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11, 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.)

E[Xi] = 1/??i

?? = ??1+…+??100 = 100/1000 = 1/10

expected_time = 1/(1/10)
expected_time

## [1] 10

14, Assume that X1 and X2 are independent random variables, each having an exponential density with parameter ??. Show that Z = X1 ??? X2 has density

fZ(z) = (1/2)??e^(????????z|)

Answer: Answer: 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)). #So when we 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|).

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1, Let X be a continuous random variable with mean ??=10 and variance ??2=100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

P(|X?????| ??? k??) ??? (??^2)/[(k^2) * (??^2)] = 1/ (k^2)

mean (??) = 10, variance (??^2) = 100/3, Standard_deviation(??) = sqrt(100/3)

a) P(|X ??? 10| > 2).

Solution: the probability <- 1/(4*3/100) = 8.3, #so probability is about equal 1.

b) P(|X ??? 10| > 5).

Solution: the probability <- 1/(25*3/100) = 1.3, #so probability is about equal 1.

c) P(|X ??? 10| > 9).

Solution: the probability <- 1/(81*3/100), #so probability is about equal 0.41.

d) P(|X ??? 10| > 20).

Solution: the probability <- 1/(400*3/100), #so probability is about equal 0.083.