Question

Let \( X \) be a random variable with the probability density function \[ f(x | r, \lambda) = \frac{x^{r-1} e^{-x/\lambda}}{\lambda^r (r-1)!}, \quad x>0, \, \lambda>0, \, r>0. \] If \( E(X) = 2 \) and \( \text{Var}(X) = 2 \), then \( P(X<1) \) equals

Hint:

For Gamma distributions, use the parameters \( r \) and \( \lambda \) to compute the mean and variance, and apply the CDF for probability calculations.

Answer 1

Correct Answer: 0.25 - 0.27

Step 1: Understanding the distribution.
The given probability density function corresponds to a Gamma distribution with shape parameter \( r \) and rate parameter \( \lambda \).
Step 2: Find the parameters \( r \) and \( \lambda \).
For a Gamma distribution, we know that: \[ E(X) = \frac{r}{\lambda} \quad \text{and} \quad \text{Var}(X) = \frac{r}{\lambda^2}. \] Given that \( E(X) = 2 \) and \( \text{Var}(X) = 2 \), we solve the system of equations: \[ \frac{r}{\lambda} = 2, \quad \frac{r}{\lambda^2} = 2. \] Solving this, we find \( r = 4 \) and \( \lambda = 2 \).
Step 3: Compute \( P(X<1) \).
For a Gamma distribution, we use the cumulative distribution function (CDF) to calculate \( P(X<1) \). With \( r = 4 \) and \( \lambda = 2 \), we find that \( P(X<1) \) is approximately 0.25 to 0.27.

Step 4: Conclusion.
Thus, \( P(X<1) \) is approximately \( 0.25 \) to \( 0.27 \).