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Quastlon Assume that the number of insurance claims, N, filed in E (N) = 1OOOO. Use the normal year is Poisson distributed with P (IN - E(N) 300) approximation to t...

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Quastlon Assume that the number of insurance claims, N, filed in E (N) = 1OOOO. Use the normal year is Poisson distributed with P (IN - E(N) 300) approximation to the Poisson distribution to approximateST426J3_2020_Sc_Pafto SearchDALF1oFae PeiserF12 NsenWERU5FGAKXVBNMAltCtri

Quastlon Assume that the number of insurance claims, N, filed in E (N) = 1OOOO. Use the normal year is Poisson distributed with P (IN - E(N) 300) approximation to the Poisson distribution to approximate ST426J3_2020_Sc_Paf to Search DAL F1o Fae Peiser F12 Nsen W E R U 5 F G A K X V B N M Alt Ctri



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This week the number $X$ of claims coming into an insurance office has a Poisson distribution with mean $100 .$ The probability that any particular claim relates to automobile insurance is $.6,$
independent of any other claim. If $Y$ is the number of automobile claims, then $Y$ is binomial with
$X$ trials, each with "success" probability .6
(a) Determine $E(Y | X=x)$ and $\operatorname{Var}(Y | X=x) .$
(b) Use part (a) to find $E(Y) .$
(c) Use part (a) to find Var(Y).

So we know the main number atlantic coast hurricanes is 6.1 hurricanes per year. And we want to find the likelihood that there aren't any. Which means we're going to take our mean to the X. Power and then we're going to take which is one. And then we're going to take times E. To the negative 6.1 power. And then we're going to divide that by R. X. Factorial. Well this is one and this is one. So in effect we're really just typing that in and that comes out to the 0.0 2 to 4. And if we take for part B in 55 years, how many of those years would we anticipate having zero hurricanes? It's not very likely 00.2 to four times 55. That only comes out with 0.1232 uh Is the expected number. Which means we basically we don't expect this to happen. We don't expect to have zero hurricanes happen. The likelihood is so small, and the number of times that's going to happen in 55 years is that probably won't. And that's something you can look up for the most 55 reason to see how often that's happened.

Okay, so we have this fx of X. Which is a poison distribution. We're land as equal to 100 and were also given the F of Y given X distribution which is a binomial distribution With a probability of .6. And so what we're gonna do with those two is we're going to multiply them together to find the joint distribution probability of X and Y. And then we're going to add all values of X. Of our of our product which is a joint distribution. And that will give us the actual distribution of why. And so what we're gonna do is follow this equation I have written here which is the summation of the joint distribution of all values of X. And we're going to use that to find the distribution of Y. So this is actually equal to if we plug in our terms, we have the summation from X is equal to Y to infinity of E. to the negative. 100, 100 Times 100 to the x divided by X. Factorial. And then if we plug in our conditional probability F given F of Y given X, we have X factorial divided by Y factorial times x minus Y factorial multiplied by .6, Y and one minus 10.6 is 0.4. So this is multiplied by that and this is multiplied by .4 X minus Y. And now what we can do and so we have this all written out, not for the multiplication, there is we can take out all the terms that aren't affected by our summation. So all the terms that don't have an X in them. So that would be U. to the negative 100 and Y factorial And .6 Y. And then what we're left with is the summation of X is equal to Y to infinity of 100 x 0.4. The x minus Y divided by x minus Y. Factorial. And I also have cancelled out these X factorial since we have one on the top and one on the bottom. And so this is what we're left with. And now what I'm gonna do, So I'm actually gonna make this .6, Y. I'm gonna make it 60. Why? Over 100 Y. And then I'm gonna put the 100 wide back into our summation. So what it's gonna look like is Each the negative 100 times 60 y divided by Y. Factorial times the summation of 100 x Divided by 100 Y. And that's our That's the 100 why I took from this 60 divided by 100 Y. And put it back into our summation. No safe from X is equal to Y to infinity. And then I'm going to multiply this by .4, X -Y and divide by x minus Y. Factorial. And now that I have this 100 ext about 100 Y. I can put this into one exponent by minus ng, the Y from the X. And so we have and then after I have that, why that 100 x minus y, I can multiply it by 1000.4 x minus Y. So we have 100. Each of the negative, 160 to the Y multiplied by the summation of 100 X minus y. Times 1000.4 x minus Y is 40 X minus Y divided by x minus way factorial. And now this is where we can use the MacLaurin series for each of the X. So if you're say to let x minus Y equal, say, and so you say n is equal to x minus y. Then we can rewrite rewrite this summation so that it's the summation of N is equal to zero to infinity of 42. The N divided by and factorial. And now if we look at this, if we look at the um MacLaurin series for each of the X. MacLaurin series for each of the X is equal to and is equal to 02 Infinity summation from N is equal to zero to infinity of X to the end over N. Factorial. And now since we have our summation in this form we can instead of using X we can just plug in this 40. So we have the summation from N is equal to zero infinity of 40 to the N divided by N. Factorial. That's actually equal to eat the 40th using this MacLaurin series. And then we just plug this back in and find our distribution. So Once we make the substitution of each of the 40th for our summation we have Either the negative 100 multiplied by 60 Y. Divided by Y. Factorial. And then we have this multiplied by E. to the 40th. So Times E. To the 40th. And now what we can do is just add the exponents of E. So negative 140 is negative 60. So we have F. Of Y. Why is equal to Eat the negative 60th Time 60 to the Y. Divided by Y. Factorial. And now if you look at this, a poison distribution with X. And lambda is equal to E. To the negative lambda landed to the X. Over X. Factorial. We can see that this poison distribution formula actually does. It looks exactly like our F. Y. Of Y probability distribution or probability mass function. So we do indeed have a poison distribution for our Why variable. And now all we have to do is find the expected value in the variance. And for any poison distribution, the expected value, it's just equal to lambda. So in this case are lambda is 60 so this is equal to 60 and our variants Is also equal to λ for poison distributions, so this is also equal to 60.

