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Given f(x) = x?/3 , F1<X<2 a Prove that f(x) id a pdf b. find P(x 7 1)...

Question

Given f(x) = x?/3 , F1<X<2 a Prove that f(x) id a pdf b. find P(x 7 1)

Given f(x) = x?/3 , F1<X<2 a Prove that f(x) id a pdf b. find P(x 7 1)



Answers

Show that the given function is a pdf on the indicated interval. $$f(x)=x+2 x^{3},[0,1]$$

So this question we know there are function after is going to be a probability function. If it satisfies dishonorable equals one. So we can rewrite what we started with our original interval. Yeah, with me. Zero of 70. The power of seven axe. Yes. Which we're now gonna put is the limit the limit as he goes to infinity. This is now our interval t zero dx witch. Well, actually, just we're actually just revising this portion now as negative e to the power of negative seven acts still cheesier. So now, All right, What we have now is our entire updated limit. The limit is she goes to infinity of one minus Z to the power of negative 70 which, at this point, let me evaluate me. Realize is just gonna be one. So the answer to this question we have shown now this is probably function because we have satisfied that we have here is equal

So here we want to show that the function f of X equals three over X times X squared is, ah, probability density function on the interval from 0 to 2. So So one way we could write that is that, um does the probability from zero lesson or equal to X less cynical toe to equal one, which would have to equal the integral from 0 to 2. And let me just make sure I get this exactly right of three over x times X squared d x X squared. Okay, so is that true? And if so, well, then we know that we have an appropriate probability density function. Now, before we integrate this function, let's do a little simplification. It's gonna make our actual integration a good deal easier. And so we're going to rewrite this by multiplying out three divided by X Times X squared. So it's really the integral from 0 to 2 of three. Will x squared divided by X. Really? It's gonna be simplified. Let's, uh, we could write that of we could multiplies out three x squared over X de acts Well, that's gonna equal the integral from 0 to 2 of just three x oops, just three x dx. And so if we're going to evaluate that well, this becomes three times X squared over to evaluated from 0 to 2. Let's do that simplification. And let's see if we end up with a value of one. So down here, we're gonna replace X with two. So we have three times two squared, divided by two minus three times zero squared, divided by two Well, two squared is 43 times four is 12. 12 divided by twos. This is actually equal a six minus zero, which is not equal toe one. So this will not be a probability density function over this specific interval.

So were given a specific function. Um, and let's let's just write it down. Were given the function. Um f of X is equal to four x cubed, and we want to evaluate it on the final interval from zero to one. And the idea is we have to determine whether this function is a probability density function on this interval from 0 to 1. So keep in mind that a probability, um, can be anywhere from zero. And will you use the value? P zero lesser goal api lessner goto one. So something can happen anywhere with a likelihood from never toe always or somewhere in between. But you can't have a probability that's greater than one or a probability. Let's less than zero. So, really, what we have to determine is we want to integrate the function from 0 to 1 of four x cubed d X, and so sorry, no dx of four x cubed. And so, if we're going to do that, um, my apologies. We do want the d. X. So if we're gonna do that, well, first we have to find integral of four x cubed. And so to keep in mind when we find the integral well in general, and they will write it over here of a simple function like this. It becomes X to the n plus one over, um, and plus one. So, really, this is going to be four times X rays to the fourth over and plus one. Well, that's four. It's gonna be evaluated from 0 to 1. Okay, so this is going to be equal to Well, look at that four divided by four. That cancels out. So, really, it's gonna be equal to one to the fourth minus zero to the fourth, which is going to equal one. And so hence the area under the curve of this function four x cubed does, in fact, equal one. And so, yes, it will be a probability density function.

So here we have to evaluate the following function F of X equals CO sign X on the interval from zero to pi halfs. And we didn't want to determine whether it's a probability density function. So if the inner girl from zero to hi, Hafs uh, co sign X DX. If this turns out to be a probability density function, well, then the area under this curve on its interval from zero to pi halfs has to be equal to one because probability has to be anywhere between zero, including upto one inclusive so you can have a probability greater than one or less than zero. Oh, yes, this is gonna be equal. Well, the integral of coastline of X is sign of X, and we're going to evaluate it. Still from zero to pie have and so if we're gonna value it that well, first we need to find sign of pie halfs. Okay, this is gonna be equal to sign of pie halves up. Sorry. Sign of I have minus sign of zero. So if you can picture the unit circle Awesome. If not, you're gonna wanna pull it up or look at it in the textbook chapter, but sign of pie halfs is the y. Coordinate when we're when we've moved Pie half radiance around the unit circle. So sign of my ass that Why coordinate? Turns out it is the value one when we move zero radiance. Well, that Why, cordon? It's actually just sitting right on the x axis. So we know that sign of zero is zero. And so the integral from zero to pi half of coastline X d x is equal to one and the left with


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