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2.21 Web access pattern: Consider a web server that receives web access requests from large number of users. We assume that the request arrival mechanism is charac-...

Question

2.21 Web access pattern: Consider a web server that receives web access requests from large number of users. We assume that the request arrival mechanism is charac- terized by the following simple model. We divide the time axis into contiguous segments of 4 seconds, and 4 is chosen sufficiently small that the probability of receiving more than one request is negligibly small We also assume that the arrivals in different seg- ments are statistically independent events_ Let p be the probability th

2.21 Web access pattern: Consider a web server that receives web access requests from large number of users. We assume that the request arrival mechanism is charac- terized by the following simple model. We divide the time axis into contiguous segments of 4 seconds, and 4 is chosen sufficiently small that the probability of receiving more than one request is negligibly small We also assume that the arrivals in different seg- ments are statistically independent events_ Let p be the probability that randomly chosen segment interval observes an arrival of a request_ (a) Assume an observation interval of T = 54_ and let p = 0.2. What is the probability that at least two requests arrive during - this observation interval? (b) For any integer 1,2,3_ find the probability that T =nA will be the observation interval required to see the first arrival: (c) Suppose that you are informed that there are k successes in n Bernoulli trials. Obtain the conditional probability that any particular trial resulted in a success_



Answers

During the first round of enrollment, students begin registering for classes at the top of each hour.
There's a mad rush at the beginning of the hour, and then logins taper off. Let $X(t)=$ the number of logins $t$ minutes into the hour, and suppose $X(t)$ can be modeled by a nonhomogeneous Poisson
process with intensity function $\lambda(t)=500 /(t+1)^{2}$ for $0<t<60$ .
(a) What is the expected number of students that will log into the registration system in the first
5 min of the hour? In the last 5 min of the hour?
(b) What is the probability that no students log in during the last 5 min of an hour?
(c) The registration system will crash if more than 450 students log in during any 5 -min period.
What is the probability that this occurs in the first 5 min of an hour? (You will need to use
software or a Central Limit Theorem approximation to determine this probability.)

Hello. This is problem 88. We know that the random variable X is equal to the number uh Type one call meat. Um uh Certain hotel X is distributed as a person With λ equal to 1 7. Such that X is the 201 and so on. The probability that X is equal to one is equal to London which is 1.7 to the X. Which is one of these kids. E To the negative Orlando. So negative 1.7. And they were gonna divided by X which is one in this case factorial. And we use Excel to figure that out. And this is A for B. We want to find the probably that X is less than equal to two which is going to equal to adding up the probability that X zero plus the probability The Xz grew one plus the probability that X is equal to. Okay. And then reports see you want to find the probability the tax is greater than you could too. Using the compliment rule is the same thing as one minus the probability But X is less than two. And so we look for those probabilities. So in Excel we use person dot this which is probably mass function. And we use the value of X just zero as an example and then 1.7 is Orland. And we write down first because we're not dealing with the cumulative distribution function. So her part a we want to know what is the probability that X is equal to one Which is .311 0.3. And now we need to figure out since we want to know what is the probability that X is less than equity to. We need to add a the probabilities um up to two. So we're going to be adding up All three of them. And if you do this you'll get is your answer 0.757. And foreseen we need to do one minus the probability The next 01 and two. But um we already wow in partly we did some of their rights only this part. So we're going to add up Probably the exact zero and executor one which are those two numbers And we get 0.493. Used to attract them you'll get 0.5 07 No for pardee We're going to interpret the mean which is 1.7 a motel or make 1.7 type one Of course on average. And for e we want the standard deviation standard deviation short and we know that that is the square for a person. And where is the square of London? Which is Squirtle 17 Using Excel who just used as security for the number. We'll get 1.304

