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Specifications for an aircraft bolt require that the ultimate tensile strength be at least 16.SkN. It is known that 8% of the bolts have strengths less than 18.SkN ...

Question

Specifications for an aircraft bolt require that the ultimate tensile strength be at least 16.SkN. It is known that 8% of the bolts have strengths less than 18.SkN and that 5% of the bolts have strengths greater than 20.6kN. It is also known that the strengths of these bolts are normally distributedFind the mean and standard deviation of the tensile strengths of the boltsWhat proportion of the bolts meet the strength specificationFind the 8Oth percentile of the sample mean breaking strength:

Specifications for an aircraft bolt require that the ultimate tensile strength be at least 16.SkN. It is known that 8% of the bolts have strengths less than 18.SkN and that 5% of the bolts have strengths greater than 20.6kN. It is also known that the strengths of these bolts are normally distributed Find the mean and standard deviation of the tensile strengths of the bolts What proportion of the bolts meet the strength specification Find the 8Oth percentile of the sample mean breaking strength:



Answers

A manufacturer of bolts has a quality control policy that requires it to destroy any bolts that are more than 2 standard deviations from the mean. The quality-control engineer knows that the bolts coming off the assembly line have a mean length of $8 \mathrm{cm}$ with a standard deviation of $0.05 \mathrm{cm}$. For what lengths will a bolt be destroyed?

Yeah, I've already drawn us a dot plot of our nine data points and now I need to calculate the mean and the standard deviation. So for the mean, I'm going to add our eight date or nine data points together, which will give us a sum of 19.56 and we're going to divide by nine, Which will give us a sample mean of 2.173 repeated. Now. For our standard deviation, I need to find out how far each one of these data values is from the mean, square them, Divide by 1- the sample size -1 and then take the square root. So I'll record here all of the differences between our data points and are mean and now I'm gonna square each one of those differences and find the sum and the sum of all these squared differences is 3.44- eight. We will divide by one sample size minus one, which will be eight and we'll take the square root of that and that will be our sample standard deviation, which equates out to be .656

In this question, we know that a rivet manufacturing company estimates the shearing strength of the rivets to be normally distributed with the mean of £925 and a standard deviation of £18. So it's a normal distribution with mean shearing strength of £925 standard deviation of £18. So if X is the strength is the shearing strength of the rivet. It's approximately normally distributed. I mean 925 standard deviation of 18 and for part a who were asked what it what percentage of rivets would have a sharing strength greater than £900. So what is the probability that the sharing strength ST greater than 900? So to convert to a standard normal distribution we're using that is equal to you X minus mu over Sigma and raising that throughout this problem. So this is equal to the probability that said is greater than 900 minus 925 over 18 which is equal to the probability that said is greater than minus one point 389 is equal to one, minus the probability that Zed is less than minus 1.389 and that comes out to a probability of 0.918 So the probability. So it's 91.8% of the rivets are expected to have a sharing strength of greater than £900 and for B, whereas what the upper bound on the sharing strength for the weakest 1% of rivets is. So let's assume this is our normal distribution for the sharing strength of rivets. So the mean is 925 and it has a standard deviation of 18. And so we're asked, There's some value down here. Let's call it X equals K, for which the one percent of the weakest rivets exists. So in that lower tail that represents the bottom 0.1 cumulative probability. So the question is asking, What is that value K such that the probability that the shearing strength being less than K is equal to 0.1 and converting to a standard normal distribution. This is the problem. The probability that said is less than K minus 925 over 18 and we confined this from the table. So we're looking for an area of 0.1 What Zed score corresponds to 0.1 We will equate that with this fraction here. So going to the standard normal table 0.1 exists about here. That's an area of 0.1 And that corresponds to a set score of minus 2.33 Approximately. So we can say this is equal to the probability that said is less than minus 2.33 And so now we can equate these two factors. These two numbers here we have minus 2.33 is equal to K minus 925 over 18 which tells tells us that K is equal to you. K is equal to you. 883 0.6 So we can we can call that 883 cents. Well, we can call it 883.6 because that's the that is the sharing strength in pounds. So for part C were asked, if Iran, if a rivet is randomly selected from all the rivets, then what is the probability that it's shearing strength would be at least £920. So that's the probability that X is greater than or equal to 920 which he was. The probability that said it's greater than 920 minus 925 over 18 is equal to the probability that said is greater than minus 0.278 It was one minus the probability that said is less than minus 0.278 which is equal to one minus 0.391 which equals zero point 609 So that's part C. In the last part, party is saying that, assuming that this probability calculated in part C is correct. So that's probability that the shearing strength of a randomly selected, selected rivet is at least £920. Assuming our answer in part C is correct. What is the probability that in a sample of 10 that at least three of the rivets will break at a force less than £920? So in a sample of 10 for each rivet there is a binary of outcomes. It can either fail or not fail. So that is, we're looking at a binomial distribution with a size 10 and the probability p. So let's if we're looking for the probability that three fail, let's call failure of a sheer success. So the failure of this year is one minus 0.6 year old nine because 0.609 is the probability that a sheer can withstand at least £920. So failure of this year is has a probability of one minus 0.609 which is 0.3 91 So the sheer feeling is our definition of success here. So now we're looking for the probability out of the sample of 10 that exactly three of them fail and we can use software or calculator to calculate this. I'm going to do this in many tab. So in many tab you can go to you calculate probability distributions by no meal number of trousers. 10. The event probability is 0.391 and we want to be sure to check probability because this is not a cumulative probability. We want the probability that exactly three successes occur and the number of successes that we're looking for is the input constant. So we can hit. Okay. And we get a result of 0.223 for the probability. This question asked us to round it to the nearest 10th so we can say the probability of X equals 30.2.

