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# Find the cumulative distribution function for the probability density function in each of the following exercises.Exercise 1

## $F(x)=\frac{1}{18}\left(x^{2}-x-2\right), \quad 2 \leq x \leq 5$

#### Topics

Continuous Functions

### Discussion

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##### Top Calculus 1 / AB Educators ##### Heather Z.

Oregon State University  ##### Michael J.

Idaho State University

### Video Transcript

for this problem were given the probability density function represented by little F of X and were asked to find the cumulative distribution function, which is typically represented by capital of X. And the relation between these two is that the cumulative distribution function is the integral of the probability density function and therefore, the probability density function is the derivative of the cumulative density function. So to start out, find cap f of X, we're just gonna take the integral off 1/9 times X minus 1/18. I said simple. We're going to get 1/18 X squared, minus 1/18 X uh, To be clear, that X is in the numerator, Uh, plus c okay, because this is, uh, an integral with no bounds on it. So we would get a plus e on the end of it here and so to actually get our final answer where you need to figure out what that C is going to be, there's a couple of ways we can find that the way that I prefer is poking in the right end point of our interval, which, in our case, it's five and keeping in mind that when we're on the very right hand side of our distribution, we should have accumulated ah, 100% of the probability up to that point. Um, so actually, let's write it. And it's it is supplying five into our cap. Affects will give us the following 25/18 minus 5/18 plus c. And I said that we want that to be equal toe one because we want to have accumulated 100% of the probability A Z, we move all the way to the right hand side of our distribution. So doing the subtraction, we're gonna get 20/18 plus C equals one, and from there it's clear to see that C equals negative too over 18 which is the same as negative one overnight. So what are final answer for this question should look like we should say cap F of X is equal to the following in the middle here, we're gonna put our function that we integrated to with our constancy plugged in. So it's gonna be won over 18 X squared minus 1/18 X minus 1/9, and you could factor out the 1/18 here. If you wish on this function is going to be only for our X. That is, on our interval X between two and five inclusive. Anything that's below for any X that is less than two or X, that is greater than five. We're gonna have different things to find eso for the less than two portion, it's just gonna be a zero. And for the greater than five portion, it's just gonna be a one. And this is because since our function is defined only to have probability between two and five and this capital ffx represents how much probability we've accumulated up to a point for all of the points before, too. There is no probability. So we just want zeros there. And for all the points after five, we've already accumulated all 100% of the probability. So we just have a one there. So our final answer would be this whole section here University of Nevada - Las Vegas

#### Topics

Continuous Functions

##### Top Calculus 1 / AB Educators ##### Heather Z.

Oregon State University  ##### Michael J.

Idaho State University