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The management supermarkel anls adopt new Dramauana policy giving gift t0 every customer who spends more than certain amount per visit this supermarket . The expect...

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The management supermarkel anls adopt new Dramauana policy giving gift t0 every customer who spends more than certain amount per visit this supermarket . The expectation of the management that after this promotionae policy advertised_ the expenditures for all customers this supermarket will be normally distributed with mean of SIO0 and standard deviation of 533.71. If the management decides give free gifts all those customers who spend more than S160 at this supermarket during visit, what percen

The management supermarkel anls adopt new Dramauana policy giving gift t0 every customer who spends more than certain amount per visit this supermarket . The expectation of the management that after this promotionae policy advertised_ the expenditures for all customers this supermarket will be normally distributed with mean of SIO0 and standard deviation of 533.71. If the management decides give free gifts all those customers who spend more than S160 at this supermarket during visit, what percentage of the customers are expected get free gifts? Round your answer two decimal piaces % of the customers are expected get free gifts, the tolerance +/-596



Answers

Coffee and doughnuts At a certain coffee shop, all the customers buy a cup of coffee; some also buy a doughnut. The shop owner believes that the number of cups he sells each day is normally distributed with a mean of 320 cups and a standard deviation of 20 cups. He also believes that the number of doughnuts he sells each day is independent of the coffee sales and is normally distributed with a mean of 150 doughnuts and a standard deviation of 12.
a) The shop is open every day but Sunday. Assuming day-to-day sales are independent, what’s the probability he’ll sell over 2000 cups of coffee in a week?
b) If he makes a profit of 50 cents on each cup of coffee and 40 cents on each doughnut, can he reasonably expect to have a day’s profit of over $\$300?$ Explain.
c) What’s the probability that on any given day he’ll sell a doughnut to more than half of his coffee
customers?

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

Hey store has overstocked a certain item, so therefore they're going to run a special promotion. And customer one would pay $100 And then customer to would pay half of that, which would pay 50. And then customer three would pay 25, half of 50, and then customer four would pay half of that, which would be 1250 and so forth. So this is an example of exponential decay, and our cost Starts at a 100 And then decays by 1/2 for every value. So we're going to say to the X. Power. Uh So in this particular problem, they're using the letter. Why? So we can change it to why and why is going to represent the number of customers who purchased the item? Now, they also say that um the number of customers follows a Poisson distribution, and the mean is too, And when we're working with Poisson distributions, we are talking in terms of lambda. So lambda would be too. Now we need to find the expected cost. So we're looking for the expected cost And according to theorem 3.1, you know that the expected value of why is equivalent to the sum of why times the probability of why which? Because it is a Poisson distribution, we are saying that it is the sum of why As why starts at zero and approaches infinity of why times. And we're going to use the formula for the probability of why using a Poisson distribution, which is lambda to the Y power times E to the negative lambda power all over why factorial. So, we've got to use that to solve this problem. So, we are trying to find our expected cost at the end of the day. So are expected cost is the same thing as the expected. And since cost is equivalent to this function, we could say E of 100 times one half to the y power. And when we now use the expected Cost formula or the theorem 3, 1, we could say that it is the sum As why goes from 0 to infinity Of that function. So we're gonna say 100 times why? Or sometimes 1/2 to the wide power. And then we're multiplying it by the formula lambda. To the white tower. Uh huh. To the negative limb. The power over why? Factorial. Now keep in mind that lambda is going to be two. So that means we could now substitute to in where we see lambda. So we'll have the summation As why goes from 0 to infinity of 100 Times 1/2 to the y power times two to the Y power times E. To the negative to power over why? Factorial. And if you think in terms of algebra, since both of these are to the same Y power, we can Merge them into one. So we'll have 100 times To over 2 to the Y Power times E to the negative to power over why Factorial. Now keep in mind what to over two is so we can clean that up a little bit and we could say Why going from 0 to infinity of 100 Times 1 to the Y Power times E to the negative to power all over. Why Factorial. And we could Basically factor out the 100 and the E to the negative to power. And we could say 100 E to the negative to power, multiplied by the sum of The function as why changes from 0 to Infinity of one to the Y power over why victoria. And I want you to think in terms of the fact that anything or one to any power no matter what it is, is still going to be one. So we could say 100 E. To the negative to power, multiplied by the summation as Y. Goes from zero to infinity of one over why. Factorial. And this here is our definition of E. That is how E. Was calculated. So therefore we now have 100 E. To the negative too, times E. Which is the same thing as saying 100 E. To the negative one power. Which is the same as saying 100 divided by E. And if you pick up your calculator, When you do 100 divided by E, you're going to get approximately 36.79. So therefore the expected cost when the average customer Who purchases this item is to would be $36.79.

