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Research Randomizer is a free service offered to students and researchers interested in conducting random assignment and random sampling. Although every effort has been made to develop a useful means of generating random numbers, Research Randomizer and its staff do not guarantee the quality or randomness of numbers they generate. Any use to which these numbers are put remains the sole responsibility of the user who generated them. a. Go to the website http://www.randomizer.org/ about.htm and generate one set of 20 random numbers from 1 to 9, where each number can repeat (select "No" for each number to be unique). (Use your computer, calculator, or Table 1 in the back, if you do not have a Web connection.) b. Test your set for randomness above and below the median value of $5 .$ Use $\alpha=0.05$. c. Conduct the test again with the same parameters. d. Test your new set for randomness. Use $\alpha=0.05$ Did you get the same results? e. Solve part d using the standard normal distribution. Did you reach the same conclusion?

So in part a of this question, you have to use the Kai Square test on basically, if to verify whether or not to read the number generators satisfies. The 99% confidence level on can be used in the electronic gaming devices. In casino games, it's for 99% conference level. So the aspect of remnants be tested using the runs test mentioning part one. You'll see runs just, and that is the random number. Generator does not produce statistics, which regards to producing patterns of occurrence and can be used in the Electronic Gearing Devices Casino. So each real position is considered random, each really position. It is a random if each real means not less than 99%. Conference level three Good save interests percent offense level. They run test. So in part C aspect of readiness were tested. Using the correlation announces method mentioning Part three. Correlation announces three that is just verifying their under number. Generator produces numbers that independent chosen. It's regarding any other symbols used to date. So if it inspires in 99% compensate, we know that they can be used and, uh, now party these gaming rules I return using the phrase 9% confidence level instead of German Jew on level of significance, it's not sure significance as a hypothesis test uses. This is appropriate because random event is given a set of possible outcomes, and the outcomes given probably parents is defined in percentage terms that 95% 6% percent.

