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2.29 A museums open at 10.00 AM After opening; Visitors per hour arrive through the Poisson process_ This museum has guide tour program; which is open every hour In...

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2.29 A museums open at 10.00 AM After opening; Visitors per hour arrive through the Poisson process_ This museum has guide tour program; which is open every hour In other words_the museum gathers people who want this program and leads it every hour (from 11.00): and the total viewing time is 30 minutes Those who do not want this program can watch it themselves The probability that someone arriving at this museum wants this program is p independent of all others_ In the case of 'viewing by o

2.29 A museums open at 10.00 AM After opening; Visitors per hour arrive through the Poisson process_ This museum has guide tour program; which is open every hour In other words_the museum gathers people who want this program and leads it every hour (from 11.00): and the total viewing time is 30 minutes Those who do not want this program can watch it themselves The probability that someone arriving at this museum wants this program is p independent of all others_ In the case of 'viewing by oneself; the viewing time of one person follows exponential distribution with an average of Whether person 1S watching as guide or by himself; everyone who finishes the visit- Will visit the souvenir shop with probability q- + Calculate the distribution of the number of people participating in the guided tour starting at 12-00.4 Calculate the distribution of the total number of people who finished viewing with guide until 13.00.+ Find the probability that any one of the guests who arrived at the museum by 13.00 and watched alone will finish viewing by 13.00.+ Find the expected value and variance of the total number of people who stopped at the souvenir shop until 13.00.



Answers

Another important discrete probability distribution is the Poisson distribution, named in honor of the French mathematician and physicist Simeon Poisson (1781-1840). This probability distribution is often used to model the frequency with which a specified event occurs during a particular period of time. The Poisson probability formula is $$P(X=x)=e^{-\lambda} \frac{\lambda^{x}}{x !},$$ where $X$ is the number of times the event occurs and $\lambda$ is a parameter equal to the mean of $X .$ The number $e$ is the base of natural logarithms and is approximately equal to 2.7183. To illustrate, consider the following problem: Desert Samaritan Hospital, located in Mesa, Arizona, keeps records of emergency room traffic. Those records reveal that the number of patients who arrive between 6: 00 P.M. and 7: 00 P.M. has a Poisson distribution with parameter $\lambda=6.9 .$ Determine the probability that, on a given day, the number of patients who arrive at the emergency room between 6: 00 PM. and 7: 00 PM. will be a. exactly 4. b. at most 2. c. between 4 and $10,$ inclusive.

Yes. Mhm. It is a problem. The setting is the time series but we can set them as define as the minutes. So it's a 30B 10 record that we actually calculate them in minutes. So So actually B -A. is 90 minutes. Yes. And following this setup we can calculate them function. The committee distribution function would be Yeah. So it will be zero F X. Is smaller than a 30. And it's going to be 1/90 X -830. Right? When Ex ranged from 8 30 to 10. Okay. It's also going to be one if access Great. Very good. End 9 30 10 30. So this is a problem. A problem. B We have the expected value would be a plus B over two and it's gonna be 9 10 50 The variance of it should be 90 square or towel and it's going to be 675 Problems. See so we have three time period because daily in here and we got 10 minutes and over the probability should be one of three. For problem there we have three but here we have 10 minutes and it's over 90 then probably to over three. So that would finish the

For this exercise, we are told that a visitor's arrival at the gate for a movie is uniformly distributed on the interval from 8:30 a.m. To 10 a.m. But for the question, we're interested in the number of minutes After 8:30 AM. That the customer arrives or the visitor arrives. So if they arrive at 8 30 that zero minutes after 8 30 if they arrive at 10, that's 90 minutes after 8 30. So this is how we would define our uniform random variable. It's the time in minutes of the arrival after 8:30 a.m. At the gate and it's uniformly distributed on 0 to 90. Now, the CDF for any continuous uniform random variable is of this form. So for our specific variable A is corresponds to zero and B corresponds to 90. So we can say that the CDF is zero. If X is less than zero it's X divided by 90 Forex is at least zero, But less than 90 and it's one Forex at least 90 minutes. So that is the CDF for the random variable for part B. Were testifying the mean and the variance or a uniform or continuous uniform random variable. This is a plus B over to. So that's 45 minutes. And the variance of a uniform random variable is given by b minus a squared over 12 is 90 squared over 12 And that is 675 for part C. We're looking for the probability That a visitor weights less than 10 minutes for a show. So we are told that there are shows at nine AM 9:30 a.m. And 10 a.m. Once the show starts, you cannot get into that show. So to get into that show, you have to arrive by the start time. We have, she'll start at 30 minutes after eight AM 8:30 a.m. 60 minutes after 8:30 a.m. In 90 minutes. So we're looking for the probability that X is creator than or equal to 20 minutes, But no more than 30 or X is at least 50 minutes, But no greater than 60 or X is at least 80 minutes, But no greater than 90. These are mutually exclusive domains. So this is equal to the probability, it's equal to the sum of these probabilities. Now, we can also note that the probability mass function or the probability density function for a continuous uniform random variable is one over b minus a forex between A and B. So in our case It's 1/90 For X between zero and 90. So our distribution function has a height of 1/90. So these probabilities are the relevant areas of the distribution To an area of 1/90 times A length of 10 And the 2nd and 3rd terms both have lengths of 10 as well. So that's times three And we end up with a probability of 1/3 and part D is a similar type of problem. We want the probability that a visitor weights more than 20 minutes for a show. So that is the probability that X. Is less than pen. So basically if they arrive at 8:40 or sooner they have to wait at least 20 minutes for the nine o'clock show or X is greater than 40 Rather that's greater than 30. So they missed the first Joe But less than 40 Or greater than 60 where they missed the second show But less than 70. So they still have to wait more than 20 minutes. So once again these are mutually exclusive outcomes. And so there this probability is the sum of the individual probabilities. So these are the individual probabilities I'm talking about. So we can see it's 1/90. And again these are Width of 10 minutes, All three of them. So we multiply by three And we once again get a probability of 1/3.

