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 life is 1000 hours so lambda = 1/1000.
If 100 are bought then it becomes 100* lambda = 1/10.
The expected minimum is 1/1/10 which is 10 hours.
The answer is 10 hours.
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) = (1/2)lambdae^(−lambda*|z|)
X1 and X2 are defined as f(x1) = lambdae(−lambdax1) f(x2) = lambdae(−lambdax2)
convoluting them means multiplying f(x1)f(x2) lambdae^(−lambdax1)lambda*e(−lambdax2) this becomes lambda^2e(-lambda(x1+x2))
subsitute Z = x1+x2 lambda^2e^(-lambda(z+2x2))
integrating the function from -z to +z to find the density becomes: f(z) = 0.5 * lambda * e^(lambda*|z|)
Let X be a continuous random variable with mean μ=10 and variance σ2=100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
the inequality is P(|X-mu| >= k*sigma) <= k^-2 sigma = sqrt(100/3)
2 = k*sigma so k = 2/sqrt(100/3)
the probability is k^-2 so 1/(4*3/100) = 8.3. so 1.
5 = k*sigma so k = 5/sqrt(100/3)
the probability is k^-2 so 1/(25*3/100) = 1.3. so 1.
9 = k*sigma so k = 9/sqrt(100/3)
the probability is k^-2 so 1/(81*3/100) = 0.41
20 = k*sigma so k = 20/sqrt(100/3)
the probability is k^-2 so 1/(400*3/100) = 0.08