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1 point) The number of customers who use a drive-thru automated teller machine (ATM) varies from one hour to the next with an average rate of 8.72 customers every 1...

Question

1 point) The number of customers who use a drive-thru automated teller machine (ATM) varies from one hour to the next with an average rate of 8.72 customers every 15 minutes_PART A 14 random variable X is to measure the number of customers who use this drive-thru ATM in a 15-minute interval. What is the orobability that between 6 and 10 customers (inclusive) use this ATM in a given 15-minute interval?Answer:(round to at least four decimals if necessary)PART B Nhat is the probability that the dur

1 point) The number of customers who use a drive-thru automated teller machine (ATM) varies from one hour to the next with an average rate of 8.72 customers every 15 minutes_ PART A 14 random variable X is to measure the number of customers who use this drive-thru ATM in a 15-minute interval. What is the orobability that between 6 and 10 customers (inclusive) use this ATM in a given 15-minute interval? Answer: (round to at least four decimals if necessary) PART B Nhat is the probability that the duration of time passing between successive customers using this ATM is between 2 and 5 minutes? Answer: (round to at least four decimals if necessary) PART C 1 customer has just driven away from this ATM and there are no vehicles in line behind that car: If at least 3 minutes pass until the next customer enters the drive-thru ATM line, what is the probability that in total, no more than 6 minutes pass until the next customer enters he drive-thru ATM line? Answer: (round to at least four decimals if necessary) PART D Over the course of a five-day week, 52 non-overlapping 15-minute intervals were randomly chosen and the number of customers using he drive-thru ATM is to be observed in each of these 15-minute intervals. What is the probability that the average number of customers Jsing the drive-thru ATM in these 52 15-minute intervals will be between 8.32 and 9.52 customers? Inewe_ NN to at loact four docimalc inec



Answers

People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs.
$$\begin{array}{|c|c|}\hline x & {P(x)} \\ \hline 0 & {0.03} \\ \hline 1 & {0.50} \\ \hline 2 & {0.24} \\ \hline 3 & {} \\ \hline 4 & {0.70} \\ \hline 5 & {0.70} \\ \hline 5 & {0.74} \\ \hline\end{array}$$
a. Describe the random variable $X$ in words.
b. Find the probability that a customer rents three DVDs.
c. Find the probability that a customer rents at least four DVDs.
d. Find the probability that a customer rents at most two DVDs.
Another shop, Entertainment Headquarters, rents DVDs and video games. The probability distribution for DVD rentals per customer at this shop is given as follows. They also have a five-DVD limit per customer.
$$\begin{array}{|c|c|}\hline x & {P(x)} \\ \hline 0 & {0.35} \\ \hline 1 & {0.25} \\ \hline 2 & {0.20} \\ \hline 3 & {0.10} \\ \hline 4 & {0.05} \\ \hline 5 & {0.05} \\ \hline\end{array}$$
e. At which store is the expected number of DVDs rented per customer higher?
f. If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form.
g. If Video to Go expects 300 customers next week, and Entertainment HQ projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain.
h. Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that?

