Page 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.)

\(\ E[Xi]=1/λi\)

\(\ λi= λ1+...+λ100 = 100/1000 = 1/10\)

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

\(\ E[Xi]= 10\)

10 hours

Page 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 \(\ fZ(z) = (1/2)λe−λ|z|\) .

\(\ f(x1) = λe(−λx1) f(x2) = λe(−λx2)\)

Multiply

\(\ f(x1)f(x2) λe^(−λx1λ)*e(−λx2)\)

to get

\(\ λ^2e(-λ(x1+x2))\)

Then we substitute,

\(\ Z = x1+x2 λ^2e^(-λ(z+2x2))\)

Lastly we integrate the function from -z to +z to find the density.

\(\ f(z) = 0.5 * λ * e^(λ*|z|)\)

Page 320-321 #1.

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. 8.2. CONTINUOUS RANDOM VARIABLES 321

(a) \(\ P(|X − 10| ≥ 2)\).

\(\ 2 = k*σ\) \(\ k = 2/sqrt(100/3)\)

\(\ k^2 = 1/(4*3/100) = 8.3. ≈ 1\)

(b) \(\ P(|X − 10| ≥ 5)\).

\(\ 5 = k*σ\) \(\ so k = 5/sqrt(100/3)\)

\(\ k^2 = 1/(25*3/100) = 1.3 ≈ 1\)

(c) \(\ P(|X − 10| ≥ 9)\).

\(\ 9 = k*σ\) \(\ k = 9/sqrt(100/3)\)

\(\ k^2 = 1/(81*3/100) = 0.41\)

(d) \(\ (|X − 10| ≥ 20)\).

\(\ 20 = k*σ\) \(\ k = 20/sqrt(100/3)\)

\(\ k^2 = 1/(400*3/100) = 0.08\)