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 expected time for the first of these bulbs to burn out is 10 hours.
λ = 1/1000
Expected_value = 1 / (100*λ)
Expected_value## [1] 10Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density fZ(z) = (1/2)λe−λ|z|
Z + X2 = X1
if z > 0 \[fZ(z) = \int_{-infinity}^{infinity}fX1(X1)fX2(x2)~dX2\] \[fZ(z) = \int_{-infinity}^{infinity}fX1(X2 + Z)fX2(x2)~dX2\] \[fZ(z) = \int_{0}^{infinity}λe^{-λ(z+x2)}λe^{-λx2}~dX2\] \[fZ(z) = \int_{0}^{infinity}λ^{2}e^{-λz}e^{-λx2}e^{-λx2}~dX2\] \[fZ(z) = \int_{0}^{infinity}λ^{2}e^{-λz}e^{-2λx2}~dX2\]
\[fZ(z) = λ^{2}e^{-λz}\int_{0}^{infinity}e^{-2λx2}~dX2\]
\[fZ(z) = λ^{2}e^{-λz}(0-(1/2)(-1))\] \[= 1/2λ^{2}e^{-λz}\]
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. (a) P(|X − 10| ≥ 2).
mean = 10
variance = 100/3
standard_deviation = sqrt(variance)
#2 = kσ
k = 2/standard_deviation
1/k^2## [1] 8.333333mean = 10
variance = 100/3
standard_deviation = sqrt(variance)
#2 = kσ
k = 5/standard_deviation
1/k^2## [1] 1.333333mean = 10
variance = 100/3
standard_deviation = sqrt(variance)
#2 = kσ
k = 9/standard_deviation
1/k^2## [1] 0.4115226mean = 10
variance = 100/3
standard_deviation = sqrt(variance)
#2 = kσ
k = 20/standard_deviation
1/k^2## [1] 0.08333333Resources: * https://en.wikipedia.org/wiki/Exponential_distribution * https://llc.stat.purdue.edu/2014/41600/notes/prob3205.pdf