Let X be a discrete random variable such thatE[X]=3.7 ,E[x^3]=65.6 , and Var[X]=.9 FindE[(x-1)^3] ....

A certain radioactive nuclide has a half-life of $200.0 \mathrm{s}$ A sample containing just this one radioactive nuclide has an initial activity of $80,000.0 \mathrm{s}^{-1} .$ (a) What is the activity 600.0 s later? (b) How many nuclei were there initially? (c) What is the probability per second that any one of the nuclei decays?

In this question. Greek conduct If the X follows the exponential random variable with the right ankle to the K and it has a density function equal to the K minus K x for the expert echo zero. Also, the expected value of the X equal to one of a k and the variance on the X because you want a K square in this question, were given that that's the function f X echoed. You know, 1.5 x Bowerman is 1.5 x and then here we can identify doesn't give me the k So we have k equal to 1.5. Therefore, we should be able to find expected value of the X. My family again it could you won over 1.5. Then we got equal to the to over three and the variance of the X because you won over 1.5 square it go to the three are equal to the to over three square. Do we get to go to the far over nine? That will be the answer for the expected value and the variance

Question 83 on The goal of this question is to show this relationship with mean lifetime of the nucleus is getting by one over Lambda one of the decay constant. So this is our angle is so we'll start with defining what I what it means by meantime, mean lifetime of a particle. So here I have, um the definition of the some of the lifetimes of each particle in something from one to total of any number of particles and that divided by the total sample and not my initial state of particles. So I can rewrite my numerator as an integral now the summation over a range, which is because the tune into goal just call it teeth at the time d n in training from zero to infinity. And since it isn't decay process, um, there's a loss with every iteration. So I'm gonna This whole integral will be negative because it against decreasing rate. So that's what the thing is. I'm negative sign comes from in front. Um, yeah, I'm anybody to observe in pretty. I'm basically extending the hints that it gave in the question. We're integrate from zero T and then from t two t plus DT So continue that logic. You can take integral from zero to infinity itself. Some of my bounds come from So now I have I'm in this interval One thing I can note I have time So maybe the useful to display my integral just DT. And since I know the relationship for activity to be D and over DT is according to native land at times but in the brook particles I can rearrange. But I have here right hand side to a place my d n with a term for DT, I can substitute that into my equation. I have a native science of my native sign. It initially goes away again Interim zero to infinity. My teeth remains. I've slammed it and DT Additionally, I haven't end over and not. And if you look at our initial are not initial but our DK equation. Um, I can rearrange this so I end over and not like I have in my function there to be equivalently negative evening of Lambda T. So you replace that ratio here that again is your identity. Land is constant terms. I can pull that out. So I t explained e native landed team. Thank you. Eso have this expression out that would integrate from zero to infinity. I'm I think I could do this, but I heard this is a well known solution, so I could just livable. I just access a look up table to determine what this function would be. So my landed return means that front. So again this history Look up. Look up table for a solution to this integral from zero to infinity. Well known TV should be easy to access But my solution would be as follows My lambda comes down from the initial I ate the negative Lambda T native landed t minus one overland a square. I get interior zero What, if any. Before I apply my bounds, I'm just gonna remain just a little bit too. So it's a little easier to see what I'm integrating. So have native land Two squared t times needed there landed TV over Landis squared, minus lambda. He didn't native land et overland a squared. I get carrying my bounds through zero to infinity. Little simplification here He's landed square to cancel out this squared cancels with this term and left with negative t didn't your Lambda p Yes. One over landed in the name of land, dirty again from zero to infinity. It's all a plug in my values here, which I'll leave it as such that I can't explain my reasoning. So, Vinnie, time to eat the native land Times infinity over one to infinity. You know, minus the starting with the 00 times exponents minus one over Landau E to the zero. So we have kind of a funky expression here. Wave land, infinity times, even negative entity. Um, even negative. Infinity is a more dominating term than this infinity on front, which you confined by plotting as well. So this term would go to zero city and they a affinity dominates. One of infinity goes to zero as well. Both terms. Both these terms actually go to infinity. Those are both zero panic and represent my final term. E to the zero is, of course, just one. I'm negative. Uh, zero minus one of Belinda, which results in the higher expected answer to the solution one over Lambda, which again is our mean lifetime of our sample. So there you

Yeah, so the expectation or that's minus mu squared because the expectation wow X squared miners to its new class meal square and these vehicles, the expectation of X squared miners too, mu times the expectation acts. Remember that mu is a constant and plus meal square and the expectation of X equals meal. Hence this equals the expectation of X square miners too, mu squared class meals squared, which he calls the expectation of X squared miners meal square. And for problem 37 the variance, because the expectation of acts squared miners meal squared Vitry calls 32/7 miners two squared, and this is 32/7 miners, 28/7, which he goes for brazil. And this gives the same results as that in a proper manner, starting not