In questions. 78. We have information about a video rental store DVD rental store called Video to Go and were given a probability model in which we have X to be zero through five there, representing the number of DVD Reynolds per day for a customer and then the probability of each of those things happening. Now the four has 40.70 beside it, and looking at that, that exceeds 100% and that would not be a legitimate probability model. And then we would not be able to answer questions. Um, following this. So I think that's supposed to be a 1000.7 I think that maybe a textbook errors I'm gonna use this, um, as 0.7 for the rest of the problem. Also noticed that because they're probably model should add upto one. There's a gap without three. So we're mento. We're supposed to fund that gap. That gap is 30.12 and now each of those probabilities when you add them together, that adds up to one. Let's answer party A says. Describe the random variable x inwards. So X is gonna be the number of DVD Reynolds from video to go that is per day per customer be found The probability that a customer rents three DVDs? Well, that was the gap in the probability model. We've already figured that out. Um, that is 0.12 See, find the probability that a customer rents at least four DVDs, which that means four or five so or means ad. So four is point of 75.4 Those together may get 0.11 d. We want to find the probability that customer rents at most two DVDs, so at most two means zero or one or two. We can find each of those probabilities and the distribution above 0.3 plus 0.5 plus 0.24 That gives us 0.77 as our total. And then we're given information about a new store entertainment headquarters were given their probability distribution zero through five for DVD Reynolds and all of their probabilities. The probabilities do add upto one or 100%. Now, um, to make things simpler, I'm gonna change the X and the p of X and this one toe why? And people why, just when I calculate, um, there's a difference between video to Go and entertainment headquarters and E. It says, At which store is the expected number of DVD Reynolds Higher? Well, we're gonna find to the expected value of X, which is videos to go from the top and then the video of why, which is our new distribution. So in order to find your expected value, you're gonna take each value of X and multiply bites probability and add those together for video to go. We get 1.82 and we're gonna do that same thing for entertainment headquarters. Each value zero times it's probability, plus one times it's probability and so on, and we get 1.4 for its expected value. So the expected number is higher at the video to go in F. It says a video to go estimates that they will have 300 customers next week. How many DVDs do they expect your rent next week? So we would salute the expected value of 1.82 Well, that's the number of videos that we would x number of DVD rentals would expect per customer. If we expect to see 300 customers would just multiply those two values together and get 546. So video to go should expect 546 DVD. Reynolds Total and G says a video to Go expects 300 customers next week and entertainment headquarters projects that they will have 420 customers. For which store is the expected number of DVD Reynolds for next week. Higher. So we've already calculated, um, the expected number of DVD rentals for video to go in the previous question. Let's look at entertainment headquarters. We haven't expected value of 1.4 with a expected number of customers to be 420. We're gonna multiply these two values. Together, we'll get 588. Eso entertainment headquarters should expect Mawr DVD rentals next week. Um, 588 being greater than 546 and h which of the two video stores experiences Mawr variation in the number of DVD rentals per customer. How do you know that? So we can find the variation. We could find the standard deviation. Either one of these one. Answer the question. So I'm gonna look at the variation of X and the variation of why side note here is that variation is gonna be the summation of every single value minus the mean when I figure out how far away far away it is from the mean and square that value and then we're going to multiply it. By its probability, we're gonna add all of those together. So in X, as you can see, we have zero minus the mean, which is the expected number squared times. It's probability we're gonna then add that toe one minus the mean square times It's probability something with 2345 together. That adds up to be 1.3476 So the variation is 1.34 76 doing the same thing with why which is entertainment headquarters. We're gonna take zero minus. It's mean square that value to get rid of the negative, multiplied by the probability and do that with 1234 and five and do that all the way across. We get variation of to 0.4 so the various you can see is greater for why than it is for X, and we'll say that here, Entertainment headquarters says more variation and DVD Reynolds per week since the variation of why was greater than X 2.4 being greater than 1.3476 so why had a greater variance?

Hello. In this video we're gonna look at an exponential function. So over on the left side you can see p. Of T. Is our exponential function. And over on the right side you can see the graph of that same function. This function is about the probability of a customer entering a store. Alright so managers is trying to look at the likelihood of customers coming in a certain number of minutes after 10 a.m. Soapy. The name of our function will be giving an output of the probability in decimal form and T. Is going to represent the input of this function and it is the minutes after 10 a.m. That a customer enters the store. Right? So those are inputs and outputs here. Uh Let's go take a look at the questions that we want to work with worked with this function to solve. So first we're gonna try to find the probability That a customer will arrive one minute. All right. After 10 a.m. All right. Within one minute of 10 a.m. So store opens at 10. What is the probability that a customer will enter within the first minute of that time period? Yeah. So I've created a table. Right. Right. You could also input one into your function. You could put one in for t use your calculator, sob. I'm gonna let that graphing app do that for me and it's showing that when the time is one minute that the probability will be 0.5- 7 6. Or if we want to write that. Get my pad over here we can change that to percent for him. So the probability of somebody coming in one minute uh within that first minute after 10 a.m. Is going to be 0.5 to 76 or 52.76%. So a little bit over 50% chance that somebody's going to walk in in that first minute. All right. Second question, same kind of concept would be the probability of somebody coming in within the first three minutes after 10 a.m. All right, so you can either input three into T for your function here is your calculus solve or you can use your graphing app here. Right? You can see we can also click on the graph is going to give us these values. So For looking at the likelihood or probability that somebody's gonna come in within the first three minutes. Now we get a decimal 0.8946 or It's about an 89. A little over an 89% chance that somebody's going to walk in in those first three minutes after 10. All right. Mhm. Our last question will tie into this is what if we flipped it around? All right. We wanted to know. Okay, let me put in a y here, right, we're gonna want to know Give you a probability this time and want to know what 20 minutes it took at what time that happens. Right, So, we're gonna say pft here uh when can you be 98% sure. So it's 0.98% sure. Uh that somebody's going to come through your doors right here. So we're going to set pFT Equal to 0.98. I'll change the colors here. We'll use our intersection tool in the graph To go up and find that time. So it looks like uh at about five 0.2 minutes after 10 o'clock. All right, So a little bit after 10 05, the probability that somebody's going to walk into your store is 98%. So pretty high chance within the first five minutes opening after 10 o'clock, somebody's going to walk in those doors.

