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Q3.(a). The thickness of a particular company’s glass sheets,say company A, produced by a certain processare normally distributed with mean 𝜇 = 5.0 m...

Question

Q3.(a). The thickness of a particular company’s glass sheets,say company A, produced by a certain processare normally distributed with mean 𝜇 = 5.0 mm and variance 𝜎2 =0.25 mm. Suppose the thickness ofanother particular company’s glass sheets, say company B, producedby a similar process are also normallydistributed with mean 𝜇∗ = 10 mm and variance 𝜎2∗ = 0.25⁡mm.i. Sketch roughly their pdfs in a single graph. Don't sketch twograph separately.ii. What is the probability

Q3.(a). The thickness of a particular company’s glass sheets, say company A, produced by a certain process are normally distributed with mean 𝜇 = 5.0 mm and variance 𝜎2 = 0.25 mm. Suppose the thickness of another particular company’s glass sheets, say company B, produced by a similar process are also normally distributed with mean 𝜇∗ = 10 mm and variance 𝜎2∗ = 0.25⁡mm. i. Sketch roughly their pdfs in a single graph. Don't sketch two graph separately. ii. What is the probability that a glass sheet produced by company A is thicker than 4.5 mm? Calculate in details. Q.3. (b) The SD of a particular type of 10-mg tablets is 1 mg, while the SD of another type of 50-mg tablets is 2 mg. Which type of tablets has more variability? Show the calculation & Justify your answer by calculating standard deviation which indicates the variation of the mean value.



Answers

Cereal The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and pour one large and one small bowl.
a) How much more cereal do you expect to be in the large bowl?
b) What’s the standard deviation of this difference?
c) If the difference follows a Normal model, what’s the probability the small bowl contains more cereal than the large one?
d) What are the mean and standard deviation of the total amount of cereal in the two bowls?
e) If the total follows a Normal model, what’s the probability you poured out more than 4.5 ounces of cereal in the two bowls together?
f) The amount of cereal the manufacturer puts in the boxes is a random variable with a mean of 16.3 ounces and a standard deviation of 0.2 ounces. Find the expected amount of cereal left in the box and the standard deviation.

In this problem, we're assuming that cans of Coke are filled with a mean of 12 ounces of soda and a standard deviation of 0.11 ounces, and this problem comes in three parts. So part A is asking you to determine the probability that a single can has at least 12 0.19 ounces, so that means it's greater than 12.19 And technically, it's saying greater than or equal to and up to this point. You don't know how to solve problems with greater than or equal to, because it does require a continuity correction. So what we're going to do is we're going to solve it as if it said X is greater than 12.19 because that's going to give us a close enough answer. So what we're going to need to do is set up are bell shaped curve and the bell shaped curve is going to have the average in the center. So we're going to put our 12 in the center and we're trying to determine the area to the right of 12.19 and in order to do that, we are going to need to utilize these scores and to refresh your memory. Z is equal to X minus mu over Sigma. So we're going to calculate the Z score for 12.19 by doing 12.19 minus 12 over that standard deviation of 0.11 And in doing so, you're going to get a Z score of 1.73 so we could go back to the bell and we can put a 1.73 at the same location as 12.19 So when we are trying to determine the probability that the soda can has more than 12.19 it's no different than saying What's the probability that the Z score is greater than 1.73? And in order to solve that, we would use the standard normal table in the back of the textbook? Uh, it would be table a two in your textbook, and if you notice the standard normal table talks about the areas into the left tail of the bell rather than the right tail. So we're going to have to rewrite this problem as one minus the probability that Z is less than 1.73 And using that table, you find that the probability of Z being less than 1.73 is 0.9582 and then we would end up with an overall probability of 0418 So just recapping part a of this problem, it's asking us what's the probability that a single can that was selected is greater than or equal to or in the words they used in the problem, at least 12.19 is going to be close to 0.41 eight in Part B. We're now drawing a sample from the population of Coke cans, and we're going to use 36 cans, so our sample size is 36 and we want to find the probability that the mean of those 36 is at least 12.19 ounces. So again it's going to be greater than or equal to 12.19 And again, um, you would use continuity correction, but you haven't discussed that yet, so we're going to solve this as the probability that X is greater than 12.19 and it's going to give us a close enough probability now because we're talking sample means we're going to have to calculate the average of the sample mean and the standard deviation of the sample mean and the central limit theorem is going to help you do that. And the central limit theorem says the average of the sample means is going to be equivalent to the average of the population, and in this case it was 12 ounces. And the standard deviation of the sample means is going to be equals. Who? The standard deviation of the population divided by the square root of the sample size. So in this case, our standard deviation was 0.11 and our sample size was 36. We're going to structure part, be very similar to part A. We're going to draw that bell shaped curve, and we're going to put the average in the center and our average in the sample means was 12. And to accommodate the finding of a Z score, we're gonna have to modify our Z score formula a little bit. So this time we're going to use the formula Z equals X bar minus mu sub x bar over sigma sub X bar, and we're trying to find the probability that X is greater than 12.9. So we're going to need the Z score for 12.9. So we're going to do C equals 12.9, minus the news of X bar, which was 12 divided by the standard deviation, while the standard deviation is this a an expression. So we're going to use 0.11 divided by the square root of 36 here, and in doing so now, we're going to get a Z score equivalent to 10.36 And again, I always like to take our answer back onto the bell. So 12.9 as an ex score is comparable to 10.36 as a Z score. And we when we're solving the problem, asking what is the likelihood that um, our 36 cans have an average greater than 12.9 is no different than asking the question. What's the probability that the Z score is greater than 10.36? Again, the standard normal table in the back of the book is set up to determine areas or probabilities into the left tail, so we're going to have to rewrite this problem as one minus the probability that Z is less than 10.36 And when you go to the standard normal table again, table A to anything over 3.5 is going to be 0.9999 as a probability. And when you do one minus that, you're going to get an answer of 10.1 So again, to recap the probability that when you select 36 cans, their average of volume is at least 12.9 would be 0.1 and now we have part C to solve. And in doing part C, the question is saying, given the result from part B, is it reasonable to believe that the cans air actually filled with a mean, equal to 12 ounces? Well, we're seeing that there is a chance of having a mean of greater than 12.19 so we could say it appears that the cans contain an amount greater than 12 ounces. So is it reasonable to believe the cans are actually filled with a mean of 12? We're going to say it appears that it's more than and then the second part of part, C said. If the mean is not equal to 12 are consumers being cheated? And we could say since it's greater than 12 ounces. Therefore, since customers are getting a little more than 12 ounces, they're not being cheated because they thought they were buying 12 ounces. They're getting a little bit more for their money, so they're not being cheated.