In this exercise were presented with a Markov chain, which models students who are taking a Web based course. And there are five states, so the first date is going to the home page. The second state is looking at the syllabus. The third state is looking at the introduction. The fourth state is going to Chapter one of the course content, and 1/5 state is just exiting without going to Chapter one. And so the students were tracked until either they reached Chapter one or they exited without reading Chapter one. And we were given in the question the one step transition matrix for this Markov chain, which is as follows and we can index the states by 12 345 and, to begin with in party were asked to identify the absorbing states of the chain so qualitatively, we're told that students are tracked until they reach Chapter one, or until they exit the course without reading without going to Chapter one. So those appear to be the two absorbing states. Get to Chapter one there stopped. They stopped tracking them, so that's kind of like an absorbing state, so they just stay in that state or the exit without going to Chapter one, and we stop tracking them. So that's in resorting state as well. The other way to do this quantitatively is to look at the DAG no of the one step transition matrix and we can see. See that for states four and five, they're self replicating with probability one. So once you read state for you stay in state for And once you reach ST five, you stay in ST five. So states four and state five are the absorbing states now for part B were asked for the were essentially asked for the cumulative distribution function for state one to the absorption state. So if t someone is the number of transitions students take starting from the home page until they reach an absorption state were asked to find the probability of t one. Being less than K for K is equal to one to up to 10. So this is basically our approximation of the cumulative distribution function. And the reason why I call it an approximation is because really, this should go up to infinity if we're doing if we're doing the actual killing cumulative distribution function because it is possible to reach and absorbing state in an infinite number of steps, or at least approaching an infinite number of steps. For generally, the cumulative distribution function is given by so starting in stay state one. So I call that piece of one is equal to the probability of going from state 12 and absorption state in K transitions. So right here we have the one step transition matrix. And so we can look at the probability of going from ST one to an absorption state in one transition. And we know that our absorption states are four and five. And so all we have to do is add up these two numbers and that will give us the probability of going from state one to either state for or state five. And we can see that that is equal to zero. So this is the probability going from state one Teoh, an absorption state and one or less transitions. So we have our answer for K equals one. And now we just have to do the same thing for K equals two through 10. And to do that, we must calculate the two step transition probabilities three step transition probabilities, etcetera. So when you want to calculate petered explaining to P. Diddy explain and three all the way up to Peter Export in 10 Now, Peter the exponents two looks like this. And so making it to an absorption state in less than or equal to two transitions can be seen as this some of these two numbers. So that's going from state one to one of the absorption states, either one. So that equals zero point 46 And so then we can repeat this calculation for the cumulative probabilities of getting from ST one to an absorbing state in three or less transitions, more or less transitions and so on, all the way up to 10 or less transitions. And we can put them in a table that looks like this. So for our first entry, we know what a zero, her second entry. It's 0.46 and we can repeat the calculation and populate this table as follows and these are our results Now. This should be not t i t. One, because we're looking at the cumulative probabilities of making it from state one to an absorbing state in six or less transitions. Seven or less transitions, three or less transitions, etcetera. Now, for part C, we're asked to use our results from Part B to estimate the probability mass function for the number of steps from ST 12 absorption. So, for example, the probability I'm going from State 12 an absorption state in three steps is equal to the a cumulative probability of doing it in three steps minus the cumulative probability of doing it in two steps. And so we have these numbers right here. So if we subtract them, we get zero point 2508 and we can put these in a similar table to Part B like this. Except this time, since it's the probability mass function, it's the probability of doing it in exactly K steps rather than less than or equal to K steps. So we already have the probability of going from state one to an absorption state in three steps calculated just above thats 0.2508 and if we do the same calculation for all of them, these air what the results will come out as and so our results for the estimate for the probability mass function look like this. So we can say that the estimated probability of going from state one to an absorption state in exactly six steps is 0.385 And remember that these air estimates, because probability mess function would continue beyond him. In fact, these numbers are actually exact the way we've calculated them. But this is an estimate for the probability mass function. Because it is abbreviated, it stops at 10. Now, using this information, this estimated probability mass function ah were also asked to determine the ah approximate meantime, two absorption starting from state one. So the meantime to absorption from state one exactly is given by the summation from K equals one to infinity of k times the probability mass of K Now, since we've estimated the probability mass function by a breathing abbreviating it to Onley 10 which captures most of the probability mass. Our estimation is based on the summation from K equals 1 to 10 sk times probability mass for each K, and this is given by one time zero. So that's one time zero plus two times 0.46 plus three times zero point 2508 plus all the way up to you que It was 10 and this comes out to you 3.1457 So we have approximated the meantime to absorption from State one as being 3.1457 and now, for part, de were asked to calculate the true meantime, two absorption starting from state. One recall that the mean times two absorption from all the states is given by this calculation. We have your identity matrix minus que, in verse times a one matrix and recall that Q is found by taking the one step transition matrix for the Markov chain and crossing out the columns and rows that correspond to absorption states. So that, in this case is columns and rows four and five. So now we have a sub matrix of the one step transition matrix, and this is Q and so that was he was a three by three sub matrix. So I is also going The identity matrix is going to be three by three identity matrix, and one is going to be a three by one matrix of ones like this. And if you perform this calculation using software. Hopefully, you should get the following values, and these represent the mean times two absorption beginning in state, one state to or a state three. So for this question, we're asked the meantime to absorption beginning in state one. So our answer is 3.208 And finally, for part E were asked what proportion of students eventual reached Chapter one in what proportion eventually exited the course without visiting Chapter one. So these are our two absorption states. So we're finding trying to find along the eventual probability of reaching either of these states. We already know that the total probability of absorption is equal to one. So we're just looking for the breakdown of whether they get absorbed in the Chapter one state, or whether they get absorbed in the exit state. Now recall that thes probabilities of eventual absorption are given by this calculation. Have an identity matrix minus que, which we've already identified in part D in verse. Times are and recall that ours a matrix of the one step transition probabilities of going from the non absorbing states to the absorbing state which looks like this. So the non absorbing states air 12 and three, and the absorbing states are four and five. So from one U cannot get to either of the absorbing states. In one, Transition two was 0.5 I'm just taking these numbers straight from the one step transition matrix that was provided in the question 3.41 0.67 in 0.9 And so, if you run this calculation using software, you get the following. And so this matrix gives us the eventual absorption probabilities from each of the non absorbing states to each of the absorbing states. For the question of hand, we're interested in the student who begins in state one so we can say that there is a probability of 0.381 that they end up in the We'll call that the Chapter one absorbing state, otherwise known as state for and a 0.619 probability that the end up in ST five, where the exit state

