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Visit local retailer/restaurant etc, count number of customers visiting the retailer/restaurant for = period of 10 minutes_ count three periods_ Use your data, choo...

Question

Visit local retailer/restaurant etc, count number of customers visiting the retailer/restaurant for = period of 10 minutes_ count three periods_ Use your data, choose appropriate distribution model, help us understand the following questions. For your own safety, please wearing - mask while collecting your dataReport your data What is the probability that there are more than customers showing up within the next 10 minutes? What is the probability that there are more than 10 customers showing up

Visit local retailer/restaurant etc, count number of customers visiting the retailer/restaurant for = period of 10 minutes_ count three periods_ Use your data, choose appropriate distribution model, help us understand the following questions. For your own safety, please wearing - mask while collecting your data Report your data What is the probability that there are more than customers showing up within the next 10 minutes? What is the probability that there are more than 10 customers showing up within the next 10 minutes? Validate your findings with the data from the other two 10-minutes slot



Answers

The time between arrivals of customers at an automatic teller machine is an exponential random variable with a mean of 5 minutes. (a) What is the probability that more than three customers arrive in 10 minutes? (b) What is the probability that the time until the fifth customer arrives is less than 15 minutes?

So um were given this probability here And we want to find the probability that exactly eight customers will arrive during the 30 minute period. Um So that's going to be um Kay X. At the average rate so every four minutes. So four X. The N. Where N. is eight. The number of customers E to the negative for X. All divided by and factorial. So we have eight factorial here and we end up getting this value. As a result We see that it's going to be about 0.14 based on this maximum here, so 0.1 for their final answer.

Problem. 56. During a 30 minute period of a checkout counter, 30 minute period, one customer arrived. All right. It's required to find the probability that this customer arrived in the last five minutes, knowing that in the last five minutes, given that the 1st 10 minutes, the first them minutes has no custom, no customer arrived. Let's see how to answer this problem. The first step is to denote the arrival time of the customer as a boy. Well, why is it under variable that follows a uniform distribution over the interval or the 30 minute barrier? A 30 minute period starts at zero and ends at could. And because why follows any from distribution. It has a dentist E function that equals one divided by B minus A. Where is this? B. And this A B minus A. Over the interval from 0 to 30. Mhm. And it's defined as zero else. Where the second step is to translate this probability. In terms of what we want to calculate the probability that a customer arrives in the last five minutes. The last five minutes mean that we want to calculate the probability for boy to be between 25 30. This is the last five minutes of the 30 minute period, knowing that means given the 1st 10 minutes no customer arrived. We know that one customer arrived through the 30 minute period and no one arrived in the 1st 10 minutes. This means this customer arrived in the last 20 minutes because no one or no customer arrived in the 1st 10 minutes. We know for sure that the customer arrived in the last 20 minutes. This means why will be between then and 20. We know that 10 and 30 between then answered. We know that the custom arrived in the last 20 minutes, because no one arrived the first 10 minutes. And this is a conditional probability that equals the probability for the intersection between the those two events. That gives why between 25 30 divided by probability for this event, probability for why between 10 and said and by definition it equals the integration from 25 to 30 for every boy devoid. And if a boy is defined in step one as one divided by 30 divided by the integration from 10 to 30 for one forever void by if avoid destroyed. And if avoid is defined as owned by the by 30 the world. Let's integrate equals Y divided by 30 from 25 to 30. And why divided by 30 from 10 to 30. Let's substitute equals 30 minus 25 by the by 30 for instance equals five divided by 20 equals one quarter or 4.25 And this is a final answer of our problem.

Hey, it's Claressa when you right here. So you know the exponential density function it's f of X is equal to zero for one X is less than zero mu to the negative one e to the negative X over a meal when X is bigger or equal. Who's there? Oh, and were given that meal is equal to 30. So that makes f of X B equal to won over 30 e to the negative x over 30. Okay, we're gonna calculate being the probability the X is less than or equal to 15. This is equal to the integral from 0 to 15 for this function. One over 30 Hey to negative X over 30 t x. This gives us negative me to a negative X over 30 from the road to 15 which is around 0.393 for part B. We're going to do the same thing, but it's gonna be bigger, were equal to 16 or finding the probability for that. So this is equal to the limit. Must be approaches Affinity from 60 to be for one over 30 e to the negative. X over 30 d x. This becomes equal to around point 135 Princey, we're gonna do when p is he of X this equal to 300. So this is just the end to grow from 303 100 and then we just substitute our equation and we get zero our equation right here, one over 30 e to the negative expo. We know that anything like, since there's no interval, it's gonna be quick and zero. Now we're going to find the probability when X is bigger or equal to 1 80 because this equal to integral from 1 80 to infinity won over 30. Eat the negative X over 30 t x. This is equal to limit our be approaches Infinity for e negative e to the negative x over 30 from 1 82 b and we get around point 00 to 5 when we do one minus 10.25 and we get around 0.9975 So it is likely got all 200 customers are gonna be served within the three minutes

