Answer: 124

## Explanation This is a problem involving an exponential distribution. The probability density function is: $$ f_X(x) = ke^{-0.0056x} \quad \text{for } x \geq 0 $$ **Step 1: Find the constant k** For any probability density function, the total area under the curve must equal 1: $$ \int_{0}^{\infty} ke^{-0.0056x} dx = 1 $$ Solving the integral: $$ \int_{0}^{\infty} ke^{-0.0056x} dx = k \left[ \frac{e^{-0.0056x}}{-0.0056} \right]_{0}^{\infty} = k \left( 0 - \frac{1}{-0.0056} \right) = \frac{k}{0.0056} $$ Setting this equal to 1: $$ \frac{k}{0.0056} = 1 \Rightarrow k = 0.0056 $$ So the PDF is: $$ f_X(x) = 0.0056e^{-0.0056x} \quad \text{for } x \geq 0 $$ **Step 2: Find the median** The median is the value m such that: $$ P(X \leq m) = 0.5 $$ For the exponential distribution, the cumulative distribution function (CDF) is: $$ F(x) = P(X \leq x) = 1 - e^{-\lambda x} $$ where \(\lambda = 0.0056\). Setting \(F(m) = 0.5\): $$ 1 - e^{-0.0056m} = 0.5 $$ $$ e^{-0.0056m} = 0.5 $$ Taking natural logarithms: $$ -0.0056m = \ln(0.5) $$ $$ -0.0056m = -\ln(2) $$ $$ m = \frac{\ln(2)}{0.0056} $$ **Step 3: Calculate the value** $$ \ln(2) \approx 0.693147 $$ $$ m = \frac{0.693147}{0.0056} \approx 123.776 \approx 124 $$ Therefore, the median amount of the benefit is approximately 124, which corresponds to option C. **Key Points:** - This is an exponential distribution with rate parameter λ = 0.0056 - The median of an exponential distribution is \(\frac{\ln(2)}{\lambda}\) - The constant k in the PDF is actually the rate parameter λ

Author: Nikitesh Somanthe