Question 11

The expected time for the first of these bulbs to burn out is:

## [1] 10

This is simply the amount of expected lifetime hours divided by the bulbs.

Question 14

For \(X_2 \ge X_1\):

\[ f_Z(z) = \int_{-\infty}^{\infty}f_{X_1}(z+x_2)f_{X_2}(x_2)dx_2 \\ f_Z(z) = \int_{-\infty}^0\lambda e^{-\lambda (z+x_2)} \lambda e^{-\lambda x_2}dx_2 \\ f_Z(z) = \int_{-\infty}^0\lambda e^{-\lambda (z)} \lambda e^{-2\lambda x_2}dx_2 \\ f_Z(z) = \lambda^2 e^{-\lambda z}(\int_{-\infty}^0 e^{-2\lambda x_2}dx_2) \\ f_Z(z) = \lambda^2 e^{-\lambda z}(\frac{ -1}{2\lambda}) \\ f_Z(z) = \frac{-\lambda}{2} (e^{-\lambda z}) \]

For \(X_1 \ge X_2\):

\[ f_Z(z) = \int_{-\infty}^{\infty}f_{X_1}(z+x_2)f_{X_2}(x_2)dx_2 \\ f_Z(z) = \int_0^{\infty}\lambda e^{-\lambda (z+x_2)} \lambda e^{-\lambda x_2}dx_2 \\ f_Z(z) = \int_0^{\infty}\lambda e^{-\lambda (z)} \lambda e^{-2\lambda x_2}dx_2 \\ f_Z(z) = \lambda^2 e^{-\lambda z}(\int_0^{\infty} e^{-2\lambda x_2}dx_2) \\ f_Z(z) = \lambda^2 e^{-\lambda z}(\frac{ -1}{2\lambda}) \\ f_Z(z) = \frac{\lambda}{2} (e^{-\lambda z}) \]

After combining the two: \[ f(z) = \begin{cases} \frac{-\lambda}{2} (e^{-\lambda z}) & z < 0 \\ \frac{\lambda}{2} (e^{-\lambda z}) & z \geq 0 \end{cases} \]

Which simplifies to: \[ f(z) = \frac{\lambda}{2} e^{-\lambda |z|} \]

Question 1

a

\[ P(|X - 10| \geq 2) \leq \frac{\frac{100}{3}}{2^2}\\ P(|X - 10| \geq 2) \leq \frac{25}{3} \]

b

\[ P(|X - 10| \geq 5) \leq \frac{\frac{100}{3}}{5^2}\\ P(|X - 10| \geq 5) \leq \frac{4}{3} \]

c

\[ P(|X - 10| \geq 9) \leq \frac{\frac{100}{3}}{9^2}\\ P(|X - 10| \geq 9) \leq \frac{100}{243} \]

d

\[ P(|X - 10| \geq 20) \leq \frac{\frac{100}{3}}{20^2}\\ P(|X - 10| \geq 20) \leq \frac{100}{1200} \]