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(10 points) YOu have observed that the number of hits to your web site follow Poisson distribution at rate of per minute_ Let T be the time (in minutes) between hit...

Question

(10 points) YOu have observed that the number of hits to your web site follow Poisson distribution at rate of per minute_ Let T be the time (in minutes) between hits: (This means that T is exponentially distributed)_(10 points) What is the probability of waiting more than 30 seconds (PIT > 1/2]) (10 points) Given that you have already waited minute for hic_ what is the probability that YOu will need to wait more than an additional 30 seconds? (PIT > 3/21T > 1])

(10 points) YOu have observed that the number of hits to your web site follow Poisson distribution at rate of per minute_ Let T be the time (in minutes) between hits: (This means that T is exponentially distributed)_ (10 points) What is the probability of waiting more than 30 seconds (PIT > 1/2]) (10 points) Given that you have already waited minute for hic_ what is the probability that YOu will need to wait more than an additional 30 seconds? (PIT > 3/21T > 1])



Answers

How Long Do I Have to Wait? The number of hits to a Web site follows a Poisson process and occurs at the rate of 10 hits per minute between 7: 00 P.M. and 9: 00 P.M. How long should you expect to wait before the probability of at least 1 hit to the site is $95 \% ?$ Hint: $P(X \geq 1)=1-P(0).$

In discussion here were recorded that if we have the X follow the exponential, then we have the That's the it would equal to the guy each of the minus k x x greater than zero where we have the e on the expensively equity one of okay. And now in this question here, where are given X will be the time between arrivals and it follows the exponential here with the main equal to 20 seconds. So it means that with the mean here we have the e on the X, they will you go to the 20 also by the formula. It will get you one of a day and this implies. Then there came a sequel to one of a 20. Therefore, we will have that. That's the F X equal to 1/20 each. The Manus X over 20 x greater equity zero. Now want to find a probability than the time between arrival will be greater than 10 seconds. Andi, smaller than the 30 seconds and therefore this probability will be equal to the X will be smaller than 10 on bigger than 10 and smaller than the 30 with the density function. Yeah, we can register into the integral from 10 to the 30 one. Number 20 able minus X number 20 The X and began the untidy riveted on this one. E Coachella minus a djellaba minus X number 20 from 10 to 30. Everyone the 30 inside We got ego June the minus E minus three at the two. And then plus attendance that we have the mind able months and a half How we can register them into the e Bella minus half minus able minus three out with you and this will be the probability we're looking for.

In this question were given a Poisson process with a rate of three per hour. And in the first part of the question, what is the probability of having 10 events occur in the interval from 0 to 5 hours? Given that we've had four events in the time interval from 0 to 2 hours, So the numbers of events that are occurring in these processes are Poisson random variables, and we know that because it's been described as a Poisson process. Remember that for a Poisson random variable, the probability of so many events occurring within a given time interval is equal to the following. So for part A, we're looking for the probability that the number of events occurring within five hours is equal to 10, given that the number of events that occur in the first two hours is for so the time intervals that we're talking about our from 0 to 2 hours and from 0 to 5 hours. And as you can see these air overlapping time intervals the first time interval is a subset of the second one. So from our probability laws, we can say that this is equal to the probability of the number of events occurring between zero and five hours, equaling 10 an intersection, with a number of events occurring between zero and two hours equaling four, divided by the probability that the number of events occurring in the first two hours is equal to four now. Another way to look at this is if we are given that four events have occurred between zero and two hours, then for 10 events to have occurred by five hours, we need an additional six events to occur between the times of two hours and five hours. So we can write this as the probability of the number of events occurring between five and two hours is equal to six. So now we're dealing with two intervals. One is from 0 to 2 and the second is from two 25 And these air non overlapping intervals and for a personal process, the number of events that occur and non overlapping intervals eyes independent, which means that we can write the intersection of the two events in the numerator as the product of two probabilities. So we could say that the probability of the number of events occurring in three hours being equal to six times the probability that the number of events that occurs in two hours is equal to four and then divided by the probability at the number of events that occurs in two hours is equal to four. And obviously this gets a lot nicer now that we can factor these out. So now we can just solve for the probability of having six events in three hours, given a process rate of three per hour. So this is e to the negative Lambda T, which comes out to negative nine. So Lambda is three and t history times lambda t to the exponents six over six factorial. And this comes out to 0.91 and then next for B were asked to consider a more general case where we have to fix times. So we have from zero until s hours and also from zero until T hours and two entered Your number of arrivals where we have m is smaller than M. And so were asked to find the probability that we receive that there are any events by Time t given that we have seen em events by time s, we can write this as the probability that by time t there have been any events given that by time s there have been m events as with part a weaken split this into two non overlapping time intervals. So let's say this is time s. And by time s we have seen m events. So in order to have any events by time T we must between time s and time t see and minus m events this way at time as we have had em events and by time t we will have a total of any events and so we can write this as follows. So the probability of having em events by time s intersected with the probability intersected with having and minus M events in the time interval from S t. T. So, just in case it wasn't 100% obvious what I just did here I replaced this event with the probability of having and minus am events in the interval from s to t. So that is T minus s hours. Number of events is and minus M so these two probabilities are equivalent. If we know if it's given that we had em events in s hours. The probability of having any events in T Ours is the same as the probability of having n minus M events in the next T minus s hours. And now, because the time from zero to us and the time interval from STT are non overlapping time intervals, we can rewrite the probability as follows and as you can see weaken, cross these out and then this probability is simply e to the Lambda T or each of the negative Lambda t. So lambda is three and he is t minus s. That's the duration times lambda t to the exponents and minus m over n minus m factorial.

