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Homework: 4.3 Assignment Score: 0 of 9 of 11 (1 complete)HW Score: 9.09%,pts4.3.19Question HelpFind the indicated probabilities using the geometric distribution_ th...

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Homework: 4.3 Assignment Score: 0 of 9 of 11 (1 complete)HW Score: 9.09%,pts4.3.19Question HelpFind the indicated probabilities using the geometric distribution_ the Poisson distribution; or the binomial distribution: Then determine if the events are unusual: If convenient, use the appropriate probability table or technology to find the probabilitiesA major hurricane is hurricane with wind speeds of 111 miles per hour or greater: During the last century; the mean number of major hurricanes to st

Homework: 4.3 Assignment Score: 0 of 9 of 11 (1 complete) HW Score: 9.09%, pts 4.3.19 Question Help Find the indicated probabilities using the geometric distribution_ the Poisson distribution; or the binomial distribution: Then determine if the events are unusual: If convenient, use the appropriate probability table or technology to find the probabilities A major hurricane is hurricane with wind speeds of 111 miles per hour or greater: During the last century; the mean number of major hurricanes to strike certain country's mainland per year was about 0.52. Find the probability that in given year (a exactly one major hurricane will strike the mainland; (b) at most one major hurricane will strike the mainland, and (c) more than one major hurricane will strike the mainland P(exactly one major hurricane will strike the mainland) (Round to three decimal places as needed:) Enter your answer in the answer box and then click Check Answer: 3 Panai remaining Clear AlI Check Answer Save



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Find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial distribution. Then determine whether the events are unusual. If convenient, use a table or technology to find the probabilities.
A major hurricane is a hurricane with wind speeds of 111 miles per hour or greater. During the 20 th century, the mean number of major hurricanes to strike the U.S. mainland per year was about 0.6. Find the probability that the number of major hurricanes striking the U.S. mainland in any given year is (a) exactly one, (b) at most one, and (c) more than one. (Source: National Hurricane Center)

So we know that the mean was the 535 bombs in 576 regions. So the number of bob's per region was 0.9288 And uh then I would store that as X. And we want to find the probability that in a given region that there would be exactly two bombings. And we know that we take our mean, which I'm just going to put as mu and then we're going to take that to the X power times E to the negative main. And then we'll divide that by two factorial. And again, I have this store so I can just directly plug that in. And when I did that, I got 20.1704 And then part B as among the 576 regions, how many of those would you expect to have? Two bombings? Well, and if we take that 5 76 and multiply it by that probability, we find out the number we expect, it's about 98.1. So we would expect that about 98 of those regions of the 576 would have exactly two bombings. And then on part CIA tells us that the actual number that had to was 93. So now these numbers are not exactly the same, but they're fairly close, so it seems to be a pretty good predictor. The on poison model seems to be a pretty good predictive.

95. More than 96% of the very largest colleges and universities have some online offerings. Suppose you randomly pick 13 Such institutions were interested in the number that have offered that offered distance learning courses inwards to find the random variable X Because it tells us we're interested in the number that offered the distance. Learning courses were going to say that X is the number of colleges or universities that offer the distance learning courses be list the values that X may take home X These colleges all the way to the full sample size of 13. We're gonna say that X b 012 all the way to 13 and see give the distribution of X because we have either an online offering or not, we can consider a probability of a success or failure. So we have a binary option. We have a set number of trials, so we're going to say that X is a binary distribution or binomial distribution with 13 as our end 130.9 sixes or probability d own average. How many schools would expect offer such courses because we're looking for expect we're gonna use the formula and top speed, which is the number that we ask, which is 13 0.6. We should get 12.48 e found the probability that at most 10 offer such courses we're looking for. Here is the probability that X is less than or equal to 10. Because we're looking for multiple values. We can use our binomial CDF function or TIAC calculator. We have 13 in 0.96 and the bottom of CDF will take whatever value we plug in for this third spot and then add together all the probabilities of each value to the left of it or less than it. Well, because we're looking for 10 or less, we're gonna play again. 10 for this and it will capture the 987 all the way to zero. We can get our answer key 0.13 Yeah. Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify your answers numerically an answer in a complete sentence. Basically, what we're looking for here is is it more likely that we'll have 12? Or is it more likely that we'll have 13 offer because we're looking for single values. We could plug these into our binomial pdf functions. Pdf allows us to look for a single value instead of adding multiple together. Of the total 13 0.96 we're looking for, I have a body of 12. This ends up being 0.32 for the 13. We have binomial pdf again, where we have 13 0.96 in all 13. We're looking for a probability 0.59 so we can see that it's more likely to get 13 than it is to get 12. Weaken. Rather answer in a complete sentence, so I'll just write. It is more likely that 13 offer such courses.