In this question, we have a non homogeneous constable process with an intensity of function as defined here. So any time you see an arrival rate, that's a function of T. Then you know it is a non homogeneous Poisson process and this intensity function is defined from 0 to 60 minutes. So for part, A were asked what is the expected number of arrivals in the first five minutes of the hour? So the first thing to do is to find your expected expected number of arrivals. So that is expected number of arrivals in five minutes. And because it's non homogeneous, we have to find the integral of this function from 0 to 5. So we evaluate this over 0 to 5 minutes and it comes out to negative 500 over six plus 500 or about 416.67 So that is the expected number of arrivals in the first five minutes. And we're also asked for the expected number of rivals in the last five minutes. So this is and this comes out thio zero point 73 So the expected number of arrivals in the first five minutes is 416.67 and in the last five minutes is 0.73 And next for part B were asked the probability that there are zero arrivals in the last five minutes of the hour. So we already have the expected number of arrivals. So we could say the probability at Exit 60 minus except 55 equals zero is equal to e to the negative zero point 73 time 0.73 the exponents zero over zero factorial and this comes out to 0.48 So we could say that the probability that there are no rivals in the last five minutes of the hour is 0.48 And finally, for part C, we're asked with the probability of more than 450 arrivals in the last five minutes of the hour, or rather, that's the first five minutes of the hour. So what is the probability of more than 450 arrivals in the first five minutes? So we already have the expected number of arrivals for the first five minutes, which is 416.67 So we're looking for the probability number arrivals is greater than 450 which is equal to the probability one, minus the probability that the number of arrivals is that most 450 which can be expressed this way and then using software. Do you make this calculation You get 0.47 so the probability of having in excess of 450 arrivals in the first five minutes is 0.47 And by the way, this is a situation where you can also use a normal approximation to the personal around a variable. So one thing we can do is if you have a put some random variable with a very high rate, let's say it's 416. We can view that we can decompose that into the summation of a whole bunch of other independent and identical puts on random variables, each having a rate of one. So if you some 416 independent plus in random variables that have a rate one you will get a personal random variable. Another way to look at it is you will have a personal random variable with rate 416. And so, viewing this as the summation of a whole bunch of random variables, the Central Limit theorem tells us that the summation will be normally distributed approximately. So then we could say that x of T minus. It's mean, which is Lambda, divided by its standard deviation, which is the square root of Lambda, is approximately distributed according to a normal standard distribution. And so we could say that probability that number of arrivals is greater than 450 equals one minus probability that the number of arrivals is less than or equal to 450 is equal to one minus probability. That said, would be less than 450.5. That's using the continuity correction minus the mean, which is 416.67 divided by the standard deviation, which is the square root of 416.67 And if you do this calculation, it comes out to about 0.0 for 87 which is pretty close to the to the actual exact value that we got calculating earlier in part C.

So the question here gives us a particular scenario of the coffee and doughnut shop. So we have here that at a certain coffee shop, all the customers buy a cup of coffee here and some also by a doughnut. So I'm the shop owner believes that the number of cups, um t cells a day is normally distributed with a mean of 320 cups and a standard deviation of 20 here. And he also believes that the number of doughnuts hey, cells is independent of the coffee sales and is normally distributed with 150 with a mean of 1 50 donuts and the standard deviation of 12 year. So for far a year, it states that the shop is open every day but Sunday, so he wants us to find the probability that he will sell over 2000 cups in a particular week. So essentially, um, if we know that our mean is gonna be 3 20 per day If we multiplied by the spy six, What we're gonna get is 1920 here. So, essentially, um, we have to find the probability that he's gonna sell greater than 50%. Um every single day. So again, the probability of let's say, 60% of the cops multiplied by a six year is going to give us, um, essentially, if we multiply it altogether. So 0.6 times 0.6 and onwards, we're gonna get a probability of 0.516 or 5.16% for part B. It states that if he makes a profit of 50 cents on each cup and 40 on each donut, can he recently expecto have a days profit of over 300. So essentially, what we can do is we can essentially multiplied by 3 20 times, 0.5. So that's gonna be 1 $60. And, um, 1 50 times 0.4, which if I put in my calculator, that's gonna be $60 So this is gonna be to 20 which is less than 300. So no Onda. Lastly for see here he asked us to find the probability that on any given day he'll sell a donut to more than half of his coffee customers. So we want to find the probability of selling one half times the mean of cups here So essentially, um, as we know that on a given day, he's gonna sell over 50 Um, 50% of his well, 50% of his, um, dough nuts he's gonna sell is gonna be 1 50 so it's gonna be normally distributed. So we want to find the probability in a normal distribution curve that if this is 1 50 we want to find something that is greater than 1 60 onwards, and we find that this is gonna be 26.11% or 0.2611


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