then P(W > E(W)) equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 0.39

To solve the problem, we first need to understand the distribution of \(X_1\) and \(X_2\). Given that both have a \(X^2_2\) distribution, this means each follows a Chi-squared distribution with 2 degrees of freedom.
1. **Expectation Calculation:**
The expected value \(E(X)\) for a Chi-squared distribution with \(k\) degrees of freedom is \(k\). Thus, \(E(X_1) = E(X_2) = 2\).
2. **Calculate \(E(W)\):**
Since \(W = X_1 + X_2\) and \(X_1, X_2\) are independent, the expected value \(E(W) = E(X_1) + E(X_2) = 2 + 2 = 4\).
3. **Distribution of \(W\):**
Since \(W = X_1 + X_2\), the distribution of \(W\) is a Chi-squared distribution with \(2+2 = 4\) degrees of freedom, i.e., \(W \sim X^2_4\).
4. **Calculate \(P(W > E(W) = P(W > 4)\):**
For a Chi-squared distribution \(X^2_4\), we want \(P(W > 4)\). This probability can be found using Chi-squared distribution tables or a calculator. For \(X^2_4\), the value is approximately 0.39.
In conclusion, the probability \(P(W > E(W))\) is 0.39.