### 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?

Since the lifetime of each bulb follows an exponential distribution, and the expected lifetime is 1000 hours, we set mu = 1000. For an exponential distribution, the intensity paarmeter lambda = 1/mu. Therefore lambda = 1/1000 = 0.0001

The expected time for the first of the bulbs to burn out would the expected value of the minimum lifetime (first one to burn out means the least lifetime amongst all the 100 bulbs). This minimum lifetime also follows an exponential distribution.

Denoting the minimum lifetime as M, we have lambdaM = 1/sum[(lambda1)+(lambda2)+…….(lambda100)] lambdaM = 1/(100/1000) = 1/(1/10) = 10 Therefore expected value of minimum lifetime for 100 bulbs = 10 hours.

E[M] = mu/number of bulbs = 1000/100 = 10

### Assume that X1 and X2 are independent random variables, each having an exponential density with parameter lambda. Show that Z = X1 ??? X2 has density fZ(z) = (1/2)lambdaxe^(???lambdax|z|).

### Let X be a continuous random variable with mean ?? = 10 and variance sigma^2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities. (a) \(P(|X-10|\ge2). (b) P(|X-10|\ge5). (c) P(|X-10|\ge9). (d) P(|X-10|\ge20)\)

Chebychev’s inequality states:

\[P(|X-\mu|\ge\epsilon)\le(\sigma^2)/\epsilon\]

Setting \(\epsilon\)=2, we have: \[P(|X-10|\ge2)\le(100/3)/2\] So the upper bound is 100/6 = 50/3

Setting ??=5, we have: \[P(|X-10|\ge5)\le(100/3)/5\] So the upper bound is 100/15 = 20/3

Setting ??=9, we have: \[P(|X-10|\ge9)\le(100/3)/9\] So the upper bound is 100/27

Setting ??=20, we have: \[P(|X-10|\ge20)\le(100/3)/20\] So the upper bound is 100/60 = 5/3