We want to do is we actually want to do some investigation on what is called the normal probability distribution function. And that function um is of the form F of X is equal to one over um delta Square root of two pi E to the negative one half. And then over here we have x minus mu over delta squared um where mu is known as the mean of the distribution or also, which is called this center. And then of course, delta is what is called the standard deviation, which is also known as the spread or the scatter, which we also know that the integral from negative infinity to positive infinity of this function is actually equal to one. And so we want to do some of our own investigation. And so the first thing we're gonna do is one, we're going to kind of make this a little bit simpler. We're going to let the mean equal zero and the standard deviation equal to one. So that's going to simplify or function just a bit. So f of X Is equal to one over the square root of two pi E. To the negative one, half X squared. Okay. And we want to kind of know what this graph looks like, and so we're going to actually do that by hand by incorporating some of our derivatives and what we know about increasing, decreasing with respect to our derivative. So the first thing we need to do is take our derivative of our function. So one over the square root of two pies, just a constant number and the derivative of either the U. Is you prime. So we're going to take the derivative of the exponents, which is equal to negative X. And then E to the negative E. To the U. Which is going to be E to the negative one, half X squared. Yeah. Okay. So now where does this derivative get a zero? Well, we know that the exponential function never goes to never will equal zero. So the only time that this derivative will equal zero is when X equals zero this part right here. Okay. So we're gonna divide our interval because we we know that the domain for this function is negative infinity to positive infinity. And so we're gonna go from negative infinity is zero and 0 to positive infinity. We're going to pick a test point. So -1 and one. So my derivative evaluated at -1 is going to be positive. So I know my function is increasing over that in a role and now we're gonna take the derivative of one. Which we now know that derivative will be negative for that particular case. And so we now know that our function is decreasing. So where my function goes from increasing to decreasing. I know I have a relative max and that relative max zero comma one over the square root of two pi. Okay so that's what we have now. So now we can do kind of a quick sketch. So I know that what does this function look like? It has a relative max at zero and 1 over the screw to two pi. And now it is it will be um increasing from negative infinity to zero and in decreasing from 02 infinity. So that's kind of what it looks like. Okay and so now you can almost say that it is actually an even function, right? So it's going to be an even function. It looks pretty symmetrical. So this is actually going to be an even function. Okay, so that's the first thing what we wanted to do is just kind of take a graph of it. Now we actually want to transition to a graphing calculator and we want to actually evaluate Um from the integral from negative end to end of one over the square root of two pi E to the negative one half X squared dx. And we want to do it for in equal one In equal to and in equal to three. So that's what we're gonna do. And we're going to see does it look like as in increases um does that um approach some single value? So let's go ahead and switch over. I have this actually over here and so you know, I already have it Graff diamond gizmos. Um I like Desmond's because you actually will do the evaluation for you. Um And so if I Go from -1 To positive one, I noticed I get .6 8-7. And now if I change my limits, the two And -2 I get .9545. So it does look like it's approaching and it looks like it actually is approaching what we think it should be or what we were told that it was and that was one right now. .97300. Okay, so let's go back and put that in there. So now we know that The integral from negative 1 to 1 of one over the square to two pi E to the negative one half X squared dx is about point 6827. The integral from -2- two. Of the function was 0.9 545 And in the integral from -323 was .99 Something so .9973. Okay. Um and now what we want to do a third step is actually um try to give in support of this. Also give an argument um that the integral from negative infinity to positive infinity of that function Actually does equal one. Okay. Um and it gave us some hints um your book gave us a little bit of hints. Um The book gave us a hint um that what we want to do is for X values greater than one. We want to actually show um that um from zero less than E to the negative one half X squared is actually less than E to the negative X over two. Okay. So how are we gonna do this? And I'm just going to start with this X greater than one. So I know that X squared is going to be greater than next. So if X is greater than one, that I know X squared is going to be greater than next. Um I know that X squared over two, it's going to be greater than X over two. And now if I change my signs, I have to flip the direction of the inequality and now if I raise both sides by E, I have now shown and we're going to do it for values greater than zero. That um my original function is in fact less than um E to the negative X over two. Okay, so now what I want to show is that for Be greater than one, that the integral from B to infinity of E to the negative X over two. D. X actually does go 20 Okay. Um and so we're gonna go ahead and get that started. So we know that the limit as B goes to infinity of the in a goal from B to infinity you have eaten the next X. Oh X over two D. X. Okay. Um and so this is as B approaches infinity. Okay, so now what we want to do is if I can show that this converges, then I know that my original function since it's less than that also converges now what it converges to, I'm not sure but I know that it converges and if I actually show that this converges to zero um for B values greater than one, then I can almost estimate that my other function for X values for B values greater than one will actually approach zero as well, which means most of my um region is going to be contained in that um in equal one in equal to in equal three, if that makes sense. Um Okay, so let's go ahead and get started. So um now if I do um a use up so I'm gonna let you be one half negative one half X. Then do you? Is equal to negative one half D. X. Which means that D. X. Is equal to negative to do you? Okay. And you notice I did leave off that one over the square root of two pi. Um Which is just a constant number. So this is going to be negative too. The limit as big as infinity of the integral of be. Oh so let's change our upper and lower limits since we're changing variables. So if X. Is be then this is negative lips negative B. Over to and if X is infinity this is negative infinity each of the you do you? Okay and so now what I can do is if I switch my upper and lower limits then I changed a sign of the two. So this is going to go from negative infinity to negative B over to either the U. Do you. Okay. And so the integral of each of the you is just either the U. And so this is going to be too times the limit as B goes to infinity of either the U. Evaluated at negative B over two and at negative infinity. So this will be equal to two. This will be the limit as he goes to infinity of E. To the negative B over to minus E to the negative infinity. So um this will actually go to zero because this is one over E. To the infinity. This will this this term will go to zero and then I will do the exact same thing here. Right? So if be as infinity, this will be negative infinity over two, which is in essence negative infinity. So that goes to zero. And so what that is telling me here is that up here for X values and this is my argument for X values greater than one And be value greater than one. That this area actually goes 2 0. And so most of most of that integration is in between these two values which now we know that based off of our ends, we know that this actually kept going got closer 21 right. And so it's going to actually get to the point where it's going to go to one and then substance subsequently um this will go to zero here and then I'll just have most of this and contained as one. That makes sense. I hope this helped help you understood what I was trying to say.