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An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in...

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An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past; it has been determined that the particle size (in micrometers) distribution is characterized by the following function:8xX>f(x)elsewhere(a) Verify that this is a valid density function_ (b) Evaluate F(x): (c) What is the probability that a random particle from the manufactured fuel exceeds 2 micrometers?(a) The function

An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past; it has been determined that the particle size (in micrometers) distribution is characterized by the following function: 8x X> f(x) elsewhere (a) Verify that this is a valid density function_ (b) Evaluate F(x): (c) What is the probability that a random particle from the manufactured fuel exceeds 2 micrometers? (a) The function f(x) is a valid density function because for all and because f(x) dx = 8x dx = X<1, (b) F(x) = X>1. (c) P(X > 2) = (Type an integer or decimal rounded to four decimal places as needed.)



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An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by $$ f(x)=\left\{\begin{array}{ll} 3 x^{-4}, & x>1 \\ 0, & \text { elsewhere } \end{array}\right. $$ (a) Verify that this is a valid density function. (b) Evaluate $F(x)$ (c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?

Recommend this question. The wave function for a particle in an in finite square? Well, right. And this is that it between X equals 20 to excuse to l and is your outside off this ridge, I'm gonna find the expectation value. So expectation value to find expectation very off a we function or you have to do we need to multiply its contra gate with the favorable that you want to find an expectation value in this case is x by X and not a body were found Should get any degree across Oh, space float negative fainting to positive in Haiti No, no case, because basically, we function only it's fairly from zero to l and it's, uh, empty out of ice I'll say so we can just leave me the range just so to hell. Let me carry of entities gets just you re function square Papa X, the X I mean, so it had to be a great dish. Um, what you want to do is to first called a vote o a saint square into ST, despite he double anger formula. So science way into just consigned to eggs and or you have to do is to drew the change through oh, into go integration by parts for this to this term can be very simply integrated. This no problem. But for distance, we have to do integration by parts. All right, so this by change rule, keep the first. Um Right. So over here you have two terms. Thanks. Every article saying so we keep you first toe in degree the second term, you might as a way, uh, we differentiate your first of and we need a greedy second toe. So this the integration by parts in, uh can I continue to simplify? So the first time? Just, uh, l square or two. And then for the second half Well, sorry for pain and face it. This is an injured. Your multiple pint. This would be cereal. The center will be zero. So you know that Have you got left? The two right, which is doing it to go and by integrating gets should get a negative sign again. Right? So this last home always to have to substitute in X equals two l and x coast with zero. So an execution bell to be course I for pie. Mary, Don't co sign pie. It's true. It's, uh, indigent. Multiple off course science to buy bank assigned to pi is just one. So this is just for by square this one minus away maybe substituting experts with zero because science hero and we know once again call signs You always one So this time goes to zero and we're left with over two. So this is the expectation value or for position X not to find it probably t that the particle lie between certain range you can do the same integration but result the x variable of this time. So you find probably it he that it exists a relative to a 0.5 I'm mutati range from 0.49 No. So you still fail 2.51? No. Every integrity, probability, density It's just site square. So a very similar integration. Yes. OK, step Yeah, necks So first ill minus away Hell for pie times slain for extra veil. And but this summer, integrating between 0.4 night no to 0.51 help. So first, don't you give us your point. T 02 right? And substituting in we should it signed to play for pie. Bring us a sign 1.6 by the circuit for a spot 5.26 instead. Poll minus five. Now we do the simple assess for the probability at around two point 25 l. It's just about around over four The great Between 2.24 and 0.26 l times again he integrate. You've already testy the and already same expression with you. Except we just started you in different values for our limits, right? We should get, actually, is your point night night instead of I s too Well, at first glance, it wouldn't make sense compared with Oh, but a answer upon a sea states he expected fair do expectation. Very youthful eggs is at over two. But we see that the probability for the particle to exist it's oh, over four. It's actually much greater, Dan probably t two axis around over to How I see is the case. That is because our probability density function in sexually symmetric seeing a trick a boat over to so it actually X is kind of like that tour picks it over four in trio over four, right? So the center off this symmetric probability density function even though it is zero right? He probably to you. But your sexual zero there is the expectation value, right? On average, because off the symmetry, this will give us our to. So there's no contradiction, right? Uh, and the expectation that is does not tell you d highest probability density, but rather it tells you what you would get on average, you forgot to take many, many experimental values from the experiments because you get equal pay for equal probability of getting helpful en trio full, on average, 90 weighted average would give us well.

