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