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.)
According to the properties of exponential distribution, if \(Xi∼ Exp(λi)\) then \(min(Xi)∼Exp(Σλi)\)
as given: \(E[x]=1000\)
We know that \(1/λ=E[x]\)
Therefore, \(λ = 1/1000\)
Next solving for nλ: \(nλ=100/1000 = 1/10\)
\(P(min{X1,X2,...,Xn}<=x)=1−e^{-x/10}\)
\(E[minXi]= 1/(1/10) = 10\)
The expected time for the first of these bulb to burn out is 10 hours.
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter . Show that Z = X1 − X2 has density:
\(fZ(z) = (1/2) e− |z|\)
We know that
\(f(x_{1})=λe^{−λx_{1}}\)
and that
\(f(x_{2})=λe^{−λx_{2}}\)
So the convolution would be
$λe^{−λx_{1}} * λe^{−λx_{2}}= λ{2}e{−λ(x_{1}+x_{2})} $
Because \(Z=X_{1} - X_{2}\)
then we can conclude that \(x_{1} = z +x_{2}\)
Using substitution, we can recreate the earlier PDF as follows:
$ λ{2}e{−λ((z+x_{2})x_{2})} = λ{2}e{−λ(z+x_{2})} $
If we assume that z is positive (z>0) then:
\(\int_{0}^{\inf} λ^{2}e^{−λ(z+x_{2})}dx_{2} = (1/2)λe^{-λ|z|}\)
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.
Given:
μ = 10 σ = (100/3)^(1/2)
k = 2/((100/3)^(1/2))
1/k^2## [1] 8.333333k = 5/((100/3)^(1/2))
1/k^2## [1] 1.333333k = 9/((100/3)^(1/2))
1/k^2## [1] 0.4115226k = 20/((100/3)^(1/2))
1/k^2## [1] 0.08333333