Problem 10 in the book tells us that the density for the minimum value has mean of mu/n. In this case is 1,000 hours and n is the number of light bulbs: 100. So the expected time for the first of these bulbs is 10 hours.
mu <- 1000
n <- 100
mu/n## [1] 10\[ f_{X_{1}}(x) = f_{X_{2}}(x) = \begin{cases} \lambda e^{-\lambda x}, & x\geq 0, \\ 0, & otherwise; \end{cases} \] \[ f_{Z}(z) = \int_{-\infty}^{\infty} f_{X_{1}} (x_{2}-y)f_{X_{2}}(x_{2})dx_{2} \] subbing in these variables gives us
\[ = \int_{-\infty}^{\infty} \lambda e ^{-\lambda(x_{2}-z)} \lambda e^{-\lambda x_2} \] now we need to simplify this
\[ = \int_{-\infty}^{\infty} \lambda^2 e^{\lambda(z-x_{2})-\lambda x_2} \] \[ = \int_{-\infty}^{\infty} \lambda^2 e^{\lambda (z -2 x_2)} \] x2 = x1-z \[ = \int_{-\infty}^{\infty} \lambda^2 e^{\lambda (z -2 x_1)} \]
solving via wolframalpha gives
\[ = \frac{1}{2} \lambda e^{-\lambda |z|} \]
P( | X - 10 | >= 2)
var <- 100 / 3
stdev <- sqrt(var)
mu <- 10
k <- 2/stdev
up <- 1/k^2
up## [1] 8.333333P( | X - 10 | >= 5)
k <- 5/stdev
up <- 1/k^2
up## [1] 1.333333P( | X - 10 | >= 9)
k <- 9/stdev
up <- 1/k^2
up## [1] 0.4115226P( | X - 10 | >= 20)
k <- 20/stdev
up <- 1/k^2
up## [1] 0.08333333