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.)
excercise 10 says that the minimum value of indepependent random variable n is exponential with mean μ/n.
mean = 1000
n <- 100
mean/n## [1] 10#The expected time for the first of these bulbs to burn out is :Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = X1 − X2 has density
\(fZ(z)=\int_{-\infty}^{\infty}{{f}_{x1}(x_{1}){f}_{x2}(x_{2})dx}\)
if z=x-y z=x+(-y)
\({f}_{w}(w)=\int_{-\infty}^{\infty}{{f}_{x1}({x}_{1})f_{-x2}(z-x_1)dx}\)
\(f_{-x2}(z-x)\quad=\quad f_{x2}(x-z)\)
if z < 0 \(\int_{0}^{\infty}{\lambda{e}^{-\lambda x}{\lambda{e}^{{-\lambda(x-z)}}dx}}\)
\(\lambda{e}^{\lambda x}\int_{0}^{\infty}{\lambda{e}^{-\lambda x}}dx\)
\(\lambda{e}^{\lambda x}(-\frac{1}{2}{e}^{-2\lambda x}\int_{0}^{\infty}{)}\)
\(\frac{\lambda}{2}{e}^{\lambda z}\)
for z > 0
\(\frac{\lambda}{2}{e}^{-\lambda z}\)
Let X be a continuous random variable with mean μ = 10 and variance \({ \sigma }^{ 2 }\) = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities
\(P(|X−μ|≥k\sigma )≤\frac { { \sigma }^{ 2 } }{ { k }^{ 2 }{ { \sigma }^{ 2 } } } =\frac { 1 }{{ \sigma }^{ 2 }}\)
\(P(|X−10|≥2)\)
v=sqrt(100/3)
u=10
x=2
(1/(x/v)^2)## [1] 8.333333\(P(|X−10|≥5)\)
x=5
(1/(x/v)^2)## [1] 1.333333\(P(|X−10|≥9)\)
x=9
(1/(x/v)^2)## [1] 0.4115226\(P(|X−10|≥20)\)
x=20
(1/(x/v)^2)## [1] 0.08333333