We want to run a hypothesis test based on data from World War Two in south London. So south London was divided into regions and those regions were a quarter square kilometer, and they analyzed how Maney bombs hit the regions. Some of the region's had no bombs hit. Others had 123 or four plus. And they found 229 of those quarter square kilometer regions had no bombs. HIP 211 had one 93 head to 35 had three and eight had four or more. They then ran a Poisson distribution, and they calculated the expected number of hits and found 227.5 of the regions were expected to have no bombs hit 211.4. We're expected to have 1, 97.9 for to 35 for three and 8.7 for four. Now we're going to test a claim and the claim is going to be what becomes our null hypothesis, and the claim is that the actual frequencies of hits to each region fits the calculated puts on distribution. Now, when you run a hypothesis test, you need an alternative hypothesis to fall back on. So our alternative is going to be the actual frequencies of hits to each region. Do not fit the calculated puts on distribution. And in order to run this test, we're going to have to run the goodness of fit test because we're asking, How well did the actual numbers or did the actual numbers fit the expected numbers? And in order to run the goodness of fit test, you will need to generate or calculate a Chi Square test statistic. And to calculate that Chi Square test statistic, you are going to sum up observed, minus expected values squared, divided by expected. So let's go back up to our chart, and these are your observed values. And here are your expected values. So we're going to add a column onto our chart, and we're going to call it oh, minus e quantity squared, divided by E. So we'll take the observed value minus its corresponding expected value. The answer we get will square before dividing it by the expected value. Now the fastest approach to this would be to put the data into your graphing calculator. So I'm going to just clear out a list for a second here, and then I'm gonna go stat edit. And as you can see, I have list one with all the observed values and list, too Already has the expected values. So I'm gonna sit on top of list three and I'm going to tell the calculator. Please take all the observed values from list one. Subtract their corresponding expected values from list too square that deviation and then divided by all the expected values. Enlist to and you will get these decimals. Now, for the sake of recording them, I'm going to record them out to three non zero decimals. So the first one would be 10.989 0.757 0.245 0.664 and 0.563 Now, in order to calculate that Chi Square test statistic, I must add all of the's values together again. The fastest approach would be to come back to that calculator and I'm going to tell the calculator to add up everything in list three. So, in order to do that, I'm going to quit out of there. I'm gonna hit Second stat. I'm going to scoot over to math and I'm going to hit number five to sum up list three. And in doing so, I get my chi square test statistic to be approximately 0.976 So I'm gonna come back to our screen here, and our Chi Square test statistic was point 9761534896 Now we're ready to find our P value and R P value. We're really asking what's the probability that Chi Square is greater than that test statistic? So our test statistic was 0.9761534 896 And to get a better, better handle on that, I recommend that we draw a picture. So this is what the chi square distribution would look like. It is a skewed right distribution, and the shape of it is dependent on the degrees of freedom, and the degrees of freedom are found by doing K minus one. And K represents the number of categories we have split our data into. And if we go back to our chart, you can see we've split our data into five different categories, So our K value will be five, and our degrees of freedom will be four. Now, not only does our degrees of freedom kind of dictate the shape of our curve, it also tells us what the mean of our distribution is. So the mean of our chi square distribution will also before and you find that mean always slightly to the right of the peak of that curve. So we know the four is right here. Now, we came up with a test statistic of 0.976 ish, so that's all the way back here. So our P value is our likelihood of being greater than that. So we're talking about the area off the curve into the right tail. Now, the fastest approach is to use our chi square cumulative density function on your calculator, and the calculator requires you to provide the lower boundary of the shaded area, the upper boundary of the shaded area and the degrees of freedom. So, in our problem, our lower boundary is that test statistic and the upper boundary. Keep in mind that right tail extends infinitely far. So we're going to use a very high number. We're going to say 10 to the 99th power, and then our degrees of freedom was four. So let me show you where you confined this cumulative density function. So on my calculator you're going to find it under second. There's and it's going to be number seven, number eight. So we're going to type in the test statistic, as are low boundary 10 to the 99 as our upper boundary and our degrees of freedom, which was four. And we're going to get a P value off 0.91339 now. Another component of a goodness of fit test is the chi square critical value and that is found by utilizing the chart in the back of your textbook. So you're going to go to your chi square distribution and down the left side. It asks for how maney degrees of freedom. And in our case, it's four and across the top there, asking you for your level of significance, and our level of significance is defined by Alfa. And for this problem, we want to run the hypothesis test at a 0.5 level of significance. So we're going to look across from four and down underneath 40.5 and you should locate 9.488 So we have all the components we need to make a final decision on our hypotheses. Now, when it comes time to make the decision, we could do it one of two ways. We can either use the P value that we located up above, or we could use the critical value. And again, that's an or you do not have to do both. I will show you both, and you can decide which way you like better. So for the P value, you're going to make a comparison between your level of significance and your P value. And if your Alfa your level of significance is greater than the P value, then your decision should be to reject the null hypothesis. So let's do our comparison. The Alfa on this problem is 0.5 and our P value was 0.91339 and our Alfa is definitely not greater than RPI value. So therefore, our decision is to fail to reject the null hypothesis. Now. The other approach could have been used a light utilizing the critical value, and in doing so, I like to draw a picture to represent what's going on So we would place are critical value on our chi square distribution. And in doing so, you have now separated that curve into two regions. The region in the tail is going to be your reject null hypothesis region and the region that is to the left of that Chi square is going to be the fail to reject the null hypothesis region. And you're going to then place your critical value on the distribution and our critical value. Oh, sorry. We've placed our critical value. We're gonna place our test statistic now on our curve and our test statistic was 00.976 So that would fall all the way back here in the fail to reject region. So that's why if we used the critical value to decide our decision would be the same. So either approach, we made the same decision fail to reject the null hypothesis. So let's go back and look at that null hypothesis. So we are not going to reject this. We don't have enough evidence to say that that is not true. So therefore, let's write our conclusion the conclusion to our hypothesis test is there is insufficient evidence to reject the claim that actual frequencies of hits to each region fits the calculated Hassan distribution. Now there's one last statement here or question. Does the result prove that the data conforms to the Poisson distribution and the answer? There is no. The result does not prove that the data conform to the Poisson distribution. There's just not enough evidence to throw it away. We're not necessarily showing that it is accurate, either.


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