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Numerade Educator



Problem 20 Hard Difficulty

Find the cumulative distribution function for the probability density function in each of the following exercises.
Exercise 2


$F(x)=\frac{1}{6}\left(x^{2}-x-6\right), \quad 3 \leq x \leq 4$


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Video Transcript

okay for this question were given the pdf probability density function for a distribution here. Ah, shown us little f of X and were asked to find the cumulative distribution function to find the cumulative distribution function. Given the pdf all that we have to do it is integrated. Our pdf integral 1/3 X minus falling over six with respect to X, uh, without bounds are integral first, Uh, so to do that, we're going to get 1/6 x squared. I just want over six x plus a constant right. And the this is almost the complete answer. We just need to fill in. What are constant is and there's a couple of ways we can do that. The way that I prefer is to pick the right end point of our function and check out what are CDF at that point is gonna be equal to if we plug in four here will get 16 over six minus 4/6 plus c, which is same as 12/6 c. Oops. Getting ahead of myself plus e which is the same as to plus and we actually want our function at that right and point to be equal to exactly one eso. If we set this whole thing equal toe one and solve for C, we're gonna get that C is equal to negative one. And so we can go ahead and plug that back into our original equation up here s O f of X is going to be equal to 1/6 X squared, minus 1/6 X minus one. Uh, but that is only for X on our interval 3 to 4. So we can say three x look three lesson x less than four. And at other points before three, if X is less than three, our functions just gonna be zero because we haven't encountered. And if the probability because probabilities only spread out on this interval. And, uh, it's going to be one for all X greater than four, because after we've gone past four with already accumulated, all of the probability. So our final answer is given like this, um, a piece of his function with this middle one, calculated using the integral like we did here

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