A company buys 100 light bulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out?
Let \(X_{i}\) be the independent variable. Expected life time = 1000 Hrs
lambda = 1/1000
n = 100
Expected_Time_Hrs = 1/(n*lambda)
Expected_Time_Hrs## [1] 10Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that \(Z = X_1 − X_2\) has density
\(f_z(Z)=(1/2) \lambda e^{-\lambda|z|}\)
\(X_1 = X_2 - Z\)
When \(X_2 \ge X_1, -\infty\) to \(0\)
\(f_z(Z) = \int_{-\infty}^{\infty}fx_1(x_1)fx_2(x_2)dx_2\)
\(f_z(Z) = \int_{-\infty}^{\infty}fx_1(z+x_2)fx_2(x_2)dx_2\)
\(fx_1(z+x_2) = \lambda e ^{-\lambda(z+x_2)}\)
\(fx_2(x_2) = \lambda e^{-2\lambda x_2}\)
\(fz(z) = \int_{-\infty}^{0} \lambda e^{-\lambda(z+x_2)} \lambda e^{-\lambda x_2}dx_2\)
\(fz(z) = \lambda^2e^{-\lambda z(\frac{-1}{2\lambda})}\)
\(fz(z) = \frac{-\lambda e^{-\lambda z}}{2}\)
when \(X_1 \ge X_2, 0\) to \(\infty\)
\(fz(z) = \int_{0}^{\infty}\lambda e^{-\lambda(z+x_2)}\lambda e^{-\lambda x_2}dx_2\)
\(fz(z) = \lambda^2e^{-\lambda z}(\frac{-1}{2\lambda})\)
Which evaluates to the density below,
\(fz(z) = \frac{\lambda e^{-\lambda z}}{2}\)
Let X be a continuous random variable with mean \(µ = 10\) and variance \(σ^2 = 100/3\). 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(b) P(|X − 10| ≥ 5).
k <- (5/sqrt(100/3))
1/k^2## [1] 1.333333(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