Chapter 7. page 303

  1. 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.)

Answer:

Here, Total lightbulbs (n) = 100 lifetime (mean)= 1000 hours

Therefore, the total expected time for the first of these bulbs to burn out = mean / n = 1000 hrs / 100 = 10 hrs.

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

Answer:

Here,

X1 and X2 are defined as f(x1) = lambdae^(???lambdax1) f(x2) = lambdae)(???lambdax2)

not sure how to do this one

pg 320

1 Let X be a continuous random variable with mean ?? = 10 and variance = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities. (a) P(|X ??? 10| >= 2). (b) P(|X ??? 10| >= 5). (c) P(|X ??? 10| >= 9). (d) P(|X ??? 10| >= 20).

Answer:

# variance
variance = 100/3
bounds <- c(2,5,9,20)
# Chebyshev's upper bound
Chebyshevs_upper_limit <- variance/(bounds^2)
Chebyshevs_upper_limit
## [1] 8.33333333 1.33333333 0.41152263 0.08333333
# let's present it in the matrix
matrix(c(append(bounds, Chebyshevs_upper_limit)), nrow = 4, ncol = 2)
##      [,1]       [,2]
## [1,]    2 8.33333333
## [2,]    5 1.33333333
## [3,]    9 0.41152263
## [4,]   20 0.08333333