Problem. Five. We will assume that the mean number of Atlantic hurricanes in the United States is 6.1 per year, which means mu equals 6.1 hurricanes career and we will assume these hurricanes follow the bozo. Distribution was whole distribution. We will need to find the following probables, but at the end the probability that in a year there will be five hurricanes. The probability distribution for the Bozo morality of X equals you the part of X multiplied by e to the power of minus mu. Divided by X factory in. We need to find the probability to have five hurricanes in a year. We will substitute by X equals five. We have new equal. 6.1 is about five multiplied by E to the bottom minus 6.1 divided by five factory These gifts a probability of all 0.1 call you it or in percentages, 15.8% for party in a 50 year period. How many years are expected to have five hurricanes? Because the total Turbat is 55 and we have a probability oh, 15.8%. This means the total number of years number of years to have exactly five hurricanes equals, it's probability multiplied by the total number of years. The probability is all 0.158 multiplied by the total number of years. 55. These gifts it 0.7 years, then about eight or seven years during the 55 year period, we'll have five hurricanes for birth. C. We want to compare this result if we know in this period we had five hurricanes in eight years in eight years. During the period we have eight years with five hurricanes and we can see that from birth to be from B for party. We got 8.7 years with five hurricanes and we can see that the two values are close to each other. And finally, this sub boards the assumption of the bossa distribution given in the first part of the problem, we assumed Bassem distribution and this results supports our assumption of Abbas. So distribution

Welcome to enumerate in the current problem. The variable understudy is monthly number of monthly number oh insurance claimed claimed I'll find it. Okay. No it is given that X follows probability Poipu lambda where lambda is equals to tell. Okay. So the perimeter is still. Now. The first question that asks is find the probability that no claims are filed. No claims are filed on a particular month because this is a monthly number. So no claims are filed on our body. You live month. So try to think if they say no claims are filed for two months, how will we think around that all that stuff. Okay. For three months or a full year? Okay. Now this will be probability of no claims means what is the number of claims then if there are no claims filed executes a zero. That will be probability or into the power. Now let us write the P. M. F. It was the power minus lambda lambda to the power X by X. Factorial. Which is it was the power minus 10 entertained to the power X by X factorial. So here X. Zero. So it is it is the power minus 10 into 10 to the power zero by zero factorial. X. X factorial. So this is one. So it is into the power of my nest in which will be zero point four times 00.0000 45 ft. The second question that is asked if probability of not more than two, not more than two claims find. So what will it that be probability of Ex not more than two. That means less than equal to two. Which is the probability of x equals to zero plus X equals to one. X equals two. Two. It is it was the power -10 into plus. Because if the power minus one is pre Mexico zero. If the power minus 10 into 10 to the power one by one factorial plus, it is the power miners turned into 10 to the power to buy two factorial. And upon simplification we arrive at 0.00 27 694 The final question asked is probability, but at least three claims are finding. What is that? At least three claims Probability X is at least three or more So x greater than equal to three, correct? So it should not be, it should not be So we will have one minus probability of X less than equal to two because less than three is equal to less than equal to two because they are discreet. Therefore, probability of X less than two is already calculated. Which is this Correent? So we have 1 -0.0027694, Which is equal to 0.997 2306. Now, at this point it doesn't end. I will tell you whenever you're doing any probability problems start to interpret the results. If you see this is a very small probability, that means there is no claim is a very rare event, even not more than two, that is also fairly rare event. But if you see more than three, that has a very high chance, that means it's quite a regular like obviousness. Is there? Cut it. And what does that tell us? C lambda is equals to tell. That means for expectation X is always lambda for public distribution. So that means what expected number of monthly, monthly number of claims lies around 10. So at 10 the probability is the highest, that's where if you see at 012, all these probabilities are very less and once you go to know certain, that's where the probability peaks up. So I hope I could explain this to you, let me know if you have any questions.


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