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?
\(F(x) = e^{-\lambda n}\)
\(\lambda\) is the rate of change, so every 1000 hours or \(\frac{1}{1000}\)
\(E(V) = \frac{1}{\lambda}\)
\(= \frac{1}{\frac{n}{1000}}\)
\(E(V) = \frac{1000}{100} = 10\)
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 \(f_Z(z) = (1/2)\lambda e^{-\lambda |z|}\)
\(Z = X_1 - X_2\)
Joint density function:
\(f(z) = \lambda e^{-\lambda x_1} \times \lambda e^{\lambda x_2}\)
\(f(z) = \lambda^2 e^{-\lambda^ x_1}e^{\lambda x_2}\)
\(f(z) = \lambda^2 e^{-\lambda^ (x_1 + x_2)}\)
\(x_2 = x_1 - z\)
\[\int_{0}^{1} \lambda^2 e^{-\lambda(2x_1 - z))}\]
\[= \int_{0}^{1} \lambda^2 e^{- \lambda x_1} \frac{e^{-2\lambda (x_1 - z)}}{2\lambda}\]
\[= \int_{0}^{1} \frac{\lambda}{2} e^{- \lambda z}\]
Let X be a continuous random variable with mean \(\mu\) = 10 and variance \(\sigma^2 = \frac{100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
\(P(|X - \mu | \geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2}\)
\(\sigma = \frac{100}{3} = 33.33\)
\(P(|X - 10| \geq 2)\)
\(\frac{\sigma^2}{\epsilon^2} = \frac{33.33}{2^2} = 8.33\)
\(P(|X - 10| \geq 5)\)
\(\frac{\sigma^2}{\epsilon^2} = \frac{33.33}{5^2} = 1.33\)
\(P(|X - 10| \geq 9)\)
\(\frac{\sigma^2}{\epsilon^2} = \frac{33.33}{9^2} = 0.411\)
\(P(|X - 10| \geq 20)\)
\(\frac{\sigma^2}{\epsilon^2} = \frac{33.33}{20^2} = 0.083\)