In this question. Would you call that if we have the X following the exponential with the right you could you Okay. And then you have the density F X. You go to the day Egypt minus K x for the X greater equals zero. Also the explain and the expensive volume. The X is equal to the one of a day in this question. Here we have the X here it will stands for the time between the time between the arrival. So with time between arrival and then it follows the exponential here with the I mean you go to the 20 seconds and now with the miracle to the 20 it means then each of the AM, the X equal to the 20 and it means that it will equal to There were one of a kind here as well. Therefore, we have the K E Co 21 of 20 and it means that we go ahead against the FX because you won over 20 each. The Bauer X over 20 minus here for the X Greater e coaches. Zero. From here, we can be able to find the probability that the temperature in the arrival a time on between. Arrival is bigger than the 60 seconds. So by the it will be given to the probability I'm the ex here will be greater than 60 here and then, Yeah, excreted in the 60 on board. With the density here, we can register into the integral from 60 to infinity off the one of the 20 evil minus X over 20 the X and we found the anti derivative reasoning Coach. Humanness each of the minus X over 20 and then evaluated from the 62 The infinity. Everybody infinity inside. We got exactly equal to the zero on then. Plus 60 and seven have each of the power. 60 hours. Something could be either by ha minus three. We can run this time. Tune the one over about three on. That's gonna be the probability we're looking for.

In Problem 99. We have a traffic flow for a specific website of 12 visits per hour. The red equals 12 physics for our. We want to assume that the duration between visits has exponential distribution for birdie. We want to find the probability that the duration between two successive visits is more than 10 minutes. First, let's define the random variable X will be the duration between visits, the duration between visits. And it's in ours because we have. We have the red 12 visits per hour and we will define its distribution. We assume that it has exponential distribution, which means X follows. The exponential distribution with M equals 12 because the rate is 12 visits per hour for about A. To calculate the probability for the random variable to be more than 10 minutes equals, it's one minus. The community probability. The cumulative probability for the exponential distribution equals one minus e to the power of minus M X, where M is the rate, which means it equals it is about of minus MX. Here we have m equals 12. He is a part of minus 12, multiplied by X exit here should be in hours then it's a 10 divided by six or one divided by sex. These gifts 4.135 for about three. We want to get the time for the top 25 of the relations between visits for the exponential distribution like this. We want to determine the time after which we have the area under the normal under the exponential distribution to be at quite a low 0.25 which represents the top 25% of durations between visits, the maximum it happens at least here at X, then we have the probability for the random variable to be greater than X equals 4.25 And using this the equation we have eight to the bar of minus 12. X equals over into five. Now we can determine X using Lynn by taking blame for both sides. Then we have X equals lin Oh, 0.25 divided by minus 12. This gives and then one quarter divide by minus 12 gives 1.1155 hours and we can change it two minutes by multiplying by 60 gives 6.9 minutes. This is the time after which all 0.25% for the duration have for the top liberations happen, Marcy. We suppose that 20 minutes have passed since the last visit to the website. We want to calculate the probability that the next visit will occur within the next five minutes, which means we know that here in the distribution that 20 minutes half post we are here now we want to calculate the probability that the next visit will happen within five minutes. Then we have here 20 within five minutes, which means we will go 25 minutes. What is the probability to happen within this area? This can be written as a conditional probability. The random variable probability for the random variable to be greater than 25. Sorry, it's just listen 20 five minutes to be within 25 which means it will be less than 25. Given the random variable is greater than 20 we can use the compliment event to get it equals one minus The probability for X to be greater than 25. Given that X is greater than 20 now, we can use the memory list property, which says that it equals one minus the probability for the random variable to be equals 25 minus 20 to be greater than 25 minus 20 which is five. This two is a compliment event for the probability to be to have the random variable is less than five minutes then equals one minus. E is about, uh, minus X. We have 12 and X is five. This gives 4.632 as a probability to have, Then we have about 63% to have the next visit within the next five minutes. Given that 20 minutes, of course the final part need We want to find the probability that less than seven visits a care within. I want our we can translate or transform this into a time and terrible between two visits to have lists, then seven visits in one hour means to have more, then one divided by seven hours between durations. Then the generations between hours must be greater than one divided by seven, which means x near equals, one divided by seven hours. And we can now calculate the probability for the random variable to be more than one divided by seven. Of course, we have here a little mistake. It's five, divided by 65 by 60 because they're unavailable is defined in ours. 16 16 and here I am here is five by by 60 the same. The answer is correct, but here 60 is missing. Let's continue party. We have defined that less than seven minutes in one hour means to have more than one divided by seven hours between durations. Then we want to calculate the probability for the random were able to be greater than 17 hours, which equals E. There's about of minus M, which is 12 multiplied by one, divided by seven. This gives a point 18 as a probability or 18 person. Then this is the final answer for body. This is for birth C. This is for Barbie and finally for Bharti.


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