Hey there, guys. How's it going today? We're gonna talk about a central limit theorem in stats. So the way I like to start stats problems is by listing out the things that were given in the problem, the terms that we might use, as well as any sort of formulas that we're going to need so right off the back, what you're given that there's an average breaking strength or the mean is £2000. Oh, on average, each of these cables breaks under £2000. There's some that can break below 2000. There's some that could break up, but the average for the population is £2000 were also given that the standard deviation of the population is £100 so each standard deviation is going to be plus 100 or minus 100 per standard deviation. And then we're given our sample size ISS 20 people's, so they're going to sample 20 cables at a time and get an average of their breaking strength. Were also given that this problem is assumed to be normally distributed, that's gonna be important when we take a look at our sense of limit. There So now we have all these things. What are we being asked to do? We're being asked to find a sample mean that cuts off the upper 95% of all samples. So we need a sample. Me, which in this case, we will call on brought back sex being our sample. And if we have a sample mean usually means we're going to need a sample standard deviation, which, in this case, we will call standard deviation of X. Just so imagine is all right. So we need a sample mean and we need a sample. Standard deviation were given the population. We usually use the central limit theorem over here. Social limit. The arm has three different points. The first being that population need, um, equals the sample Mean Ampex aren't pretty straightforward, so we can come over here. You also see that the sample standard deviation. So as the X equals the standard deviation of the population divided by the square root of our sample size of 20. All right, The last thing down here is kind of the maker break. All right. Our sample size has to be greater than or equal to 30. No, maybe saying what? We can't do this problem then because we have examples. Eyes of 20. And you'd be right if the problem I'm given us that the that the population is assumed to be normally distributed. The only reason end has to be greater than or equal to 30 is because you want to ensure that your population is normally distributed. That's how this central limit theorem works best is when our populations are normally distributed. All right, So because we're given this normally distributed in the problem, we can kind of ignore that our sample is 20. All right, now we can move on and solve the problem, as is So the first thing that we can dio right off the back is fill in our sample mean due to the central limit There it is, assumed that our sample mean is going to equal our population mean of £2000. All right. Pretty straightforward. We already have one of our unknowns. The next thing that we need to do is find our sample standard deviation. Now, this is going to go ahead and take a little bit of map. But sure, we can get it done All right. So what we're gonna do is we're gonna start off with our sample standard deviation about that, we're going to start off with our sample standard deviation. Okay, Now, over here, we see that the formulas air sample standard deviation equals our regular standard deviation. Which iss 100. And we divide that by the square root our sample size of 20. Okay, so doing some radical mouth and that we multiplied by the square rates of 20. Cancel out that 20 done. They removed the radical up top because we don't like radicals in the denominator of fractions. So doing that is going to get you a radical mixed radical of 10 over five or 10 times a square of five, which is going to approximately equal 22 point three six. And in this case, it will be in pounds. All right, so now we can come up here and we can go ahead and fill in our standard deviation of 22.36 All right, so moving on, we're gonna go ahead and draw our curve for our samples. All right? With god, baseball on average, Dennis and I like to go out on my standard deviations just because it helps me visualize. I recommend that you do the same, usually drawing about three on either side to make things easy. Dan Hannan is our and that £2000 and each standard deviation will be plus or minus this. 22.36 double. Plus, you're minus 22 right? Bring. So now we can actually go ahead and start solving for the upper 95%. All right, we want upper 95%. So we are ready? No, but this right here is 50% right, because this whole curve is 100% of our 20 size samples. This right here is our 50%. Wait. Now we know that each of these has its Each standard deviation has a percentage assigned to it. So this one right here would be 34. Same thing with this. This right here is 13.5 approximately you 2% and 0.5%. All right. So altogether, that be 50%. We are asked to find 95% in the upper 95%. So if we add on another 34 we're gonna get 84 not quite enough. If we try to add 13.5 onto that, using this standard deviation, it's a little too much. So what we need is something that's going to be Oh, probably right about in here to make the rest of this 95%. Okay, so we want it. Upper 95 percent. I'm sorry. And while we're here, let's go ahead and fill in this down. I think lower. So we're looking for a value. We're looking for a value. This value right here that cuts off this 95% because we only really want this bottom 5%. We want to cut all this off. You came. So how do we find this? Well, when dealing with percentages, we have to look at our Z tables. Okay, So if we come over here, we have a Z table right here. See, tables are made up of percentages and their corresponding values. Z is the approximate of a standard deviation. Christ zero right here is 00.5. Because 0.5 is going to be are means zero zero is going to be our means. So in this case, it would be 2000 right? So we also need our RZ value, right? We need our formula for R C value. So we have C equals Rex minus. I mean, in this case, it will be our sample mean divided by the standard deviation of army There of our sample so bad, that's on formula right there. That's the formula that we're going to be using. Anybody thinking? Well, how the heck do I get a Z value? All right, so let's go back here. We have 5% right here, right. And 5% can be written as point zero. That is 5% right on. 95% will be point nine. All right, So when we look at the tables, we have to think in terms of the left hand side of the curve. All right, so in our case, the left hand side of the curve is the 5% so we don't really need this 95%. We just need t 05 We just need the 0.5 So we're looking at this. We've gotta find a percentage in these numbers. We have to find our percentage in these numbers, right? Closes. We can get So the closest we can get 2.5 is actually going to be right in here is going to be these two numbers right here, all right, cause they're both equally as far from point No. Five. So we'll start with this number right here. Negative 1.6 cream. So we'll say Z equals negative. One point six. All right, So we'll travel on our over here to these two values and point if I was probably somewhere right in the middle, right? So we come up here, we take a look, and we're given our values for that fair number. And ours is right in between my 04 and point on five. So what we can call it is 1.64 and now we can keep going with our formula. So this number is going to be equal Toe X, which is the number of the value are trying to find that's cutting off the upper 95% minus our means. Who are mean. Waas that 2000 pounds were invited by the sample standard deviation. Make sure you use the sample standard deviation. Another population is very, very important to use the sample standard deviation because we're looking at our sample percentages were not looking at our population percentage. We're looking at samples that we've got to use the sample mean and the sample standard deviation of 22 point three £6. All right, so when we go ahead and solve this out been, multiply over, we get that we get negative. 36 point 78 equals X minus 2000. There were only add our 2000 over. We get X equals along those June 963 point to two. And in this case, we have our unit as hounds. All right, so if we come back over here to our are you curve, we were looking for this X value right here. Right? That X is this X, which is that 190 or £1963.22. So this value right here is going to cut off the upper 95% off All 20 size samples, pounds 20 size samples, average

