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Problem 2 (30 points ) Do Q1.29 in Casella and Berger reproduced below: My telephone ring" 12 times each week the calls being randomly distributed Aong the day...

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Problem 2 (30 points ) Do Q1.29 in Casella and Berger reproduced below: My telephone ring" 12 times each week the calls being randomly distributed Aong the days_For this situation; enerate the ordered samples that make Up the unordered samples {4,4,12,12} and {2,9,9.12}Suppose that we had collection of six numbers, {1,2, 8,14,20}. What is the probability of drawing, with replacement the unordered sample {2,7,7,8,14,14}?Verily that An uordered sample of size k, from m different nubers repeat

Problem 2 (30 points ) Do Q1.29 in Casella and Berger reproduced below: My telephone ring" 12 times each week the calls being randomly distributed Aong the days_ For this situation; enerate the ordered samples that make Up the unordered samples {4,4,12,12} and {2,9,9.12} Suppose that we had collection of six numbers, {1,2, 8,14,20}. What is the probability of drawing, with replacement the unordered sample {2,7,7,8,14,14}? Verily that An uordered sample of size k, from m different nubers repeated K1,k2 Kmn titules_ has FSEK77 ordered components, where 61 += +km Establish that the mber of multinomial coefficients; and hence the nuber of distinet bootstrap samples is Iu other words: +m = ') K(hi+bt_tkm=b} Ax



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Using the phone numbers listed in your telephone directory as your population, randomly obtain 20 samples of size $3 .$ From each phone number identified as a source, take the fourth, fifth, and sixth digits. (For example, for $245-8268,$ you would take the $8,$ the $2,$ and the 6 as your sample of size $3 .$ ) a. Calculate the mean of the 20 samples. b. Draw a histogram showing the 20 sample means. (Use classes $-0.5 \text { to } 0.5,0.5 \text { to } 1.5,1.5 \text { to } 2.5, \text { and so on. })$ c. Describe the distribution of $\bar{x}$ 's that you see in part $b$ (shape of distribution, center, and amount of dispersion). d. Draw 20 more samples and add the 20 new $\bar{x}$ 's to the histogram in part b. Describe the distribution that seems to be developing.

So for this question, we've got a bunch of different possible dice rolls for some of two days. Uh, and we have data from an experiment where we tossed the dice 500 times and record the frequency of each song. Our first step that we want to do is write out the approximate empirical probability that each some occurs. That's all that we have to do. To do that is divide the frequency. It occurs by the total number of times we threw the dice, which was 500. And so for our first frequency, we will simply by 11 by 500 and get 0.0 to 2. And if we go down, glide and divide all these by 500 and write down the resulting decimal approximations were actually all of these air exact answers. No Robbins involved. We can fill out our list of empirical probabilities. So let's take a second. Just read all of these out and you're gonna see this stuff in the middle here of 6789 are far more likely to occur because they are bored, uh, combinations of two dice that make those numbers. And so that means that these sums are more likely to occur than the ones on the end. Sit two and 12 where we know there's on Lee. One specific way to make each of those Kansas just up down here go to. So from here, we want to think about theoretical probabilities that each of these will occur. And so we know with two dice roll, uh, here, like draw a little 123 to find six 12 3456 little grid Here, um, we can think about some of the two days roles as this grid. It's got 36 spots now here on the number of spots that we could get. A sum of two is just right here, the intersection of one and one. So one way to get it over 36 possible ways of rolling two days. If we count up all of the ways that would make each of our possible sums, we can fill in theoretical probabilities. So let's start from the outside and work in 12 is also gonna have 1/36 because the only way to get that is a 66 three, we could get a one in a two or a two and a one. So we're gonna have to over 36 we could reduce this fraction. But I'm gonna leave it for now. Settle maker. Our work a little bit easier here. 11. Same way we could have 65 or a five and a six, for we could have a one in a 33 and a one or a 212 sir. After 3/36. 10. Same way we're gonna have six and a four a 4 to 6 or 55 five weeks before the one I wanted before Attorney three or 32 So for over 36 9 should be the same way we could do a six and 3365 and 4 45 It's a four of 56 um six. We could do a five and a one I one in a five, a four to a two and four or three and a three, and so we'll get 5/36 8 Likewise, should be the same way to in 662 three and 55 and three and four and four. So that's five total ways. And now we only have seven left. We can count up all the ways we can do seven. But we also know that if we add up all of these numbers 55 is 10 plus foreign for 18. Those three and three is 24 close to, and she was 28 +11 is 30. So we've got 30 of our 36 combinations and we only have the sevens left, which means there must only be six ways to get seven. And, uh, we actually have this kind of interesting thing, the theoretical probabilities of our two dice rolls. They go. 12345654321 all over 36. That's it. These are our theoretical probabilities like that. We could some foot fractions, and we could also write them as decimals if we wanted. But I would go ahead and just leave everything over 36 because it's nice and clean, and then we can see the pattern, right? The 123456 uh, that are last up here is to get the expected frequency. That's and so all that we have to do to figure out the expected frequency is multiplayer theoretical. Ah, probability times the number of samples in our experiment. So in this case, we have 500 so we're gonna take 500 times 1/36. That's just 500 divided by 36. Um and so, actually, how I would write all of these, I would simply write them as 500 over 36. 1000 over 36 1500 over 36 and all the way down the list. And you're gonna notes that what we get for these five here for 23456 will be the word. Damn. The answers we get for 89 10 11 12 and seven will be unique on its own. There it will be 3000 over 36. And then after you've got all these, if you wanted to reduce them and compare it to the actual frequency that we got, I should say that they should all be pretty close. And so all of this together is kind of the final answer for this question.

