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.)
Exercise 10 tells us that “Let M be the minimum value of the Xj .
Show that the density for M is exponential with mean \(µ/n\)” so for us to find the expected time
for the first of these bulbs to burn out, we plug in our numbers:
\(µ\) = 1000 hours
\(n\) = 100 hours
E(M) = \(µ/n\) = 1000/100 =
10.
So the expected time is 10 hours.
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density \(fZ(z) = (1/2)λe^(−λ|z|)\)
To find exponential density, we use the formula: \(f(x) = λe^(-λx) x>0, λ>0\)
So for X1 (using x1 as x): \(f(x) =
λe^(-λx1)\)
So for X2 (using x2 as x): \(f(x) =
λe^(-λx2)\)
Then we plug into \(fZ(z)\) as Z is
X1-X2 with X2 = X1-Z so we only have x1 to work with:
\(fZ(z) = fx1-x2(z)\)
=> ∫\(λe^(-λx1)λe^(-λ(x1-z))\)
=> ∫\(λ^2e^(-2λx1+λz)\)
=> ∫\(λ^2e^λ(-2x1+z)\)
=> \((1/2)λe^(-λz)\)
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.
Chebyshev’s Inequality: \(P(|x−μ|≥a) ≤ \frac{σ^2}{a^2}\)
So, \(\frac{σ^2}{a^2}\) => \(\frac{\frac{100}{3}}{2^2}\) => \(\frac{\frac{100}{3}}{4}\) => ~8.333333
So, \(\frac{σ^2}{a^2}\) => \(\frac{\frac{100}{3}}{5^2}\) => \(\frac{\frac{100}{3}}{25}\) => ~1.333333
So, \(\frac{σ^2}{a^2}\) => \(\frac{\frac{100}{3}}{9^2}\) => \(\frac{\frac{100}{3}}{81}\) => ~0.411522
So, \(\frac{σ^2}{a^2}\) => \(\frac{\frac{100}{3}}{20^2}\) => \(\frac{\frac{100}{3}}{400}\) => ~0.083333
(Questions from the probability textbook)