And this problem we have ah, particle in a box in its ground state and equals one and were asked to find some probability is using the square of the wave function. So first asked for the probability in the first quarter of the box, then in the second quarter And then we're asked to find the total probability of the 1st 2/4 which I heavily suspect will be 1/2. But we'll prove that. So to start, we can I get the way fun the for a particle in a box which will come out to be a sign and pi x over l where a is our normalization constant racks actually find that on our own. So start now, we can get the square of the wave function because this is gonna give us our probabilities, our probability density. Rather So we have a squared sine squared and pie over X and baek several rather ran is equal to one will be left with just pie acts of rail. So to start, we need to get this normalization constant. We can do that by integrating over the entire space, sitting and having an equal to one because the problem for the entire space hasp equal to one. That's a good sanity check. So in writing from zero tale of a sine squared or a squared sine squared pi X several in X city Goto one we have to do it anyway for my part So we'll need a little bit more space We have one equals a squared X over two minus l over for pie Sign of two pi acts of rail with our bounds from zero Nelle in this case, which will come out to be a squared Elvir too minus l over for pie Sign of two pi which will come out to be zero. So the second term cancels out. We're just left with the A squared L over too equals one. Thus, our normalization constants gonna be rude to over. Well, we're gonna use this for the upcoming results. We're also going to use this indefinite integral in the upcoming results. So now we're asked to find the probability from zero to L over four. Same integral, just Ah, this time we're going from zero to l A. Before, it was just a change of bounds and we can put in our constant now to over. Well, that's a squared science word of ah pi x of rail. Same indefinitely. Role would be dealing with just a different constant new, different bounds. So when we actually do the rich, my parts and change the bounds and everything's gonna end up with, uh, 1/4 minus one over to pie, which will be approximately 0.9 it's only 9% probability of Ah, finding this particle in this first quarter is gonna only be 9%. We repeat this process for 1/4 toe 1/2 we're gonna get a considerably larger value. So I was right out the general once again, we're just changing the bounds. Same minute roll, same immigration, my parts. And in this case, we're gonna end up with, uh, 1/4 plus over to pie, which in this case, will be a 0.409 Okay, so let's take a look at these two answers. If we added these two together, it's gonna be essentially 20.5 or half. Which makes sense. The reason is because if we actually plot this probability function Thea probably amplitude or the square of the wave function, we're gonna have is a symmetric function about l over to look something like that. There's a relevant to there's well, so we would expect the area under the curve to be 50% in the first half of the 50% the other half, and it falls up pretty quickly. So we got most of the area and that second quarter in the third quarter and much less on the edges. So if we're tracking a particle in here, we're gonna expect to be much closer to the metal that on the edges. And that's really the interpretation that's important for this problem. Another thing worth noting is that this is only for the ground state. As we go up into higher energy states the page there will be a little bit more interesting, but for this case, uh, it's just centered around the middle, and that's it.

We are given the pro go to density, which is well, the probability density key to this is the same as the absolute value. Some of this squared right, Andi, since this needs to be continuous the way function, the probability density must to be continuous and then they will be needed to fulfill following X equals zero millimeters. We need to have that. So a one minus zero needs to be the same. Has being one month zero, which means a equals b. So that's a for me. Will draw the way function. Oh, no, the probability Density. Oh, over minus 2 to 2 millimeters. So it's just throwing The function dead was given waas This is being that zero The way function is just a which is the same as being and then it minus one. It's to be, um, a over to read so around here and well, it's zero here, Another one. It's just zero to okay. And at one, it also needs to be zero. And since the positive branch it's the leader functional. Get something like this and then the negative ranch. It's one over X right, So it's one over X and the X has just shifted by one unit. We'll get something like this. So this is our rowboat immensity. Now we want to know the values are A and B, which is really just one value. So we just need to take into account that the way function is normalized is that condition. So it's from minus 1 to 1. The probability they created needs to be one right, and this is just the following integral. So it's from minus 1 to 0. You need to integrate one minus size, so a over one minus X the X them from 0 to 1. You integrate B, which is really a one, My sex, the X and this will give a local rhythm in the 1st 1 right? So it's minus a times logarithms one plus x from minus 1 to 0. Then the second integral will get just a times X minus, X squared over to from 0 to 1. This whole thing will give just home. It's a times logarithms to plus a over to which can also be reading this a times longer rhythm too. Waas on behalf distance to be one right, which gives us that Hey, it's the same. Just be this one over logarithms of to was 1/2. This is around zero point 838 Okay, so finally, we want to know the probability that the particle is found, um, in the negative side with X, the X axis. So it just means from minus 1 to 0 on. Then we just need to meet the great So d is just integral from minus 1 to 0 off the probability density. Which is this branch? A over one, minus eggs. The X and this will simply give this er on. That is a longer them too. They love a rhythm to it is really longer than two over Lovering. Luxury them a two plus 1/2. So this will be around zero once you see three. No, sorry. 38 one or around 58%.

Okay. In this problem, we have ah, particle in a box in one dimension were asked to take a look at the probability density P of X and particularly in particular which X values results in a minimum value of P of X and which X values result in a maximum px and were asked, Just look at the ground state and ground state, which is an equals the n equals one state. So the only thing we need to know restrict their problem is that the wave function for an infinite square well in one D is given by a sign of and pi x of rail, where l is the width of our square. Well and are the energy levels exes location, of course. And a is some normalization constant, which we don't actually to worry about for this problem. So the other thing we need now is that the probability density function is simply the complex square of the wave function. In this case, there's no complex valued anything we need to worry about, so it ends up just being the square. So size squared. So it's actually compute that we would end up with a squared sine squared of the same stuff. So our probability density is given by a sine squared constrained for X between zero and l. Let's take a look at what this would look like. So for the n equals one case, our probability density is, uh, a squared sine of pious Terrell going from zero Dale. So we'd go replied in this this ground state would look something like this. No zero and right in the middle. This is R l over too. So remember, So if X is equal to hell over two, we have signed a pie over too. And if we just look at a normal sine function of not that I really need to refresh your memory on this Dubai Bye if I were to. So when excel over too results in, ah, sign of pi over to which would give us a maximum. Which is why we have a match from p of X when X is equal to l. We have signed up. I signed up I zero of course sign of 00 So we have Max Max probability when X is equal to L. A. For two, we have minimum probability right at the edges. So when excess either zero or al and this is consistent with the figure that they show you with a different energy levels and you notice that as the energy levels go up, you get more and more Ah, peaks in your science, of example, the first energy level would look something like that squared, and that's would be three and so on and so on, going up in end and that's it. Very simple analysis.


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