1. A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the frst of these bulbs to burn out? (See Exercise 10.)

Let M be the minimum value of the random variables as an exponential denisty with mean \(\frac{\lambda}{n}\).

n = 100 and \(\mu\) = 1000 so expected time is \(\frac{1000}{100}\) or 10 hours.

  1. 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 \(f_z(z) = \frac{1}{2}\lambda e^{-\lambda|z|}\)

The probability density function for both \(X_1\) and \(X_2\) is \(\lambda e^{-\lambda x}\), if \(x\ge0\),

If Z =\(X_1-X_2\) then \(x_2=x_1-z\) from the definition.

So \(\int_{-\infty} ^\infty f(x_1) \cdot (fx_2)dx_1\)

\(\int_{-\infty} ^\infty f(x_1) \cdot (fx_1-z)dx_1\)

\(\int_{-\infty} ^\infty \lambda e^{-\lambda x_1}\lambda e^{-\lambda (x_1-z)dx_1}\)

\(\int_{-\infty} ^\infty \lambda^2 e^{-2\lambda x_1+\lambda z}dx_1\) = \(\frac{1}{2}\lambda e^{-\lambda z}\)

  1. Let X be a continuous random variable with mean \(\mu\)= 10 and variance \(\sigma\) = \(\frac{100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

Chebyshev’s Inequality: \(P(|X-\mu|\ge k\sigma) \le \frac{1}{k^2}\)

  1. P(|X - 10| \(\ge\) 2).
    \(\sigma=\sqrt\frac{100}{3}\) \(k\sigma=2\)
    \(k=\frac{2}{\sigma}\)
    \(k=2\cdot\sqrt{\frac{3}{100}}\)
    \(\frac{1}{k^2}\) = \((\frac{1}{2}\cdot\sqrt{\frac{100}{3}})^2\) = 8.33 or 1 as the upper bound.

  2. P(|X - 10| \(\ge\) 5).
    \(\sigma=\sqrt\frac{100}{3}\)
    \(k=\frac{2}{\sigma}\)
    \(k=5\cdot\sqrt{\frac{3}{100}}\)
    \(\frac{1}{k^2}\) = \((\frac{1}{5}\cdot\sqrt{\frac{100}{3}})^2\) = 1.33 or 1 as the upper bound..

  3. P(|X - 10| \(\ge\) 9).
    \(\sigma=\sqrt\frac{100}{3}\)
    \(k\sigma=10\)
    \(k=5\cdot\frac{\sqrt{3}}{10}\)
    \(\frac{1}{k^2}\) = \((\frac{1}{9}\cdot\sqrt{\frac{100}{3}})^2\) = .412 as the upper bound.

  4. P(|X - 10| \(\ge\) 20).
    \(\sigma=\sqrt\frac{100}{3}\)
    \(k\sigma=20\)
    \(k=20\cdot\frac{\sqrt{3}}{10}\)
    \(\frac{1}{k^2}\) = \((\frac{1}{20}\cdot\sqrt{\frac{100}{3}})^2\) = .083 as the upper bound.