N <- 1000
lightbulbs <- 100
N/lightbulbs## [1] 10
\[ fZ(z) = \frac{1}{2} \lambda e^{-\lambda|z|}\] \[ \int_{0}^{\infty} \lambda^2 e^{-\lambda z} dx = \frac {1}{2} \lambda e^{-\lambda z} \]
\[ \int_{-z}^{\infty} \lambda^2 e^{-\lambda z} dx = \frac {1}{2} \lambda e^{\lambda z} \]
Let’s combine Z negative and Z positive
\[f_Z(z) = \begin{cases} \frac{1}{2}e^{-\lambda z}, & \mbox{if } z \ge 0, \\ \frac{1}{2}e^{\lambda z}, & \mbox{if }z <0. \end{cases}\]
Let X be a continuous random variable with mean \(\mu=10\) and variance \(\sigma^2= \frac {100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
(a) P(|X- 10| >= 2).
k <- (2/sqrt(100/3))
1/k^2## [1] 8.333333
Since probability cannot be greater than 1, the upper bound is 1.
(b)P(|X - 10| >= 5)
k <- (5/sqrt(100/3))
1/k^2## [1] 1.333333
Since probability cannot be greater than 1, the upper bound is 1.
(c)P(|X -10| >= 9)
k <- (9/sqrt(100/3))
1/k^2## [1] 0.4115226
(d)P(|X - 10| >= 20
k <- (20/sqrt(100/3))
1/k^2## [1] 0.08333333