Final answer:
The expected values of X1 and X2 can be calculated. The machine with the longer expected lifetime can be determined. The conditional pdf of X1 given that its lifetime is less than 3 years can be found. The conditional expected value of X1 can be computed. The probability that the lifetime of Machine 2 exceeds 2 years can be determined. The probability that Machine 2 will work for at least another 2 years, given that it has been working for 1 year already, can be found.
Step-by-step explanation:
(a) To find the expected values of X1 and X2:
The expected value of a uniformly distributed random variable is equal to the average of the minimum and maximum values of the distribution. In this case, X1 is uniformly distributed between 2 and 4, so the expected value is (2+4)/2 = 3 years.
The expected value of an exponential distribution is equal to 1 divided by the parameter λ. In this case, λ2 is 0.5, so the expected value of X2 is 1/0.5 = 2 years.
Therefore, Machine 1 lasts longer on average since its expected lifetime is 3 years, compared to Machine 2's expected lifetime of 2 years.
(b) To find the conditional pdf of X1 given that the lifetime of Machine 1 is less than 3 years:
Since X1 follows a uniform distribution, the conditional pdf of X1 given that X1<3 is 1/(b-a) = 1/(4-2) = 1/2 within the interval [2,3]. Outside this interval, the conditional pdf is 0.
The graph of the conditional pdf will be a straight line with slope 1/2 from 2 to 3, and the value at any discontinuity will be 0.
(c) To compute the conditional expected value of X1 given that the lifetime of Machine 1 is less than 3 years:
The conditional expected value of a random variable is equal to the integral of the product of the conditional pdf and the random variable over its support.
In this case, the conditional expected value is equal to ∫x * 1/2 dx from 2 to 3, which evaluates to (1/2) * (9/2 - 4) = 1/4 years.
(d) To find the probability that the lifetime of Machine 2 exceeds 2 years:
The probability that the lifetime of a exponential distribution with parameter λ exceeds a certain value x is given by the complementary cumulative distribution function (1 - CDF(x)).
In this case, the probability that the lifetime of Machine 2 exceeds 2 years is 1 - CDF(2), where CDF(x) = 1 - e^(-λx).
With λ2 = 0.5, the probability is 1 - e^(-0.5 * 2) = 1 - e^(-1) ≈ 0.6321.
(e) To find the probability that Machine 2 will continue to work for at least another 2 years, given that it has been working for 1 year already:
This question is asking for the conditional probability that the lifetime of Machine 2 exceeds 3 years, given that it exceeds 1 year.
The conditional probability can be computed using the formula: P(X>x given X>y) = P(X>x+y)/P(X>y).
In this case, the probability is P(X>3 given X>1) = P(X>3-1)/P(X>1) = (1 - CDF(3))/(1 - CDF(1)).
With λ2 = 0.5, the probability is (1 - e^(-0.5 * 3))/(1 - e^(-0.5 * 1)) ≈ 0.4866.