57 We have a chart, and they wanted to make up a frequency distribution table or a probability table. So we have a plane 185 onion bagels, 60 bry 55 cinnamon raisin 1 25 and sour dough at 75. So you total this which totals by 100 and then 1 85 divided by 560 divided by 500 55 divided by 501 25 then 75. Each of them will be divided by 500. So that's your part? A. For part B. Uh, the question is three plain Biggles right in a row. So assume they're independent, which means the person coming in next to the next person next to the next person. They don't have any, um, relationship. And there's no, um, what one person chooses does not impact with the next person choose. So the first person the probability of choosing the plane is 0.37 and the next person is 0.37 and the next person is 0.37 Typically, when we have independent events, we multiply them, so this would be 0.37 raised to the third power, which equals 0.5 one and I round that and then part see, is it really independent? And no, it's really not, um, people with similar tastes share regions. So a person in Hawaii, um, may have more of a preference for span. That's on the menu at McDonald's, where a person in the south may have amore preference or fry food. Where is the person in the north, like in Pennsylvania? Uh.

Question 57. We have a name onion, rye, cinnamon, raisins that weirdo 1 85 60 when. 20 No. 55 1 25 75 all of those who love to 500. So now I divide each one of those by 500. And then the probability for playing would be 0.37 When I take 60 divided by 500 I get one, too. When I divide 55 by 500 I get 5000.11 Divide 1 25 by 500. I get 0.25 and 75. Divided by 500 gives me 5000.15 and all of that adds up to one. So Part B says, What's the probability? Three people asking for plane assuming independence. So I would take 0.37 raised to the third power, which gives me 0.51 and part See us. Are they independent? And we even figured out mathematically, which in this case, we can, um so then we think about it logically, and here I would say no. They're probably not independent as different

In this problem, we're told that the number of visits that a person makes to an auto repair shop in a given your follows a personal distribution where the personal print parameter is based on the number of visits that they made in the previous year. So, for example, if they make if they made zero visits in the previous year, then the number of visits in the next year is a possible distribution with this parameter. And so for part, they were asked to construct the one step transition matrix for the number of repair shop visits by a randomly selected customer in the observed year. Now recall that postal distribution for X and parameter mu is given by this equation, and so the probability that except n plus one equals X Given except Ben was I visit. Is it possible with this probability distribution? So we could say that the one step probability of going from zero visits in the previous year two zero visits in the next year is equal to the probability of zero given this rate and so that rate amuse sub zero is this value and so that's equal to you, you to the miners 1.93 nine approximately times 1.939 to explain it. Zero over zero factorial, which comes out to approximately 0.144 And we continue with those count calculations for all single step transitions. Because there are five possible states, we know that the transition matrix is going to be five by five. So you would save a lot of time by calculating these probabilities using software. So that's what I did. And here's the one step transition matrix that resulted. And so this answer is part A. So for Part B were asked that if a customer makes two visits this year, what is the probability that customer makes two visits in the next year and two visits in the year after that? So it helps to index the is just Teoh. Make sure we make less mistakes. So we're looking for the one step probability that we go from two visits to two visits, and then we again go from two visits to two visits. So this represents two visits in the next year, and this represents two visits in the year after that, and they're both one step probabilities one step transitions because the first time we're just going from this year into the next year. So what? We know we went to two visits this year. What is the probability of two visits in the next year and then in the next year, Once we've gone to visits, it's just one more step to the next year. So it's going from two visits next year, 22 visits the year after. So that's one transition. So the one step transition probability from two visits to you to visit is 0.27 So this comes out to you. Zero point 27 squared and that equals 0.73 for Part C were asked if a customer makes no visits last year, what is the probability? They make a total of exactly two visits in the next two years? So we're looking for all of the ways to have a total of two visits over two transitions, assuming the current state is zero visit. So one way is to go ahead and make two visits in the next year, and then we would need to also make zero visits in the next year that is in the year after the next year. So next year we make two visits and the year after we make zero visits for a total of two visits and another way is to do one visit in each of the next years. So that's the problem. One step probability of going from 0 to 1 times the single step probability of going from 1 to 1. So here again we will have a total of two visits and the third way is to have zero visits in the next year, but have two visits in the year after that. And so this comes out to 0.27 times 0.148 plus zero point 279 time, 0.333 the 0.144 time, 0.27 and that comes out to about 0.172


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