data605hw8

Page 303 #11

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?

The density of minimum value among n independent random variables with an exponential density has mean mu/n, where mu is the mean of exponential density of individual variable.

Here we have mu=1000 and n=100 (100 bulbs)

mu <- 1000
n <- 100
mu/n

## [1] 10

1/(1/10)

## [1] 10

Page 303 #14

Assume that \(X_{1}\) and \(X_{2}\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = \(X_{1}\) − \(X_{2}\) has density

\[f_{Z}(z)=(1/2)λe^{−λ|z|}.\]

Two random numbers in the interval [0,\(\infty\)) with exponential density has the following sum: W = X + Y

\[ fx(x) = fy(x) = \left\{ \begin{array}{ll} \lambda e^{−λx}, & \quad x \geq 0 \\ 0, & \quad otherwise \end{array} \right. \]

\[\left[fw(w) = \int_{-\infty}^{\infty} f_{x}(x) f_{y}(w-x) \; dx\right]\]

If we apply this same logic to Z = \(X_{1}\) − \(X_{2}\) or even Z = A + (-B)

\[\left[fz(z) = \int_{-\infty}^{\infty} f_{a}(a) f_{-b}(z-a) \; da\right]\] \[\left[fz(z) = \int_{-\infty}^{\infty} f_{a}(a) f_{b}(a-z) \; da\right]\]

\[\left[fz(z) = \int_{0}^{\infty} \lambda e^{−λa} \lambda e^{−λ(a-z)} \; da\right]\]

\[\left[\lambda e^{λz} \int_{0}^{\infty} \lambda e^{−2λa} da\right]\]

\[\left. \lambda e^{λz} (-1/2 e^{−2λa)} \right|_{0}^{\infty}\]

Page 320-321 #1

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.

cheby <- function(e,variance)
{x = variance/e^2
return (x)
}

(a) P(|X−10|≥2).

variance = 100/3
e <- 2
p <- cheby (e,variance)
p

## [1] 8.333333

(b) P(|X−10|≥5).

e <- 5
p2 <- cheby (e,variance)
p2

## [1] 1.333333

(c) P(|X−10|≥9).

e <- 9
p3 <- cheby (e,variance)
p3

## [1] 0.4115226

(d) P(|X − 10| ≥ 20).

e <- 20
p4 <- cheby (e,variance)
p4

## [1] 0.08333333