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Qumruen 40 (10panbs) Anempl | 0 1 0 In 3im Remztern9 3 Sar- EneHory mud is in that cn? machine that Ills bevcrage cans supposed [Nt . 12 ouriceS bevcrage each cn: F...

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Qumruen 40 (10panbs) Anempl | 0 1 0 In 3im Remztern9 3 Sar- EneHory mud is in that cn? machine that Ills bevcrage cans supposed [Nt . 12 ouriceS bevcrage each cn: Following dotplot Ior simplee random sample ewght cans- reasonable t0 assurne (lat tle conditions for performing hypothesistest are satisfied? Explain.11.95IU1215The dotplot shows that there (Choose: one)outliers.The dotplot shows that there (Choose one)evidenc strong skewness(choose one)assume tiat the population approximately normal:

Qumruen 40 (10panbs) Anempl | 0 1 0 In 3im Remztern 9 3 Sar- Ene Hory mud is in that cn? machine that Ills bevcrage cans supposed [Nt . 12 ouriceS bevcrage each cn: Following dotplot Ior simplee random sample ewght cans- reasonable t0 assurne (lat tle conditions for performing hypothesistest are satisfied? Explain. 11.95 IU 1215 The dotplot shows that there (Choose: one) outliers. The dotplot shows that there (Choose one) evidenc strong skewness (choose one) assume tiat the population approximately normal: (Choose one} Tcasonablc Jssume that the conditions are satisted. Haon



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Soup A machine is supposed to fill cans with 16 oz of soup. Of course there, will be some variation in the
amount actually dispensed, and measurement errors are often approximately normally distributed. The manager would like to understand the variability of the variances of the samples, so he collects information from the last 250 batches of size 10 and plots a histogram of the variances:
a) Would a Normal model be a useful model for this sampling distribution? Explain.
b) The mean of this distribution is 0.009 and the SD is 0.004. Would you expect about 95% of the samples to have their variances within 0.008 of 0.009? Why or why not?

This question covers drawing normal distribution plots. So in order to draw a normal distribution plot, you must first know how much entry CSR there's 22 entries and we find the corresponding table three for 22 entries which begins with negative 1.91 from here. We can construct a table with these values sorted in order. So I would and put these in a calculator or some application that can sort numbers from highest, lowest. So the lowest point 17 and the highest I believe is 7.6. So you would sort this and find the corresponding um table three values. So the lowest value of table three corresponds to the lowest value from this dataset and so is the highest corresponds to the highest. And in between they also have the same pattern. Yeah, so once we have that we can figure out, we can plot it in an X. Y axis. Okay. Yeah. Once we plotted in an X Y axes we can determine two overall pattern of where the majority of the data is. And once we determine the overall pattern, it is pretty clear what data points are. Outliers. Yeah. So in this case there are two outliers right here and there are outliers because they don't follow the general trend, Right? So the outliers are 6.7 and 7.6 um Mimos per square meter per day of effusive oxygen uptake. Um And after this we can determine whether it is normally distributed or not. Yeah. So nose we normal since the majority of the data um follows this trend. If we just eliminated the outliers, then we know that it's mostly normal, so it's normal they distributed.

