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Question #11 The percent of fat calories that person in America consumes cach day normally distnbuted with mean of about 36 and standard deviation about ten. Suppos...

Question

Question #11 The percent of fat calories that person in America consumes cach day normally distnbuted with mean of about 36 and standard deviation about ten. Suppose that 16 individuals are randomly chosen:TRUE or FALSE: The random variable definedthe average percent of fat calories consumed each day by person America:For sample of size 16,state the mean of the sample MeanFor a sample of size 16,state the standard deviation of the sample mean (the "standard error of the mean"}d) Suppo

Question #11 The percent of fat calories that person in America consumes cach day normally distnbuted with mean of about 36 and standard deviation about ten. Suppose that 16 individuals are randomly chosen: TRUE or FALSE: The random variable defined the average percent of fat calories consumed each day by person America: For sample of size 16,state the mean of the sample Mean For a sample of size 16,state the standard deviation of the sample mean (the "standard error of the mean"} d) Suppose sample of size 16 taken Find the probability that the average percent of fat calories consumed more than 30. For part (al: Enter TRUE or FALSE For parts (b} through (d}, enter answer in decimal form rounded t0 nearest thousandth (3 places after decimall Examples correctly entered answers: 0.00O 0.004 0.080 1.200 0.594 13.000



Answers

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let $\overline{X}=$ average percent of fat calories.
a. $\overline{X} \sim$ _____(____,____)
b. For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined.
c. Find the first quartile for the average percent of fat calories.

The question your has given the data that the sample sizes 6 25 households and they've given the average or who has 2.3 persons. Poor sample Household understanding division, for example, is 1.7 fire. So you have to take everything, given statements are true or false and explain them. The first part states that the same Sam standard later for the sample averages 0.7 So to start with the Force bottom. But I know the data that we have and is 6 25 of the sample sites standing division is given us Oh, 1.75 Ando. The first part I'm going to check into started editor is 0.7 So if I apply the formula standard editor is given by the stand in division on that is divided by n. So if I substitute devalue here 1.75 that is divided by look off 6.5 Sophie value in this, we're getting this as 0.7 So therefore we see that the statement is too no, we're going to the next part. The question states that on anybody wasn't confidence in total for the average household size and the sample will be 2.162 point 44 So we need to check if this is correct or no. So then I can say that the 95% confidence in Doha sure got responds to toe standard errors. All right, so in this case, I can therefore say that this will begin by the average, which is given. It wasn't a question plus or minus. We take to winter the standard deviation. So then if I substituted, the value will have to go in 30 lesson minus. This is too into the standard deviation is uh oh, yeah, just one connection. This is supposed to be the started at our understand indignation. So this is a standard endorse. If I take this as 0.7 because you just calculated in part a. So then the value that I'm getting here. If you add and subtract after modifications, we have 2.30 plus or minus 0.14 Softer adding and subtracting. We're getting the Rangers 2.16 Omar who 0.44 So begin. Therefore conclude that this street mond what have given so nine if it wasn't confidence in Dover for the average household size and the city is 2.16 2.44 But here we can also say that a particular are confidence in door will Some radio robotic killer confidence in double Gore's the population evident, actually. And it turned not ours for the sample. So what are you here? It is actually governing for the sample that like what we've calculated. So therefore we conclude that given statement or hearers falls now the next part is the part. See? So for the park, see, the question is have to check on 95% confidence in trouble for the average household size and city is 2.16 to 2.44 So now, as you can see in the part be, you already proved this part. So the statement we can say is therefore true. So this is over here in the part be bigger than the solution. So we condemn force, eh? The statement was sure that the 95% off the class in double for the average household, what size in the city is 2.162 point 44 We approve that again here in Barbie. So I will Definitely The statement here given is true moving folder for the body People check. Uh, if 95% of the households in the city contain between 2.16 and 2.44 persons. So if I don't warn that part again, here's we've you will see that confidence in double off 95% says something about the sampling procedures we see it is saying about the sampling procedure. So therefore we can see that it could be duplicated that 95% off the household in the sample may contain between 2.162 point 44 persons. So that is why then therefore, we say that the loan statement over here is force now moving for the The next part is of even talk about the part e. So for the party, the question states that the 95% confidence interval is about right, because household size for was the normal call. So, yes, when they talk about the normal car, you're going to say that the individual household size may not follow the normal course, so if I talk about. But you know, the sample averages of the household size will follow normal call So I can see that the 95 persons off the glass in Dublin, um is a bold right, because the probability hist o gram for the average, it actually draws the normal call. Mean draws. We can see it follows the normal call and therefore times concluding that the statement that they've given us false or here now we want to the next part. For what If the question states that the 90 the night that wasn't confidence in total is about right, because but 6 25 draws from the box that probably he's took them for the average of the drawers before the normal coughs. Oh, yes, we've just spoken about it in the part E. So the statement is actually true. So we can see that, uh, the 95% of the confidence in double is actually about right. Because 6 45 draws from this box as I have mentioned it and party that we the probabilities, took down for rose the normal calls. Therefore, we say that the part if the statement, if it's true,