Hello, this is number three. Section 5.1 point three. The question gives us some information about some coating thicknesses of paint, and it gives us the data in the problem. Um, and then it asks us to find find a few things. Um, it also says that we should assume the distribution of the coating thickness is a normal probability curve, so that's gonna help us farther down the line. Okay, um, the first one says, calculate a point estimate of the mean value. Okay, so that's going to be, um, you could do that a couple of separate ways, but let's go ahead and do it. And when I think the most, obviously it would be. And that would be using the sample mean? So just do part A. I would find the sample mean which is written that way. Or since it's part of the sample, you could write mu carrot on top, right? And that is equal to this is our typical average. Right? So we have some of all the little X is there in the data set, and then the number of entries in the data set, um, real quick, just for reference I made this. Um, this has a bunch of our symbols And what they mean, it also has a couple of formulas that has sort of been there. But this is kinda helpful to go back and look at, um we may need them all here at various points. Okay, So when we go through and figure that one out, um, we end up with this from you, which is the mean, right? I love saying, mu reminds me of Pokemon 1.34 8125 But we're going around it. I usually feel OK going out for decimal places. Okay, so that would be part a in green blocks. That one up. Okay. C o part B, part B. It says, Could you maybe, you know, use. Ah, there's multiple methods. Of course, we could do this, but it says, could you calculate a point estimate of the median and then we have to kind of talk about it. Um, well, I mean, one way to do it would be to actually find the sample median right, which you can certainly do. Just by looking at the numbers in the set and finding the middle value of values, right? Uhm, but the back of the book is giving you the same. Same you here. So what they did logically was they said that you could go ahead and just say the sample mean would pretty accurately represent it, right? So their their reasoning here Is that the point? The sample right. The sample median is going to be equal to essentially Jose was. Example me, which see? So here's thistles What I would like. I mean, they ask you for the media. You could certainly do this, but that's not what the back of the book has. So they just go and use the sample mean which would be inaccurate representation. So they're just gonna go ahead and say, Hey, it's 1.34 way one, same as the other 11.3 481 Just like the other one. So you can use either one. I guess it argue for it or against it. Um, I think the media is a very special indicator, and I like to use it for things. Just dio All right, next up part C um, Let's see. Okay. Yeah, Part C has Ah, a couple of different parts to it. So they want us to find a point estimate of the value. That is It's not like, you know, you can do like, you know, the core tiles, right? And the media, um, they wanna find the value that separates the largest 10% of all the values from the other 90% of the values. Right? So your top 10% what are they type thing? Eso You can write that in terms of mu and the standard deviation. If we want Thio, we certainly can, um, in general, right, we can find out if we know what the standard deviation is. Uh, you know, what I mean is we can use those things to figure out our answer, but we also can use the table, and I like being able to use that. So here's ah little fact. Um, if we have our mean new right, um and we add 1.28 times the standard deviation to it right, which in our case, I guess would be s because we're dealing with a sample, right? So in our situation, we would end up with a new the mean just of our sample And we'd add this 1.28 and you're asking yourself, Where the heck did that 1.28 come from? Yeah, it took me a while to realize this to back when I took a class. And, um, there's a There's a chart and then probably in the back of your book, honestly, but it's also available online. Then it's all over the place. It's the Z score table, right? And if you have a normal distribution, you confined those percentile, like what that number is for whatever percent you want. Um, in this handy dandy chart, what you do is you go down. Hey, look through all the values and to come across 90% like the ones that say zero point, okay? And when you find that there's actually there's two of them here, they're pretty close. But you can like assume it's gonna be closer to probably