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

In this exercise were given the joint distribution of the discreet random variables X one and X two for part A were asked for the probability that X one is equal to one and x two is equal to one. Now we can re express this as the probability that X one equals one intersected with X two equals one. So this is the case where both X one and X two are equal to one. So that is 0.15 for part B. We're looking for the probability that X one and X two are equal. So the probability masses where X one and X two are equal Are these entries along the dad? No, you could express this as the probability that X one equals I and X two equals. I summed over all possible values of I. This is only going up to three because X to only goes up to three. So this is equal to points here late plus 0.15 plus 0.10 plus 0.7 to give us a probability of 0.5 For part. C event is defined as being at least two more customers in one line than in the other and were asked to express a in terms of X one and X two and then calculate the probability of a A is simply the union of X one being at least two greater than X two with X to being at least to greater than X one. And so the probability of a Is this some of the individual probabilities that's satisfy either of these conditions. So for the first condition, it would be these probabilities. I would just group these to keep it organized. Plus the probabilities of the second condition, which are these three probabilities and the soul comes out to 0.22 for Part D were asked for the probability that the total number of customers in the two lines is exactly for and at least four. So for the first one, we're looking for the probability that X one plus X two equals four. So this is the sum of the probabilities for any situation where X one plus X two equals four. So that's one x one is equal to one, and x two is equal to three when both of them are equal to two. When X one is three and x two is one and when x one is four and X two is zero and this comes out 2.17 are the probability that X one plus X two as at least four. That's equal to the probability that X one plus X two is equal to four. I'm only breaking it up this way because we've already calculated this probability, plus the probability that X one plus X two is at least five. So now I'm just finding all the probabilities where some of X one and X two is at least five, and this all comes up to 0.46 for a party were asked to find the margin OPM F of X one and then calculate the expected number of customers in line at the Express check out. So all we have to do here is for each possible value of X 10 through four, with some each role. So we some oh, we some for all possible values of X two. So, for example, X one equals zero would be 0.19 and we can put that in tabular format. Yeah, so we have 0.19 Looking at the other roles for the second rail, we have these all sum up 2.3 0.25 0.14 and 0.12 So this is the PMF for X one. The expectation of X one is given by the following summation, and this gives us an expected value of 1.7. Part F is to find the marginal P M. F of x two. So it's similar to Part E. But now we're doing for X two. So instead, now, for each possible value of x two, we some down the column and right the total in the margin below the table and we get the following marginal probabilities. So this is the PMF forex, too. And now for G. We're asked the question by looking at the probabilities of X one being four and of X to being zero, as well as the joint probability of X one being four and X to being zero. What do we conclude about the independence of the two variables? Independent variables would mean that the joint probability of X one and next to is equal to the product of the probabilities of x one and x two For all possible pairs of X one and x two Not looking up at the table, the joint probability for X one, equaling four and x two equaling zero is zero. And from our marginal distributions, the probability of X one being four is 0.12 and the probability of X to being zero is 0.19 And as we can see these air not equal, and since we have therefore at least one pair of X one and x two for which this does not hold, then we can say that the random variables X one and X two are in fact dependent.


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