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Suppose the amount of rainfall in one region during a particular month has an exponentialdistribution with mean value 3 in., the amount of rainfall in a second region during that samemonth has an exponential distribution with mean value 2 in, and the two amounts are indepen-dent of each other. What is the probability that the second region gets more rainfall during this month than does the first region?

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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University of St. Thomas

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So we're told that in two different regions there's two different mean values of rainfall. In one region we have three inches as our mean value. In another region we have two inches. And we're told also that they are both exponentially distributed. So what we're gonna do first is find these two distributions. So how we do this is this lambda here is just equal to one over our expected or are mean value. So for the first one for F. Of X, we can say that lambda is equal to one over a mean value which is three, so one third E. To the negative one third X. And for our 2nd 1 we can say F F. Y is equal to Well I mean value is two. So 1/2 eat the negative one half. Why? And now what we can do is we can say that their joint probability function is equal to the product of fx and Fy. And the reason we can do this is because they're independent of each other. So if we take ffx F of Y and multiply it by F. Of X. We get this is equal to one half. I'll do this in black actually Is equal to 1/2 E. to the negative one half. Why? Multiplied by 1 3rd E. to the -1 3rd. X. And if we simplify this, you can see that this is equal to one half times one third is 1/6. And we can just add the exponents here. So we get negative one FY one third X. And now to find the actual probability that the second region has a higher amount of rainfall in the first region we're going to set up a double integral, define this using this joint probability and it's gonna look like this. So we have a first. Our outer integral. Our outer integration is from 0 to infinity. And our Inner one is from 0 to why. And this is defined our why value being greater than our X. Value or our Our amount of rainfall to be greater in the 2nd region than it is in the first region. And then we just do the integral of our joint probability minus one third X. And we want to have dx then dy we're going to integrate with respect to X first and then integrate with respect to Y. And so what we can do here is we can set this up a little bit differently so that is equal to this. We're going to put this in parentheses so that it's a little bit easier to look at. And I'm gonna cut this into 1/2 times 1 3rd again. So we have eat the negative E. To the negative one third X. And then this is multiplied by 1/2 Each. The negative one half. Why the X. Dy. And the reason I'm going to do this is because since we're only integrating with respect to X. And actually this dx should be and here is um the the because we're on the integrating with respect to X. Where you can look at this one half, either the negative one half, Y is a constant. So we can just do this stuff in parentheses first. So the integral of each of the negative one third X is negative three either negative one third X. So if we If you multiply that by 1/3 just get negative E. To the negative one third X. And this is from zero to why? So at zero we have negative E. to the zero which would be -1. So we have one because we're also minus sing it. That negative one will just become one. And then at why we have negative E. to the -1 3rd. Why? So are integral. Will now look something like this Times 1/2. -1/2. Why? The Y. And now we can do is multiply this through. So we have From 0 to Infinity of 1/2. and actually we'll take that one half and we'll put it out here in the front. So we have E. To the negative one half. Why minus E. To the negative one third plus negative one half is negative 5/6 negative 5/6 Y. Mhm. And this is all with A. D. Y. And now if we integrate these both separately we can do it separately and then just minus them. Or just do it together. It's both. Both are doable. So if we take the integral of E. To the negative one half. Why? Well we just need to take this exponents reciprocated and multiply it by a negative one since it's a negative. So we have negative one half. We need to bring down a -2. We actually don't have to multiply by -1. We just have to bring down the reciprocal. So we have negative to E. To the negative one half. Why? And the reason this works is if we were to take the derivative then we would have either the negative one half Y. Times negative one half, which gives us disappear. So for the second part we need to bring down the reciprocal of 5 6. So we have minus negative 5/6 sorry negative 6/5 Each, the -5 6th y. And this is all From 0 to Infinity. And then We are multiplying by 1/2 out here in the front as well. So if we look at this equation, you can see that we're only going to be putting either the zero and the infinity into the exponent. So when you put zero into this exponents, the heat of the negative one halftime zero, that just becomes one. So we have negative two times once we have negative too minus. And then the same thing happens over here. So we're just Left with this positive 6/5 and this is at the zero part, so this is going to be minus. And then for the infinity part, if we put infinity into these exponents can see that any positive number and and Any positive number above one that has a negative infinity in the exponent is going to go to zero. So we're gonna have negative two times zero which is zero and negative 6/5 time zero which is zero. So this is just gonna be zero in the front and then this all is multiplied by one half. And now if we just do this addition and multiplication, we get negative two times negative. One is two minus 6/5 is 4/5 so we have 4/5 inside the parentheses in this one half still, So this ends up being four tense.

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