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The density function $$ f(x) = \frac{e^{3 - x}}{(1 + e^{3 - x})^2} $$is an example of a logistic distribution.(a) Verify that $ f $ is a probability density function.(b) Find $ P (3 \le X \le 4) $.(c) Graph $ f $. What does the mean appear to be? What about the median?

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A. $=\frac{1}{1}-\frac{1}{\infty}=1-0=1$B. 0.23C. mean and median at $x=3$

Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 5

Probability

Applications of Integration

Campbell University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

Lectures

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The density function$$…

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The function $f(x)=x^{3}$ …

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\begin{equation}\begin…

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(a) Explain why the functi…

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Let $X$ be a continuous ra…

Okay. And this question were asked about a probability distribution function and we have f of x equals E. to the 3 -1 Divided by one plus e 23 minus X quantity squared. So the first part asked us to see if this is in fact a probability distribution function. What that means is is the integral equal to one. Because when you have probabilities you need all the probabilities to add up to one. So the question is if this equals one. And so the first thing that we're gonna want to do is use a U substitution. So I'm going to use U equals one plus E to the three minus X. To make the denominator a little bit simpler. Then using the chain rule you get D U equals negative E To the three -X. So we can write this integral from X equals negative infinity to x equals infinity of the numerator becomes negative to you. The dominator becomes U squared. And the integral that using the power rule is going to be one over you. Now we have to evaluate this with X values. So I'm going to replace this with X now, one over one plus E to three, modest X. And evaluated from negative infinity to infinity. When you plug in infinity you get e raised the negative infinity which is going to be zero. So the first part becomes 1/1 plus zero. And then when you plug in negative infinity what that does is the negative and the negative cancel out. So you actually get 1/1 plus infinity. And anything over infinity is going to be zero. So this comes out to be equal to one. So that means that it is a pdf The second part asked what's the probability that X is in between three and 4? Mhm. And you can evaluate this by setting up an integral From 3 to 4 of this fx dx and from the previous part we know that this is one divided by one plus 8 to 3 minus X Evaluated at 3: four. So when you plug in four you have one divided by one plus 2 to 3 miles four which is one divided by one plus E to the negative one. And when you plug in three you get one divided by one plus E to the zero And each of the zero is 1. So you could simplify this to be one half if you wanted to one plus one that and in decimals this is approximately 0.23. Okay. The next part asked to graph the function and to approximately what the mean and median is. So down here, I have a picture of the graph and from here we can see that the I mean and the median is going to be around here Okay, is going to be around here which is at X equals three And the reason why I say that is because this function is pretty symmetric so it looks like everything to the left of it will be equal to everything to the right, And so you can use that to justify that, the mean and the median is X equals three.

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