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?
(lambda=1/1000) #the expected life is 1000 hours## [1] 0.001(b<-100* lambda) #100 are bought## [1] 0.1The expected minimum is 1/1/10 which 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)lambdae(−lambdax2) this becomes lambda^2e(-lambda(x1+x2)) subsitute Z = x1+x2 lambda2e(-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| >= ksigma) <= k^-2 sigma = sqrt(100/3) a.P(|X − 10| > 2). 2 = ksigma so k = 2/sqrt(100/3) the probability is k^-2 so 1/(4*3/100) = 8.3. so 1.
b.P(|X − 10| > 5). 5 = ksigma so k = 5/sqrt(100/3) the probability is k^-2 so 1/(253/100) = 1.3. so 1.
c.P(|X − 10| > 9). 9 = ksigma so k = 9/sqrt(100/3) the probability is k^-2 so 1/(813/100) = 0.41
d.P(|X − 10| > 20). 20 = ksigma so k = 20/sqrt(100/3) the probability is k^-2 so 1/(4003/100) = 0.08