The life (in months) of a certain computer component has a CDF given by \( P(x)=\left\{\begin{array}{ll}1-e^{-z / 4} & \text { if } x>0 \\ 0 & \text { if otherwise }\end{array}\right. \) Let the random variable \( X \) represent the time (in months) that a randomly selected component lasts. a) Find \( P(X \leq 11) \). b) Find \( P(X>16) \). 0.0183 Round your answers to 4 decimals.

To solve the problem based on the given cumulative distribution function (CDF) for the life of a computer component, we will calculate the probabilities as requested. ### a) Find \( P(X \leq 11) \): Using the CDF, we can calculate: \[ P(X \leq 11) = 1 - e^{-11/4} \] Calculating this step-by-step: 1. Calculate \( \frac{11}{4} = 2.75 \). 2. Now, we compute \( e^{-2.75} \). Using a calculator for \( e^{-2.75} \): \[ e^{-2.75} \approx 0.0633 \] Now substituting back into the equation: \[ P(X \leq 11) = 1 - 0.0633 = 0.9367 \] So, rounding to four decimals: \[ P(X \leq 11) \approx 0.9367 \] ### b) Find \( P(X > 16) \): To find \( P(X > 16) \), we will use the relationship: \[ P(X > 16) = 1 - P(X \leq 16) \] Calculating \( P(X \leq 16) \): \[ P(X \leq 16) = 1 - e^{-16/4} \] Calculating \( \frac{16}{4} = 4 \). Now, we compute \( e^{-4} \): Using a calculator: \[ e^{-4} \approx 0.0183 \] Now substituting back into the equation: \[ P(X \leq 16) = 1 - 0.0183 = 0.9817 \] Thus, \[ P(X > 16) = 1 - 0.9817 = 0.0183 \] To summarize: **Answers:** - \( P(X \leq 11) \approx 0.9367 \) - \( P(X > 16) \approx 0.0183 \)