Everyone. So in discussion there, given click on the site uh http open stacks college organic. Uh and select the mixer sort of stand hired the president composition and average atomic masses boxes. And then select the element boron. A right the symbol for boron that are shown. Uh It's naturally occurring insignificant amounts. So is a sort of Oberon. I set off some blood Iranian cleared. Yeah I served dogs of boron includes. Then we're on and alone we're on. So so the average atomic mass. Mm Yeah. Mhm. Mhm. Then cannot be you can say I sawed off the next atomic mass. I am you you can stay here percentage. So I saw Tapie's for you there on time. I am you 10 points 0129 19 x nine then For you were on 11 I am you 11.0 93 percentage 80.1. And yeah man see bob. Yeah yeah yeah. And I sort of to the block box to make a mixture that matches their operations. You may write a sort of from the veins on the, click more and then move the sliders to the appropriate um on soc part. Uh It is foolish. One ratio. Okay forest to one that is shelf he learned we're onto then we're on then depart reveal the percentage composition and average atomic mass boxes. How well does your picture match with their operations if necessary. They're sort of amount to mind your prediction so be part. Yeah. Mhm. Yeah. Mhm actual abundance of I should of one dance of we're on no for your 10 is 19 point 9%. And that off. Yeah and that off For you were on 11 it's People in one average atomic mass average atomic moss. Okay my chest. My chase with periodic devil. Mhm mm. And then he part select the nature's mixture of I. Set up and can vary eight to your predictions. How well does your protection compared with the natural perimeters? Explain if necessary unless you're amounts to make them match nature's amounts as closely as possible. So naturally so tops natural. I saw dobbs. Mhm. I sort of and our predictions and our predictions are identical. Uh huh identical with naturally occurring. Which wait naturally, I'm cutting my picture. Thanks a lot.

Everybody. My name is Colin. Let's go ahead and look at this Snow mobile and environmental clubs problems. So in court they were asked to find the probability that the individual and the first part is a snowmobile owner. So to do that, we're going to look at this table here. We're going to see that the total number of snowmobile owners is to 95 and the total number of respondents was 15 26. So we know that the probability that the respondent was a snowmobile owner is equal to 0.1933 They were going toe some part two of this problem. We're looking for the probability that the person belongs to an environmental organization. Or is this new move you honor? So to do that, what we need to dio is we need to take the probability that the person that the respondent is a member off an environmental club. So 305 Oops, that is 3 50 30 5/15 26 and added to the probability of that person is a snowmobile owner. However, we've gotta be careful here because it would be easy just to say to 95 or 15 26. But we've got to take into account that 16 of those 305 people who belong to an environmental club also own snowmobiles, so we don't want a double count them. So the number that we need here is actually to 79/15 26 and what that does is that gives us that, not the stove ill owners who were not already previously counted. And when you do that calculation, we're going to do 5 84/15 26 or about a probability of 260.383 and the last part of this first problem, as it's asking us for the probability that they have never used a snowmobile, given that they belong to an environmental organization. And so for that, we're going to look at the people who answered Yes, they have never used a snowmobile, which is 212 and the total number here that we're looking for is the total number of people who are a member of an environmental club which is 305 and so that number right there is about 0.6951 and that is the probability that the person has never used a snowmobile, given that they belong to an environmental organization. Now we're asking Part B, whether or not belonging to an environmental organization and owning a stone will be our independent. And the answer here is a resounding no. You would think that maybe because there are so few snow mobile owners that are also environmental club members, that you could maybe go ahead and say they are independent. But in fact they are not, because not all snowmobile. Not all snowmobile owners do not belong to environmental club, and some folks who own snowmobiles do belong to environmental clubs. So there is some overlap. Those two categories are not independent, so now we'll move on the part, see where were asked to find the problem. If we were asked to choose, turn to survey respondents at random, and then we're going to find some probabilities based on assuming a random selection of to survey respondents. So the first part assets the probability that if we choose to respondents at random, what is the problem? They both own a snowmobile. So to solve that, what we're going to do is we're going to look at the probability that we've got to respondents and this first guy his first responded owns a snowmobile. And that probability is Aziz. We calculated in part a Tu 95/15 26. And since this guy is the same, also owns a snowmobile that's 2 95 15 26 and is your calling. We're doing this. We've got to take both of these responses into account. We're going to multiply those two numbers together, and we're going to end up with a final probability that both people who responded in this case own snowmobiles as 20.374 And now part two of Part C asks us the probability that at least one belongs to an environmental organization matter than doing this captive. Probably that one is. And then the added to the probably that to our I'm just going to have to go ahead and do the shortcut here. I'm going to find the probability that no, you long and then what I'm going to do is I'm going to subtract it from one. So I realize now that I wrote that in a little bit of a confusing way. But What we're going to end up doing is if we say that X equals the number of people, the number of respondents who belong. And of course, our options here are 01 or two because we're choosing to response at random. But I'm going to do is take one. It's attract the probability that none of those respondents belong, because that will give me the same you'll remember as the probability that one is a member and the probability that, too, is a member. So to calculate the probability that none of them are, remember what I'm going to do is look at, uh, I'm going to get the probability that the first respondent is not a member and the second respondent is not a member. So 12 21/15 26. And I'm going to multiply those two together, and I'm going to get 20.6402 And so then I'm just going to subtract that number from one. Like I said, I'm going to get to 0.3598 And this right here is the probability that at least one of the respondents


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