So we're looking at the data and we can see that the data goes from 27 all the way up to 34. And if we look at the history, Graham Instagram, uh, looks to be, you know, pretty bell shape. So we would think that this is approximately a normal distribution. Then part B were to look at the normal Quanta will plot and remember, for a normal quantum plot, you're probably not ever asked to make one long hand. I never had my students do that. But you are looking at those values and you're hoping that you're seeing some linearity and so approximately linear. Therefore, the tendency is for the distribution to be approximately normal, not perfect, but not too bad. Then we want to look at on part, see what the inter quartile ranges and, uh, we need We have 50 numbers, so if you have them all listed down, um, if you put them all on your calculator, your calculator will give it all to you with 11 should put one variable stat. But if you have 50 numbers and then you need to count to find the 25th number 26 number, and that number will be your median and that number ends up coming out to be 30. And then we have 25 numbers who are below and so 25 numbers below. If I take 25 divide it by two, I get 12.5. So there are 12 numbers here, 12 numbers there. We need the 13th number in the list counting this way, and we also will need the 13th number counting this way, this one will be our Q one and Q one comes out to be 29 and therefore Q three counting comes out to be 31. And so our inner core tell range is the difference between Q three and Q one is two. And so, in order to find where those whether we have outliers or not, we want to take the Q one, and we want to subtract away 1.5 boxes. So 29 minus three anything below 26 is going to end up being an outlier, and there are none. No low outliers, and then we need to take Q three and add on 1.5 boxes, box wits or recurs. So 31 plus three is 34 anything higher than 34 is going to be an outlier. And our highest number is 34. So there are none. There are no outliers. Yeah, we don't count those that are right at that limit. And then we need to find the Pearson uh, index. And there are two methods for finding that one is the ski Eunice, where we take the mean and subtract away the mode and then divided by the standard deviation. And I have that the, uh I mean, when I calculated was this my mode most frequently occurring number was 30 and then dividing it by the standard deviation which I had had that day to put in my calculator. And I get this as the scariness. So it is positive it's pretty close to zero. So if it's skewed, it's just a little bit skewed to the right. And if we use the second index formula, that's three times we take the mean minus the median divided by the standard deviation. And in that case, it ends up that the median and the mode are the same. So I just finished writing here. So basically three times that quantity that we had up here and this Kunis number would be 30.265 So again, if there is Kunitz, it's skewed just a little bit to the right. And so it appears that this distribution is pretty close to being a normal distribution. Not too bad. You can tell that really from the 22 earlier plots that there's not too much skin this.

In Program 29. We have the lettuce it off the loads off aluminum cans, the aluminum cans off the thickness, or point all 111 inch or 11, 11, 11 cans in the deficit. You want to construct the frequency distribution, including the out lawyer and another frequency distribution, which does not include out lawyer, which is the load off 504 bounds. The first step is to sort the data in an ascending order in an ascending court. The second step is to construct the closest we will have here. The clothes, which is the load bones and the first close starts at 20 starts at 203 and we will use the close with off £20 which means that the next to close we started at 220. We've added 20 to 200. Then we will go further until we reach five 104 bones for the distribution that includes 504. Then we have 240 200 60 207 80 300 around 20 around 40 360 400 380 400 120 on 40 460 180 500. And we will stop here because the hour limit off this close will be 520 minus one, which is 519 which means that we now have 504 in our close. The bravest close will be 480 minus one here around 79. And we have here 500 minus one, 499. And so on. 99 300 339 and so. And we have here the frequency for the first distribution, which includes the out lawyer. And we will have here the frequency which does not include the outlaw. For the first close, we have from 200 toe 219 after sorting the data, the account for the first close is six. The second close starts a 200 20 toe, 239. We have just only five on. Then we have 242 159 which has 12 counts then from 260 which is included in this close until 279. 36 count is 36. Then we have from 220 299. The count is 87 from 300 319. We have 28 then the frequency is the Europe zero zero for all these classes 000000 and the year we have only one. Let's check that the total number is the total number is 175. Because this is the deficit. We have 170 five numbers. Six plus 5, 12, 36 plus 87 28 plus 1, 175. Just to make sure that we have not missed any number, we can see that the distribution for the deficit that does not include that. Would our lawyer no out life? It's the same 65 Well, 36 87 28 and we will stop here. This is the number off classes. Four. This distribution, which says that it's obviously very ridiculous to include all these classes just to write this frequency, which can be neglected as it's another ploy. This will be very, very and the end Very obvious toe. Represent that that just we need only 123456 We just need here six classes and we don't need any other clothes. And this is the final answer off our problem.

Consider a binomial experiment with our equals 3139 successes and and equals 5792 trials we want to construct a 99% confidence interval for the population proportion P. We proceeded the following three steps below to complete first step a We check the requirements to use a normal distribution to approximate the confidence interval. P So the testes is 60 hat is all over and equals 600.5419 mp. Had equals 31 39 to have equal 2063 are both greater than five to the requirements have in fact met next. We construct the interval, we have margin of error equals critical Z score rupee hat you had over end Since we are 99% confidence the Z score is 2.58 which means our margin of error is 0.169 which means that our confidence interval is P hat minus e. 4.2 point 5 to 51 is less than P is less than 510.5588 Finally, we interpret this to mean that we are 99% confident. The probability of success in a random trial is between P hat plus or minus.


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