In exercise dot team. We're considering a scenario where a thread manufacturer Test the sample of eight lens of a certain type of threat made of blended materials and obtains uh mean tells a tensile strength of £8.2. With standard deviation of £0.6. Now from that statement we can extract three sets of information there, the first one being the sample science. So the sample and sample size and equals eight because the manufacturer is using eight length and also the mean Tensile strength X. Bar. It's a sample name is equal to £8.2. And we also given that The sample standard deviation is 0.06 lbs. Now using this information and assuming that the tensile strengths are normally distributed, we're going to obtain a 90% confidence interval. Mhm. For the population means now to proceed we have to determine the formula that is needed. And the formula for the confidence interval is X. Bar. Class of minus the critical value of T. For the given level of significance multiplied by S. Advair with this cultural event. Now for 90% confidence interval, The level of significance often is one when a 0.9 which he calls 0.1 and when we have that it's going to be 0.05 fought. The degrees of freedom corresponding to the T distribution here, N -1 equals 8 -1 and that equals seven. No, the next thing we need to get is the critical value for the T. V. The T here being um Having 7° of freedom in the us, Europe and 05 level of significance. And that From the table equals one 895. We're now ready to substitute all the values that we have been given and what we have just obtained to get their confidence interval. So expo is £88.. 1.895 times 0.06, divided by the square root of eight. So when you work that out using a calculator, the margin of error Becomes 0.04. So in pounds, this is the 80 90% confidence interval for the mean 10 style strength of this thread.


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