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.
Based off the question it seems we need to find the expected value to figure out when the bulbs will no longer work.
Expected value = \(\frac{\mu}{n}\)
In this case 1000 will be equal to \(\mu\) while 100 will be equal to \(n\)
cat("The Lightbulb will Malfunction in the Following Hours","\n")## The Lightbulb will Malfunction in the Following Hours1000/100## [1] 10Assume 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.
\[fz(z)=(1/2)\lambda e^{-\lambda|z|}\] ##Exercise 1
Let \(X\) be a continuous random variable with mean \(\mu=10\) and variance \(o^2 = 100/3\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
Chebyshev’s Inequality gives an upper bound to the probability that the absolute deviation of a random variable from the mean will exceed a stated amount.
Wolfram MathWorld @ www.mathworld.wolfram.com/ChebyshevInequality.html gives the following formula
\[P(|X-\mu|\geq k)\leq \frac{o^2}{k^2}\] Based on the example we know \(\mu\) = 10 \(o^2\) = 100/3
\[P(|X-10|\ge2)\] In the following solution the value of k will have the corresponding problem letter as its prefix; for example a the value of k will be ak
ak <- 2/sqrt(100/3)
(ubpa <- round(1/ak^2, digits=3))## [1] 8.333\(\le8.333\) is greater than 1, therefore the upper bound of the probability is 1
\[P(|X-10|\ge5)\]
bk <- 5/sqrt(100/3)
(ubpb<- round(1/bk^2, digits=3))## [1] 1.333\(\le1.333\) is greater than 1, therefore the upper bound of the probability is 1
\[P(|X-10|\ge9)\]
ck <- 9/sqrt(100/3)
(ubpc<-round(1/ck^2, digits=3))## [1] 0.412\(\le0.412\) is less than 1, therefore the upper bound of the probability has been solved.
\[P(|X-10|\ge20)\]
dk <- 20/sqrt(100/3)
(ubpd<-round(1/dk^2, digits=3))## [1] 0.083\(\le0.083\) is less than 1, therefore the upper bound of the probability has been solved.
(exercise_a <- paste0("The upper bound probability for Exercise A is 1 because ",ubpa," is greater than 1"))## [1] "The upper bound probability for Exercise A is 1 because 8.333 is greater than 1"cat("\n")(exercise_b <- paste0("The upper bound probability for Exercise B is 1 because ",ubpb," is greater than 1"))## [1] "The upper bound probability for Exercise B is 1 because 1.333 is greater than 1"cat("\n")(exercise_c <- paste0("The upper bound probability for Exercise C is ",ubpc))## [1] "The upper bound probability for Exercise C is 0.412"cat("\n")(exercise_d <- paste0("The upper bound probability for Exercise D is ",ubpd))## [1] "The upper bound probability for Exercise D is 0.083"