n <- 1000
litbulbs <- 100
n/litbulbs
## [1] 10
\[\begin{equation} fZ(z) = (1/2)\lambda e^{-\lambda|z|} \end{equation}\]
\[\quad \int _{ 0 }^{ \infty }{ { \lambda }^{ 2 }{ e }^{ -\lambda z }dx } =\frac { 1 }{ 2 } \lambda { e }^{ -\lambda z }\]
\[\quad \int _{ -z }^{ \infty }{ { \lambda }^{ 2 }{ e }^{ \lambda z }dx }=\frac { 1 }{ 2 } \lambda { e }^{ \lambda z }\]
1, Let X be a continuous random variable with mean µ = 10 and variance ??2 = 100/3. Using Chebyshev’s Inequality, ???nd an upper bound for the following probabilities.
k <- (2/sqrt(100/3))
1/k^2
## [1] 8.333333
k <- (5/sqrt(100/3))
1/k^2
## [1] 1.333333
k <- (9/sqrt(100/3))
1/k^2
## [1] 0.4115226
k <- (20/sqrt(100/3))
1/k^2
## [1] 0.08333333