Assignment_8

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? (See Exercise 10.)

Exponential lifetime = 1000 hours n = 100 (number of lightbulbs)

minimum value = μ/n

so 1000/100 = 10 hours

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|

knitr::include_graphics("./AMoyse_Assignemnt8.jpg")

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

Givens:

mean =10

variance =100/3

standard_deviation = sqrt(100/3)

P(|X - μ| ≥ kσ) ≤ 1/k²

P (|X − 10| ≥ 2)

kσ = 2 k = 2/sqrt(100/3)

1/(2/sqrt(100/3))^2

## [1] 8.333333

P (|X − 10| ≥ 2)

kσ = 2 k = 2/sqrt(100/3)

1/(2/sqrt(100/3))^2

## [1] 8.333333

P (|X − 10| ≥ 5)

kσ = 5 k = 5/sqrt(100/3)

1/(5/sqrt(100/3))^2

## [1] 1.333333

P (|X − 10| ≥ 9)

kσ = 9 k = 9/sqrt(100/3)

1/(9/sqrt(100/3))^2

## [1] 0.4115226

P (|X − 10| ≥ 20)

kσ = 20 k = 20/sqrt(100/3)

1/(20/sqrt(100/3))^2

## [1] 0.08333333