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S}polncPreuousDeroreSlalyRoirtA Your Teachysuppose sma alrcraft arlve certaln airport according Polsson process with rate (Round your answers three decimal places W...

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S}polncPreuousDeroreSlalyRoirtA Your Teachysuppose sma alrcraft arlve certaln airport according Polsson process with rate (Round your answers three decimal places Whar the probability that exactly smal aircraft arnive during T-nour period?per hour; so that the number cf arrivals during tme perod of [ hoursPoleconwith parameter0124What is the probability that at least small aircratt amive durng (0 407hour Ferod?What the probability that least 10 283aircralt arnve during J-hou period?what is the e

S}polnc Preuous DeroreSlaly Roirt A Your Teachy suppose sma alrcraft arlve certaln airport according Polsson process with rate (Round your answers three decimal places Whar the probability that exactly smal aircraft arnive during T-nour period? per hour; so that the number cf arrivals during tme perod of [ hours Polecon with parameter 0124 What is the probability that at least small aircratt amive durng (0 407 hour Ferod? What the probability that least 10 283 aircralt arnve during J-hou period? what is the expected value and standard devlatlon of the number of small aircralt that Jnrive during 45-min perlod? exected value standard devarion 24 What 0,003 probability that least 27 smab aircraf arnve durlno 2 5-cour Derod? What the probability tnat most small aircraft arnve during 5-hour pericd? 0 363 You may need use the appropriate table the Appendix Tables answer this question; Need Help?



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Changes in airport procedures require considerable planning. Arrival rates of aircraft are important factors that must be taken into account. Suppose small aircraft arrive at a certain airport, according to a Poisson process, at the rate of 6 per hour. Thus, the Poisson parameter for arrivals over a period of hours is $\mu=6 t$ (a) What is the probability that exactly 4 small aircraft arrive during a 1 -hour period? (b) What is the probability that at least 4 arrive during a 1 -hour period? (c) If we define a working day as 12 hours, what is the probability that at least 75 small aircraft arrive during a working day?

Here we are told that the arrivals of small aircraft at a at a certain airport Is exponentially distributed with the mean of one hour. So let's let X be the time between arrivals for part A were asked for the probability that more than three aircraft arrived within an hour. So that's probability that number of aircraft arriving Is greater than three. Now, if X is an exponentially distributed random variable and it is the time between arrivals, then the number of arrivals in a given time is a person random variable. So this is equal to 1 -1 The probability that end is at most three. And for the person random variable this would be the summation or K equals 0- three. Ive eat the negative lambda X and linda X to exponent K over K factorial and the rate lambda is equal to one over the main, Which is one arrival per hour. This is going to be one minus the summation. This comes out to eat the -1 because lambda is one And the time is one hour. We're interested in the number of arrivals in one hour. So it will be one to the exponent K over K factorial. And this comes out 2.01 90. Now for part B were asked that if If we consider 30 separate one hour intervals, what is the probability that None of these intervals contained more than three arrivals. Now for a given interval of one hour we know that the probability that it contains More than three rivals is .0190 And we're looking at 30 of these one hour intervals. So let's say why is the number of 30 Minute intervals or sort of one hour intervals? Now, each one hour interval is independent from the others. So each interval, each one hour interval can be seen as a Bernoulli trial. So why is the binomial random variable? It is the number of successes Out of a sample of 30 where the probability of success Is constant at .0190. So we want the probability that none of these intervals contain more than three arrivals. This will be equal to 1 -2 To the exponent 30. And this comes out to the probability of approximately point 5626 and last for part C. We are asked to find the length of an interval in ours. Such that the probability that no arrivals occur During the interval is .1. This can be seen as the probability that we go more than x hours before having an arrival Is equal to 0.10. It means that one the CDF at X Is equity .1 means that eat the negative lambda X. Where lambda is one, so E to the minus X 1.10 means that minus X Equals a natural algorithm .10, solving for X gives two point 30 to six hours. So there is a 10% probability that we go more than approximately 2.3 hours before our first arrival.