the first one of these two that are boxed in red on. Then you go across and say, Okay, well, that's 1.2 and then this 0.8 You say it's 1.28 So they came from here. The 1.2 you came from here. They're adding those. So it's 1.28 If I wanted to do 30% I'd go find 30 somewhere here on this, this nifty see chart. Um, but we're gonna go ahead and use that 1.28 because we're looking for the divider between the 90 and then the top 10. So that's where that came from. So go ahead and use that. I always took a post it, and I stuck it on the Z chart page of my textbook. So if it's got it back there, I'd say you'll probably be referencing it quite a bit. Or maybe you already have. Okay, so the actual math behind this ends up being 1.3481 Right, which we got earlier. Plus this this multiplier 1.28 times 0.338531 And this is our standard deviation are Actually it's our sample deviation. Pardon me. And how do we get that? Well, you can compute it using the formula in the book. But honestly, every time I have a problem like this, I always go to my calculator and I type in all the values, like, literally all the values. And then I could do this handy calculations thing. This is the P 89 simulator. Um, the similar things that you can do on the other calculators. Anyway, So I press a five. I say I'm dealing with one variable. Say, take C one, and then when they enter, it just kind of does. Ah, couple of key statistics for me, right? These things are all really, really useful. And it allows me to double check, right? What my sample mean waas x with the bar Little X with bar on top of it. It also allows me to go through and see what the standard deviation is of this, which is the S with X right here. So that 0.38531 I literally took from the calculator. I mean, that's what the calculators four. But if you want a calculated by hand, you certainly could. I wouldn't stop you. Um, but that gives us an answer. In the end, that is equal to just about. And these were kind of been they've been rounded their approximations, but whatever. Uh, 7844 All right, so that's the answer toe, part C, part D, part D. So they're really asking this, right? And whenever you see something like this 1.5, you can kind of read that like, um, the proportion of all the values that are less than whatever, Never in this case, 1.5, right. And we can't do this for the whole entire, like, really like larger set. We just have a sample of it, right? But we can We can estimate them right so we can go ahead and do it this way. Um, it's kinda like the sample proportion of all the successes. That's how they define it. I actually have this back on this handy dandy page, right? It's this one over here. Um, so let me go ahead and I'll show you how I got. I went ahead and said, P and those little cat means it's every time you see the pointer considered. It's like a point sample. It's part of whatever our sample is. So this is little P regarding just the sample that we have. So here we have the number of successes in the sample, so I'll just put successes. Always good to know how to spell successes. I hope that's right over the number in the sample, which is end and the math behind this is exceptionally simple because it's really just four divided by 16, which is equal to 1/4 which is equal to a decimal of 0.25 And I'm gonna go ahead and box this one up, all right? That's what part D was asking. And the last very last part was Part E. And they're asking, really, for the standard error here, the way they phrase it is. What's estimated Standard error of the estimator you used in B S O. This applies specifically to what they had in the book, which waas um they just went ahead and gave, you know, the sample mean not the simple medium. So, you know, you might have done it differently if you did. You have to come up with a different formula because this is the one that they actually have in the book. So our standard error, this is estimated it's going to be this guy I love making that and that acute symbol. I just love it, uh, which is equal to RS, which we know divided by the square root of N. Hey, that's handy. Okay, so in this one, we've got R s, which ours are standard deviation of the sample 0.33 8531 all divided by the square root of 16, which we know is equal to four handily enough. But I'm gonna put that my calculator anyway, because what the heck? I'm gonna have to do some division, and the answer will be. Get back is 0.846 All right, that's part E. So you have now all the different parts and where they came from. Thanks for following all following along with me.


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