In problem 18. We have data on me number of calories, but bugger in three popular passport restaurants, and we're going to see whether these difference in the mean number of calories for this buggers from one front chase to the other. Now the thing if you need to do first is to get the not like where this is, which is given by meal equals you, too. He called the three Those actually means so. A three different restaurants, restaurants, uh, franchises. Next. The authority for this is that at least one mean is different. Um, critical. Run new. You have to run their analysis of variance test on Excel. So Preston detail, and then determine Alice is in another single factor. No need to select about detail, which is 33 columns and each other 30. If I set at 0.0 fire and then we rest OK, so the critical value depends on two new greater degrees of freedoms and tell d 90 degrees of freedom. Andi on our farm 0.5 and that is 4.10 The test Bella is the calculated value f, which is obtained from the I mean square for the factor developed better means graph for error, and in this case, we get 14.2. Now that we have those two bodies, we can you can compare them so that we can make the decision about rejection for idiot rejected. Now for the system. No, the critical value is four, and the test Molly is 14. This shows that the test value is within the critical region, and this means that we can make the decision to no. So now I produces. We see that there is enough evidence to conclude that at least one mean is different from they have a. So this is to say that when we compare these three popular fast food restaurants, we'll see that one of them would have higher or lua number off calories per bugger. So the summary is that the means in calories difference from from one to the other, from one francese to the other.

Okay. This is for number 78. In this question, we have a random sample of 24 laboratory rats. So simple size will be 24. Yeah, we got unknown. You percent of confidence. Interval with a mean blood cholesterol level. So 95% confidence and a little for me. This government which says on 88.2 to 89.8. So they're asking which of the falling could caused a Mr Worry about about validity of Angela. Question A. There is a clear outlier in the data. Yes, out. Liar is what you should have warned the minister for a confidence until because once there is an outlier, the whole distribution could be could be changed on the T distribution or the pediatrician that you used a can along longer be, um validate it for construction of confidence in the book. So, um, choice A is gonna meet most worrisome question. Be a stumble out of the data show a mild right skew of minor rights, curious fine questions that you do not know the publish in standard deviation, stigma, while stigma is okay because even if you don't know sick about you can still under the sample. Standard deviation asked D. The publishes. The distribution is not exactly normal. It's fine as long as it's practically wanted normal. We can use tea into the war Z interval E. None of these would be a a certain problem. Because of tea. The T procedures are robust. Not really. Um, not until they were certain that wire when there is no out wire, that's fine. But if there's any outlier than the the T distribution cannot long have used, so we should pick the choice, take a liar will be most worries up.

The percent of fat calories that a person in America consumes each day is normally distributed with the mean of about 36 a Senate deviation. 10. Suppose that one individual is randomly chosen. Let X equal percent fat calories a X could be represented by a normal distribution with the mean of 36 a standard deviation of Tim fee. Find the probability that the percent of fat calories of person consumes its more than 40. Graft the situation shade in the area to me. So because we have a center 36 36 year Senate aviation is 10. So we can mark 46 on the right, 26 on the left and what we're actually looking for here is the probability of more than 40. Well, shade in to the right because it says more than in order to solve this, we could use a normal CDF function. You go from 40 to infinity, which I'm gonna do 9999 30. I mean, of 36 with the standard deviation of 10 and I get an area or probability 0.3 for 46 I could see find the maximum number for the lower quarter of the percent of that calories, sketched the graft and write the probability statement. So again, I have a normal distribution center at 36 with 10 being extended deviation. And I want to know which value represents the first reporter of what's is my maximum value for the, um, for the lower quarter, which is just another way to say it was my portal one quartile. One is the value that represents the 25th percent marks 25%. The data falls to the left. So ultimately, I'm looking for which value left of K his 0.25 And I'm looking for Kay here so I can use my inverse norm feature in which I answered the area to the left of that boundary value. Insert the mean in the standard deviation, and I found out that value is 29.26


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