Here we are told that the arrivals of small aircraft at a at a certain airport Is exponentially distributed with the mean of one hour. So let's let X be the time between arrivals for part A were asked for the probability that more than three aircraft arrived within an hour. So that's probability that number of aircraft arriving Is greater than three. Now, if X is an exponentially distributed random variable and it is the time between arrivals, then the number of arrivals in a given time is a person random variable. So this is equal to 1 -1 The probability that end is at most three. And for the person random variable this would be the summation or K equals 0- three. Ive eat the negative lambda X and linda X to exponent K over K factorial and the rate lambda is equal to one over the main, Which is one arrival per hour. This is going to be one minus the summation. This comes out to eat the -1 because lambda is one And the time is one hour. We're interested in the number of arrivals in one hour. So it will be one to the exponent K over K factorial. And this comes out 2.01 90. Now for part B were asked that if If we consider 30 separate one hour intervals, what is the probability that None of these intervals contained more than three arrivals. Now for a given interval of one hour we know that the probability that it contains More than three rivals is .0190 And we're looking at 30 of these one hour intervals. So let's say why is the number of 30 Minute intervals or sort of one hour intervals? Now, each one hour interval is independent from the others. So each interval, each one hour interval can be seen as a Bernoulli trial. So why is the binomial random variable? It is the number of successes Out of a sample of 30 where the probability of success Is constant at .0190. So we want the probability that none of these intervals contain more than three arrivals. This will be equal to 1 -2 To the exponent 30. And this comes out to the probability of approximately point 5626 and last for part C. We are asked to find the length of an interval in ours. Such that the probability that no arrivals occur During the interval is .1. This can be seen as the probability that we go more than x hours before having an arrival Is equal to 0.10. It means that one the CDF at X Is equity .1 means that eat the negative lambda X. Where lambda is one, so E to the minus X 1.10 means that minus X Equals a natural algorithm .10, solving for X gives two point 30 to six hours. So there is a 10% probability that we go more than approximately 2.3 hours before our first arrival.

All right, So this question is asking us, uh, suppose a small aircraft arrives in an airport according to some of the process that the rate of eight per hour so that the number of arrivals during the time period I'm sorry, operates under the parameter mu equals eight t A. What is the probability that exactly six small aircraft arriving during one hour period at least six or at least 10 B. What are the expected him? Ah, number of small aircraft that arrived there. United minute, period. And see, what is the probability that at least 20 small aircraft arrived during a 2.5 hour period? All right, get into that here. All right. So question their problem. A for fasting probability six. Do that probability. Six to meal. This is eight, which we get from, um u equals eight p just eight times one here. This play in the equation and you're gonna get e negative eight, which is just that when I go over four times 8 to 6. Over six factorial. I didn't match. This comes to you. I want to very easy enough. So the next part heart a, um asks us. It's probably the least six arriving in what appears rates and find this were pretty much the same thing we got. Probability X is greater than or equal to six. And to do this all we do is one minus the probability if X is less than six. Right? Because this is essentially saying the same thing just hard to where it's actually impossible to get the embalm the probability by itself. So you just use all the numbers between zero and six to get your answer in that equation. Looks like this. You've got one minus the summation from 0 to 5. The Probability X again toe eight that we calculated from a first nest equals one mine. It's summation 0 to 5. Sorry. My finger here eats the negative eight times, Bates to the X over X factory. Then you plug in your ex and you get 0.8, you know? So I'll circle numbers here. The interest here? Yeah. The last part of part A asks us for a probability that X is at least 10. Read the same thing here. Um, use one minus estimation. Thanks. 0 to 9. Um uh, right. So it turns out to be one minus summation from zero. It's and I to the negative times Eat to the X over x factorial. And that is you. 0.283 All right, clear enough. Yeah. So what part b privies pretty quick. Here. Um, B is asking us, um, more the expected amount of the number of small aircraft that arrived during a 90 minute period. All right, so what do you do here? You issues that because a t did they give you? And this goes to eight. Time is one hour? Yes, with 90 minutes to hours. Uh, this gives us right. And then all this is is data. Why? Well, the answer they were looking for pulls a square root meal. Mathematical square equals three. Be easy now for C. See is asking us to find a probability that at least 20 small aircraft arrived during a 2.5 hour period. Okay, so on. And then at least, um, and then it's also asking us to find the probability, that probability that at most, 10 for Audrey, five hour period, right to the first part there. First, we have to find mu, which equals 2.5 times eight hours using for using this equation right here. Do you Down here. That you. So the probability that Z is greater than 20 inches from variable here, um, equals gambling to use that same principle instead of finding the probably that it's greater than we're gonna take one minus the probability z smaller than to find this dysplasia right here. One of minus estimation. Oh, zero. Uh, didn't teams giving 19? I don't want 20. All right. Oh, ability. So, um, Dems before this is just for the year equation, you get one minus summation 0, 19 times 22 z over the factorial and doing that Matthew summary right there for you. Sorry. 0.5. All right. And then the last part of this entire problem baskets Probability Z is at least 10. So resting or equal to 10. Distanced. Asking for less than or equal to. We don't have to go through the intermediate step of one minus a difference. Formacion. Um, is that simple as summation 0 to 10. Um oh. Probability of Z 20 equals. It's equal estimation. Summation zero negative. 10. The pictorial. This is an equal 01 All right. That was clear for free. You know,

