p.303_#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? (See Exercise 10.)

The minimum of n exponential random variables with mean μ has an exponential distribution with mean \(\frac{\mu}{n}\), give n = number of lightbulbs (100) and \(\mu\) = s the mean lifetime of each bulb (1000 hours) \[\frac{\mu}{n}=\frac{1000}{100}= 10 hrs\]

p.303-#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

\(fZ(z)=(1/2)λe−λ|z|\)

\(Z = X_1 - X_2\) , given: \(f_Z(z) = \frac{1}{2} \lambda e^{-\lambda|z|}\), we can use the convolution theorem.

Given: \(X_1\) and \(X_2\) are independent random variables, each having an exponential density with parameter \(\lambda\), their density functions are given by:

\[ f_{X_1}(x_1) = \lambda e^{-\lambda x_1} \quad \text{for} \ x_1 > 0 \] \[ f_{X_2}(x_2) = \lambda e^{-\lambda x_2} \quad \text{for} \ x_2 > 0 \]

\[ f_Z(z) = \int_{-\infty}^{\infty} f_{X_1}(z+x) \cdot f_{X_2}(x) \, dx \]

\[ f_Z(z) = \int_{0}^{\infty} \lambda e^{-\lambda(z+x)} \cdot \lambda e^{-\lambda x} \, dx \]

\[ = \lambda^2 \int_{0}^{\infty} e^{-\lambda z} e^{-2\lambda x} \, dx \]

\[ = \lambda^2 e^{-\lambda z} \int_{0}^{\infty} e^{-2\lambda x} \, dx \]

\[ = \lambda^2 e^{-\lambda z} \left[ -\frac{1}{2\lambda} e^{-2\lambda x} \right]_0^{\infty} \]

\[ = \frac{\lambda}{2} e^{-\lambda z} \]\[ f_Z(z) = \frac{\lambda}{2} e^{-\lambda |z|} \]

which matches the given density function \(Z = X_1 - X_2\) has the density \(f_Z(z) = \frac{1}{2} \lambda e^{-\lambda|z|}\)

p.320 #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.

Chebyshev’s Inequality states that for any continuous random variable \(X\) with mean \(\mu\) and variance \(\sigma^2\), and for any \(k > 0\):

\[ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} \]

Given \(\mu = 10\) and \(\sigma^2 = \frac{100}{3}\), find \(\sigma = \sqrt{\frac{100}{3}}\).

  1. For \(P(|X - 10| \geq 2)\), \(k = \frac{2}{\sigma} = \frac{2}{\sqrt{\frac{100}{3}}}\).

  2. For \(P(|X - 10| \geq 5)\), \(k = \frac{5}{\sigma} = \frac{5}{\sqrt{\frac{100}{3}}}\).

  3. For \(P(|X - 10| \geq 9)\), \(k = \frac{9}{\sigma} = \frac{9}{\sqrt{\frac{100}{3}}}\).

  4. For \(P(|X - 10| \geq 20)\), \(k = \frac{20}{\sigma} = \frac{20}{\sqrt{\frac{100}{3}}}\).

  1. \[P(|X - 10| \geq 2),k = \frac{2}{\sigma} = \frac{2}{\sqrt{\frac{100}{3}}}\]. \[ k = \frac{2}{\sqrt{\frac{100}{3}}} = \frac{2\sqrt{3}}{10} = \frac{\sqrt{3}}{5} \]

\[P(|X - 10| \geq 2) \leq \frac{1}{k^2} = \frac{1}{(\frac{\sqrt{3}}{5})^2} = \frac{25}{3}\]

  1. \[P(|X - 10| \geq 5)$, $k = \frac{5}{\sigma} = \frac{5}{\sqrt{\frac{100}{3}}}\]. \[ k = \frac{5}{\sqrt{\frac{100}{3}}} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2} \]

\[P(|X - 10| \geq 5) \leq \frac{1}{k^2} = \frac{1}{(\frac{\sqrt{3}}{2})^2} = \frac{4}{3}\]

  1. \[ P(|X - 10| \geq 9), k = \frac{9}{\sigma} = \frac{9}{\sqrt{\frac{100}{3}}} \] \[P(|X - 10| \geq 9), k = \frac{9}{\sigma} = \frac{9}{\sqrt{\frac{100}{3}}}$. \] \[k = \frac{9}{\sqrt{\frac{100}{3}}} = \frac{9\sqrt{3}}{10} \]

\[P(|X - 10| \geq 9) \leq \frac{1}{k^2} = \frac{1}{(\frac{9\sqrt{3}}{10})^2} = \frac{100}{27}\].

  1. \[ P(|X - 10| \geq 20), k = \frac{20}{\sigma} = \frac{20}{\sqrt{\frac{100}{3}}} \]

    \[ k = \frac{20}{\sqrt{\frac{100}{3}}} = \frac{20\sqrt{3}}{10} = 2\sqrt{3} \]

\[ P(|X - 10| \geq 20) \leq \frac{1}{k^2} = \frac{1}{(2\sqrt{3})^2} = \frac{1}{12} \]