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 density of minimum value among \(n\) independent random variables with an exponential density has mean \(\frac{\mu}{n}\), where \(\mu\) is mean of exponential density of individual variable.
Here \(n\) = 100
\(\mu\) = 1000
So the expected time for the first of the bulbs to burn out is:
\(E(M)=\frac{\mu}{n}\)
\(\frac{1000}{100} = 10\) hours
Let X be a continuous random variable with mean \(\mu\) = 10 and variance \(\sigma^2\) = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
# set variables
var <- 100/3
mu <- 10
e_a <- 2
# compute
(e_a_upper <- var / e_a^2)
## [1] 8.333333
e_b <- 5
(e_b_upper <- var / e_b^2)
## [1] 1.333333
e_c <- 9
(e_c_upper <- var / e_c^2)
## [1] 0.4115226
e_d <- 20
(e_d_upper <- var / e_d^2)
## [1] 0.08333333