The joint density of Y1 and Y2 is f (y1, y2) =((1/8)*y1*e ^(−(y1 +y2)/2) where y1 > 0, y2 > 0, - and the function is 0 elsewhere The cost C of replacing the two components depends upon their length of life at failure and is given by C = 50 + 2Y1 + 4Y2. Find E(C) and V (C).
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
Hi I am having some trouble with this statistics question.
Y1 and Y2 denoted the lengths of life, in hundreds of hours, for components
of types I and II, respectively, in an electronic system.
The joint density of Y1 and Y2 is
f (y1, y2) =((1/8)*y1*e ^(−(y1 +y2)/2) where y1 > 0, y2 > 0,
- and the
The cost C of replacing the two components depends upon their length of life at failure and is given by C = 50 + 2Y1 + 4Y2. Find E(C) and V (C).
I undrestand that this is an exponential distribution. Do we use an mgf table to find the alpha and beta or is there a better way to compute the E(C) and V(C).
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