Mhm problem. 13 Question A. We need to calculate the probability that X is bigger than 1 18, which is corresponding to access bigger than 18. 181 80.5. So, uh, we need to calculate a value corresponding to this X value. So the city is equal to X, which is 1 18.5 minus N b. So N is equal to 317 times p, which is a property of success. So 3170.6 over square root of 317 times 4.6 minus times one minus 4.6, which is equal to negative 1.11 So you think table five the probability of that is bigger than negative. 1.11 is equal to one minus, provided that there is smaller than negative 111 And this can be reminded within two with five, which is one minus 4.1335 which is all 0.8665 Okay, for question, be question B. Um, we need to calculate the provocative acts is smaller than 200. So, uh, it's corresponding to value of 9199.5. So, uh, asset is equal to, uh, corresponding to this value, which is equal to X minus n b over square root of envy times one minus B, which is equal to one point or seven. Uh, so the probability that that is smaller than one point oh seven is equal to 0.85 77. So using Table five and this can be your mind using Table five for questions, see for questions saying, Um, it's given that probability that by given sample is oh, 0.37 probability of sample is, oh, 0.6. So we generally say that the probability that symbol and by is equal to probability of same build times probability of by given sample, which is, oh 0.6 times 0.37 which is equal to a 0.22 question d. We need to calculate the probability that 16, uh, is bigger than 16. The X lost between 60 and 80 which, corresponding to probably that X, lies between 15 59.5. Uh, so we need to find the school for 59.5. We see the school is equal to negative 1.47 and 80.5. That is equal to 1.37. So the probability that negative 1.47 that's 1.37 which is equal to probability that that is smaller than 1.37 1 minus probability. That said, it's negative 1.47. So this is equal to from Table five. You can get up on the two probabilities, which is equal to 0.9147 minus four point oh seven oh eight, which is equal to 4.8439

So we have a 10 minute shopping spree and we're assuming that the main amount spent is $20 with the standard deviation of $7. And again the shape is unknown. But now we take a sample of size 100 customers. And the central limit theorem tells us when this number gets larger and larger and larger that the sampling distribution of X. Bar becomes closer and closer to an approximate normal distribution. It doesn't matter that the original distribution is not normal because that sample size is relatively large. The sampling distribution of X. Bar will be approximately normal and the mean of X. Bar will be the same as the individual. Mean, it will still be $20 but the standard deviation of X. Bar will be approximately equal to the standard deviation of the individual, divided by the square root of N, which is 7/10 or 0.7. So what's the likelihood that you sample 100? And get the mean to be between $18 and $22. And that's an average of 100. Uh Different shoppers. Well it's converted to a Z. 18 minus the mean of 20 divided by that standard deviation. And then the 20 to minus 20 which we know that this one is there are going to just the opposite. So once we no one will know the other and let's see what that comes. We have to well negative two divided by 20.7 which comes out to be negative 2.8 roughly. Six. And then this one is positive 2.86 And I'm going to use my normal CDF for this and use the lower limit as negative 2.86 We can use a table in the book as well, look up the one the larger value and then the smaller value and then subtract them. But my C. D. F. Can do that for me directly. And it's very very high probability it's very very likely that the mean of 100 is going to be between 18 and 22. Now on the other hand, if our distribution of excess is normal or approximately so than the likelihood that an individual shopper will spend 18 to $22 that Z value is going to be very different because it's standard deviation is so much larger for an individual it's 10 times larger. And so we take this and divide it by the seventh and the 20 to minus 20 divided by the seven. And so again are our Z. Value is going to now be negative to seventh, negative two divided by seven. It's only negative point to call it nine. This will be positive 0.29 And again I'm going to use that normal CDF. We can also once again look it up in the table, and this probability is not as likely to happen because that standard deviation is so much bigger. That's only a 20 to 23% chance that an individual person will spend that much. So when the marketers are looking at not just individual shoppers looking at averages of shoppers, they could be quite certain that the average shopper is going to spend between 18 and 22 but individuals will vary far more than that.


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