Okay, So with this problem, we have a basan process with the rate of eight per hour. So that parameter is m equals 18. And the first part of this question is asking us to fund probability of six ships arriving in one hour on the fund probability of at least six, and then the probability of at least 10. So do that. We got our first part here, So we got the probability other ships arriving equals six. All right? And to do that Gap Probability six, eight Thetis, God, just plug it into the equation. Oh, yeah, E negative eighth. Sorry. I made a little clear you could be, uh, times eight to the sixth 86 Sorry. Um uh, there we go 8 to 6 over six. Sorry about that. My screen exactly were for a second, and then this answer comes out to be 0.1 to. So that's your answer to the first part of part of the next part. Asks us to find the probability of at least six ships arriving. Sort of do that. Find a probability that at least six we'll arrive. We've got one minus probability of less. Uh, West and six arriving. Alright. Right. And that makes sense because we can easily find a probability of less than six. But we will have no way of calculating the way of more than six taking this initial step. Right? Okay, to do this, they got one Linus summation, zero five a probability eight. Right for them. This equals, um, one minus summation E to the negative. Eighth times, eight to the X over X factorial. And this math comes out to be zero point. One of the hardest part about this is first making this initial step. That's a key step right there. And then you also just put numbers in the equation, right? This eight, in case you're wondering, comes from the, um, initial parameters that we were given. That is what? We're here in blue over here. Yeah. Look, we got em equals 80 against his t equals one. We got em equals eight. And that's where the eight that you're seeing that's coming from in all of these equations. All right, so that's your answer. The second part and then the last part of part pay possess, Mr Find a probability that a T least 10 ships arrive during this time, right? And we're gonna take the same step. It was enough for one should do one, minus the probability that West and 10 ships arrived for the same reason. Because we cannot possibly calculate the number between 10 and Infinity. So how? This is a very similar problem. One minus the summation. 0 to 9. Right here over nine. We don't include 10 because we're going less than, um, of, uh, okay. And ness equals one minus. Summation from 0 to 9. Off E to the negative beef times A to the X over X factor. Right. So if you look between this problem in the one in part B, the only thing that changes the parameters home of a summation, right, You're 509 Other than that, the same problem right here. Sorry about that. All right. And then this part. Sorry. This part comes out to 0.28 story. Okay. Okay. So that's party for you. Uh, let's go down. Well, tapping a part B heartbeats. Pretty self explanatory. Um, Harvey is actually find the expected value and standard deviation of ships arriving in a 90 minute time period. Right? Do this you still need. You need to use that same equation that we had written up there and blue. All right, again. Um, that equation is m equals eight c. And since you got a 90 minute window, Got 90 minutes times 60 minutes. Uh, sorry. 90 minutes. Time. One hour, over 60 minutes. Those units cancel, and here you get 1.5 hours case you're gonna put that in for tea, you get em equals Pete. 1.5 that you get. That's an important number for this part. Okay? Because the rest of this problem but early, the only thing to do is this equation right here. Why equals square root M, which equals a split of 12 which he calls 3.46 That is your answer for part B. So we don't think you have to do in that problem is apply the time that they give you into the promise they gave you. Take a square with that number and you got your answer. All right, so last part of this here, See, um, parte si is asking us to find the probability of 20 ships arriving in 2.5 hours. Tom and then probably of at most 10 arriving in 2.5 hours. So again, we have to start with that equation that they gave us. People ate tea, and they give us 2.5 hours. So this equals eight times two 0.5 in this equals 20. That's not what we're going to be using. And so we've got the probability 20 ships arriving in 2.5 hours and to do this year, we just do. Probability X is greater than 20. Matt equals one minus the probability that packs is less than 20 right? Samaj abuse before. Um, so this equation goes to one minus X summation 0 to 19. Um, probability of Pakistan 20. All right, we're going to use that same equation we did before in the problems above. And that looks like this. God. One minus the summation from 0 to 19 um, e to the negative 20th in times onesie to the axis over x factorial. That equals 0.5 three. Right. In case anybody's are confused. I wasn't very clear with this. The X that you're doing, it's It's these numbers here, right? X equals zero in 19. 0, I see, you're just adding, if you plug in zero for X here, you get that number. Plus the number you get from adding one in there, too. And they're always 90. You add all those up, and that is where you get here. 0.5 through Sorry for any confusion that I could have cleared up, uh, more transparent from start in. The last thing that we have to do for this problem, we'll find the probability that Oh, at most 10 ships come into the port in 2.5 hours. All right, so to do that, he used Probability X is less than or equal to 10. And this is easy. We're already looking for less than or equal to. We're less thing. Um, so this looks like the summation. X equals 02 And since it's since you got this or equal to sign, we've got you have these all those numbers summation of the probability of Okay, Z 2. 20. All right. In that look, something like this, the equation is going to be summation. X equals 0 to 20 of e to the negative 20th times once e to the x over X factorial I'm sorry. I'm sorry. That's not even 20. That's gonna be eating to the or the summation from zero town of even a 20 times. I went to the X over X factorial. That equals zero point 011 All right, that's your answer. I'm sorry for any confusion. That last part is well again. This you're doing it from zero to 10 not 25.


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