A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out?

mu = 1000

n = 100

E = mu/n

E
## [1] 10

Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density fZ (z) = (1/2)λe−λ|z|

\[\begin{equation} f_{X1}(x1) = λe^{−x1λ} \end{equation}\]

\[\begin{equation} f_{X2}(x2) = λe^{−x2λ} \end{equation}\]

The joint probability density function is:

\[\begin{equation} f_{X1,x2}(x1,x2) = 1/2λe^{−x1λ −x2λ} \end{equation}\]

The double integral gives us the cumulative distribution function Fy(y) of Y

\[\begin{equation} \int_0^\infty \int_{x1-y}^\infty λ^{x1-y} f_{X1,x2}(x1,x2) dx_2dx_1 \end{equation}\]

\[\begin{equation} \int_0^\infty \int_{x1-y}^\infty 1/2λe^{−x1λ −x2λ} dx_2dx_1 \end{equation}\]

\[\begin{equation} 1/2λe^{yλ} \end{equation}\]

Let X be a continuous random variable with mean μ = 10 and variance σ2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

P (|X − μ)| ≥ kσ) ≤ Var(X)/k2σ2 = 1/k^2

  1. P (|X − 10| ≥ 2).
Var = 100/3
sigma = sqrt(Var)
k = 2/sigma
1/k^2
## [1] 8.333333
  1. P (|X − 10| ≥ 5).
k = 5/sigma
1/k^2
## [1] 1.333333
  1. P (|X − 10| ≥ 9).
k = 9/sigma
1/k^2
## [1] 0.4115226
  1. P (|X − 10| ≥ 20)
k = 20/sigma
1/k^2
## [1] 0.08333333