In this question. The scenario is that it is lunchtime at a given restaurant and customers are arriving at the drive thru in a random manner, and we were told that they follow a Poisson process with a rate of 0.8 customers per minute, so we can say the rate is 0.8 customers per minute. Now, when something follows a Poisson process, there's two things that we can say about it. The first is that the number of arrivals that occur in non overlapping time intervals are independent, and the second is that the number of arrivals in a given time interval follows a Poisson distribution with mean lambda T. So if we're given a certain time interval, the mean of the Poisson distribution is equal to Lamberti now. For part A were asked, what is the expected number of customers in one hour and what is the corresponding standard deviation? So if we define X as the number of customers that arrive in one hour, we know that X follows a possible distribution with mean equal to Lambda Times T. And so they mean for possible distribution is simply Lambda Times T, which is 0.8 times 60 minutes in one hour because this rate is 0.8 per minute and so that we expect that there will be 48 arrivals on average in one hour now. Another property of the possible distribution is that the variance on the number of arrivals is equal to the expected number of arrivals, which tells us that's the standard deviation on the number of arrivals is equal to the square root of 48. How and this is equal to 6.93 now for Part B. Were given information that the drive thru workers cannot handle more than 10 customers in any five minute span and were asked to find the probability that too many customers arrive for the workers to handle between the time of 12:15 p.m. And 12:20 p.m. So we're talking about a time of five minutes because it's a poison process. It doesn't matter if we're talking about the time interval from 12 05 to 12 10 or 12, 10 to 12 15 or 12 15 to 12 20. All that matters is theory length of time that passes, so if we want to find the probability that the drive thru cannot handle the number of customers that occur in that five minutes. We're looking for the probability that the number of customers who arrive in that five minutes is greater than 10 because we know that they can handle up to 10 customers in five minutes. And this is equal to one, minus the probability that the number of customer arrivals is that most 10. So this is equal to one minus. The summation from X is equal to zero to 10 of E to the minus lander times T, which is equal to minus four. I got that from tee times lambda, which is 0.8 times Lambda Times t to the exponents X over x factorial. You know, just to explain this, remember, the probability of getting X arrivals is equal to eat to the negative and the tea on celebrity to the exponents X over x factorial. So here we're finding the some of the probabilities of getting zero through 10 arrivals in the next five minutes and then subtracting that from one. And so if you calculate this, it comes out to about 0.0 28 So that is the probability that too many customers come between the time of 12:15 p.m. And 12:20 p.m. For the drive through staff to handle and moving on to Part C. We're told that a customer has just arrived and were asked to calculate the probability that another customer will arrive in the next 30 seconds. So one way to think of this question is that for a Poisson process, the inter arrival times are distributed according to an exponential distribution with rate lambda. So if we say that we define a as the inter arrival time, it's the time between subsequent arrivals. It follows an exponential distribution with parameter lambda, so we want the probability that a is less than or equal to 30 seconds. This is the cumulative distribution for the exponential distribution, and we know that our rate is 0.8 per minute. So to keep this consistent, I should probably put this in terms of minutes, so that would be half a minute and remember, for an exponential distribution, a cumulative function is one minus e to the negative Lambda Key. So if we feel these numbers in, we get one minus e to the negative 0.8 times half, and that should come out to zero point 33 with your calculator. So the probability that the next customer arrives in the next 30 seconds is about 0.33 now for Part D, starting at noon. We want to determine the expected arrival time of the 1/100 customer as well as the standard deviation of that time. So recall that the inter arrival times of customers air exponentially distributed, and we want to find the average time for the 1/100 customer to arrive. So if we define the time that it takes for the 1/100 customer to arrive, it is thesis, um, of all of the inter arrival times for the 1st, 2nd, third customer and so on up to the 1/100 customer where all of the teas are distributed according to the exponential distribution with rate parameter lambda. Then why some 100 is distributed as a gamma with parameter and equals 100 and Lambda equals 0.8. Now the expected time for why 7 100? It's simply end times one over Lambda and remember one over Lambda would be the expected enter arrival time between any successive arrivals and since Lambda equals 0.8 would simply be 1.25 minutes. That is the expected time between success of any two successive arrivals. So for the 1/100 arrival, all we're doing is multiplying that by 100. So we end up with 125. So the expected time for the 1/100 arrival is 125 minutes. And the standard deviation for the 1/100 arrival is given by the square root of end times one over Lambda squared and this comes out to 12.5 seconds. So the standard deviation on the expected time for the 1/100 arrival is 12.5 seconds s. Sorry. Actually, that's minutes. So our units are minutes. So the standard deviation has the same units as the expected Valley, which are minutes


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