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Suppose two identical components are connected in parallel, so the system continues to functionas long as at least one of the components does so. The two lifetimes are independent of eachother, each having an exponential distribution with mean 1000 h. Let $W$ denote system lifetime.Obtain the moment generating function of $W,$ and use it to calculate the expected lifetime.

$\dfrac{2}{(1-1000^t)(2-1000^t)},1500 $h

Intro Stats / AP Statistics

Chapter 4

Joint Probability Distributions and Their Applications

Section 1

Jointly Distributed Random Variables

Probability Topics

The Normal Distribution

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we're told that two identical components are connected in parallel, that the system continues to function as long as a least one of the components does. So we're also told the two lifetimes are independent of each other. Each of them have an exponential distribution with a mean of $1000. Let WB the system Lifetime, were asked to find the moment generating function of W. And to calculate the expected lifetime. Well, we have that the lifetime was 1000 hours, so we'll take a constant A to be won over 1000 so one over the lifetime. We have that because the system only fails that both of them failed. W. Is is a maximum of to exponential and very and therefore probability. Density function is FW FW is going to be the maximum. I'm sorry we actually had a result about this earlier. That's if we have a maximum of two variables. Then this is going to be the cumulative density function two times f x of w times the PdF effects of W, which is equal to because he's their exponential. Two times one minus beat to the negative A w times okay into the negative A W, which is equal to to a times e to the negative a w times one minus e to the negative a w. So this is the probability density function and therefore the moment generating function. W This is going to be the expected value of E to the T W, which is the integral over all the possible values of teeth. So from zero to infinity of E to the t W Times pdf of W, which we know is to a it's a negative a w times one minus e to the negative a W D w I'm sorry. Integrating really possible values of w night. The possible values of T It's a factoring out breaking up the integral we get. This is 28 times the integral from zero to infinity of e to the negative a minus T W d w then minus two a times the integral from zero to infinity of into the negative to a W d w well, actually gonna stick here it should be eat to the negative to a minus. T W d w and then integrating this This is pretty easy. We get to a over a minus T minus to a over to a minus T. And so combining these we get to a squared over a minus T times two a minus t which substituting back in you get this is going to be to over one minus 1000 t times tu minus 1000 t simply divide through bye a squared top on the bottom and therefore using the moment generating function we have the expected value of w is we know from previous problem This is the derivative of the moment generating function at w I mean, at zero sorry and evaluating the derivative here this could be found taking the derivative, which is easy, but we have that this is going to be 1500. So the way to see that is Navy. We have the derivative of the bottom which is going to be and evaluated it zero. So it's like negative 1000 times, too since 2000. Then one times no right, Have a good deal

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