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