Data605-Week8-HomeWork8-kamath

Home Work 08

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?

    ==> Solution:

    E[Xi] = 1 / λi = 1000

    Expected lifetime of a bulb is 1000 hours.

    λi = 1 / 1000

    Xi is exponential so

    min{X1,X2,…,X100} ∼ exponential (∑ i=1 to 100 for λi)

    E[min Xi] = 1/ (1/10) = 10

    ==> Expected time for the first of these bulbs to burn out is 10 hours

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14. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z=X1−X2 has density

\(fZ(z)=(1/2)λe−λ|z|\)

==> Solution:

.

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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.

   1. P(|X−10|≥2)

      var <- 100/3
      e <- 2
      p <- var / e^2
      paste0('Upper bound is ', round(p, 4))

      ## [1] "Upper bound is 8.3333"

   2. P(|X−10|≥5)

      var <- 100/3
      e <- 5
      p <- var / e^2
      paste0('Upper bound is ', round(p, 4))

      ## [1] "Upper bound is 1.3333"

   3. P(|X−10|≥9)

      var <- 100/3
      e <- 9
      p <- var / e^2
      paste0('Upper bound is ', round(p, 4))

      ## [1] "Upper bound is 0.4115"

   4. (|X−10|≥20)

      var <- 100/3
      e <- 20
      p <- var / e^2
      paste0('Upper bound is ', round(p, 4))

      ## [1] "Upper bound is 0.0833"

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