This question covers the relative frequency, hissed a grams. And how to make those and determining if it's relative if it's normally distributed or not. So relative frequency. Yeah. Okay. Hist a grams. And just know that relative frequency is the same as percentage percent. So the first to construct the instagram we must have axes. These axes might not be big enough but we can extend it always. Um Then we have to determine the X. And Y. So it's wrong. Relative frequencies. Yeah. So for the X. Axis we have for wrongs zero 12 3 4. All the way to eight. Of course we could extend this. 67 eight. Yeah. And for the relative frequencies we have, the highest number Is 30.5%. So 35% is good enough. And we can increment by fives. So and afterwards we just graph the relative frequencies. So for zero it's um 17.2 35. Doing rough estimate. But there's technologies for these things. 36 15.1. It's about there. We just want a quick type of sketch to determine if it's normally distributed. Um 7.3. Mhm. And some very oh numbers. So as you can tell if we make the approximate shape and compare it with a moral distribution, we know that this is right skewed so it is not normally distributed. Yeah.

So in number 86 A. It looks to define the random variable, so the random variable X would just be the length of a long distance phone call through the length of a long distance phone call. In Minutes B is ex continuous or this screen on exponential graphs. They are always continuous. See, just wants us to write the distribution. So we're told it's an exponential and we're told the mean is eight. So the parameter would be 1/8 one over the meat, so usually it's written as a decibel could be written as fraction, though that would be 0.1 to 5 for your parameter d asked for the mean the mean is eight. We were given that information in the problem f the standard deviation, the standard deviation and the mean are always equal in exponential sze. So that makes the standard deviation ain't as well f draw the graph labeled the axes, So this would be about 0.1 to 5. So you have minutes down here. Probabilities here This graph would look something like that. G. Find the probability that a phone call let last less than nine minutes, so less than nine minutes. So we're looking at all this being shaded. So since it's to the left, you would want to use one minus e to the negative 0.1 to 5 times nine. You can do that right in a T I calculator, and that gives you 0.67 53 h says last more than nine minutes. Well, the quick way would just be one minus g, and you could figure that way. But if you're looking for a formula since it shaded to the right, we would just use E to the negative 0.1 to 5 times nine, and that would be 0.3247 I asked. Find the probability that a phone call last between seven and nine minutes. So between seven and nine so e to the negative 0.1 to 5 times seven. This will find seven or Maur, but I only want 7 to 9, so I'll subtract out e to the negative 0.1 to 5 times. Not, and this will subtract out everything overnight. And that is 0.9 22 And for Jay. If 25 phone calls are made one after another on average, what would you expect the total to bay? Well, the mean was eight for each one. If we're doing 25 calls in a row, it's just eight times 25 and that equals 200 minutes.

So this question they have told us that they're not in total 30 components right now. All of this started components. They have said that eight out off off suppliers one then, uh then out off supply. It'll then we have our 12 which is from Supplier three. And now from the 30 gamble in the older select six components such that they've said that oh five will be from, say, suppliers one and two and, uh, one will be from Supplier Tree. And, uh, we're here to simplify this Father, We're going to consider our variables there, Which because, except by had already given to us in the questions were taking up component. Is that where it denotes the Come on, Roland from the supplier tree. So vision this this this available ex missus for why and for supplier three days there. So, uh, since we know that x y and there are independent components here, so we can use the product rule and apply the formula now simplify the form letter who applies for support? I have a number of values say 32 because they've told that they want three competence affects and two off. Why so we can write this as equal do and off acceptable to three because we cannot ignore the compliments from suppliers. Create into this will be an off why is equal to do. And we also considered an off that is equal to one. So to start finding the probabilities, I would therefore say probability off three toe will now be actually equal to you know that in total there are eight supply one components from it were selecting three. Correct then for the supplier we have total 10 are off, which for years letting too. And for supply three we have 12 out of which were selecting one and divided by the total is 30 on from thought Davia selecting six. So to simplify this part, then the probability of getting killed a 0.509 moving on to the but be This is what we've sorrowful about E. Now the next part is where you're going to use the hyper dramatic distribution. So I would say that using hi Bush German trick distribution so we can see that the process we have used in party the formula can build dinners. It's using the party that we just sold. We can ride on the formal off, be off X right that can be written as, oh, so there are eight components of supply run out of it. We might select X values then from by we might summon from 10. We will select by values from 12 levels, Elect said. Values divided by from 30. We were selecting six values. Now this said, is actually going to be out off six. So we can say that if ah, out of the six components. Okay, if I say excess for supplier one, then rise for supplier toe and it's just taking this thought is there. I can also say this is for Supplier. Three events can be six, which is the daughter to six minus X minus wife or supplier Clea. So, because of which I can change this formula and I can read I Desai's eight X, then begins it. Then why and who I can date 12 which is sure to be changed to six minus x minus. Y you don't want to introduce it orally level and is upon from top TV a selecting six components and we know that addition of excessive I cannot exceed six year. So therefore, the condition Here it is at zero less than equal to X less Why is there's an equal do six. So this over here What if Britain is the probability? As for this hypothermic of distribution So if I want to write the final pmf then the final pmf over here she got like 30 c six. Then that would change too dealing 5 11 7 93,007 fights. So the final answer But I can write down here has one up 159 cream 775 and then we have eight. Next mixtures then why then we have 12 six minus x minus y. So this is between the value 0 to 6 sand it is zero otherwise so this is the PM If that we've just found


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