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Ooo H2O11:10 PMmath-wwserver hartford.eduSetsChap4Part1: Problem 10PreviousProblem ListNextpoint)The top-selling Red and Voss tire is rated 70000 miles which means ...

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Ooo H2O11:10 PMmath-wwserver hartford.eduSetsChap4Part1: Problem 10PreviousProblem ListNextpoint)The top-selling Red and Voss tire is rated 70000 miles which means nothing In fact, the distance the tires can run until wear-out is a normally distributed random variable with a mean of 83000 miles and standard deviation of 5000 milesWhat is the probability that the tire wears out before 70000 miles?ProbabilityB. What is the probability that tire lasts more than 91000 miles?ProbabilityNote: You can

ooo H2O 11:10 PM math-wwserver hartford.edu SetsChap4Part1: Problem 10 Previous Problem List Next point) The top-selling Red and Voss tire is rated 70000 miles which means nothing In fact, the distance the tires can run until wear-out is a normally distributed random variable with a mean of 83000 miles and standard deviation of 5000 miles What is the probability that the tire wears out before 70000 miles? Probability B. What is the probability that tire lasts more than 91000 miles? Probability Note: You can earn partial credit on this problem



Answers

The useful life of a particular make and type of automotive tire is normally distributed with mean 57,500 miles and standard deviation 950 miles. a. Find the probability that such a tire will have a useful life of between 57,000 and 58,000 miles. b. Hamlet buys four such tires. Assuming that their lifetimes are independent, find the probability that all four will last between 57,000 and 58,000 miles. (If so, the best tire will have no more than 1,000 miles left on it when the first tire fails.)

The lifetime of a tire is modeled on a normal distribution, with the mean lifetime being 22,000 miles and the standard deviation being 4000 miles and were asked two things first rest out of 4000 tires. How many tires do we expect to last more than 18,000 miles? So what we're asking for is the probability that X is greater than 18,000. And again, with the normal distribution, you have to use a computer or look it up in a table, and you should get roughly 0.841 So that tells us that about 84% of the tires will last longer than 18,000 miles, so 84% of 4000 is about 300 3365 tires. So that's about how many we can expect will last longer than 18,000 miles. And then we're asked, What is the minimum number of miles we can expect that the tires will last for for 90% of the tires. So what we're asking for here is something slightly different. Here were have something like number of miles m the probability, but that the tire last for longer than em, miles. We want that to be 90%. Okay? And now the issue here is we want to solve for em and so right, We can't necessarily do this exactly, But we can at least get a good estimate again either by. So what you can do, for instance, is so we see 18,000 miles gives us roughly 84% chance. So next thing you might try Well, 17,000 miles, there should be more. Uh, we're tires that will last for greater than 17,000 miles. Because we're getting further away from the mean. You can try that and you can basically keep by trial and error. You confined value for M, which gets you pretty close to nine out of 10. And so if you experiment with a few values, I think you might find that's 16,850 miles. Gives you a good estimate. There

Problem 55. The function f of x presents a probability dynasty function for the random variable X. Where X is the thread life of a certain nick of time. And it's in thousands of mines. Thousands of miles for party. We want to find the probability that tire of this make will have a trade life at most 30,000 miles. Then we want to find the probability for X at most means let's then equals 30,000 miles. It equals the integral from the minimum value of X which is zero to 30,000 for every X. Dx. If X. The X let's integrate the integration of E. And we divide by the differentiation of the book. The substitute from 0 to 30,000 equals we remove this with this we have minus E. It was about negative 4.02 multiplied by 30. We have said that X is in 1000 miles. Then we compare it with Thousands of miles away. Then we change the limit to be 30,000 To be to be 30 not 30s. Um Then we have minus 0.02 multiplied by 30 minus it was about a zoo. Then it equals 1 -1. To the power Of -4.6. Which equals all point for 51 Who bought me. We want to find the probability that selected toil. We'll have a trade life between 40,000 and 60,000. You want the The probability for X to be between 40 and six Equals the integration from 40 to 60 for F. Of X. D. X equals the same integration as previous. We substitute from 40 to 60. We substitute by 61st. My industry substitute by four equals E. To the power of negative a 0.8 minus E. To the power of negative 1.2 equals 4.14. It Football. See we want to find the probability that selected tired. We'll have a good life of at least 70,000 miles. B probability for X at least means greater than or equal 70,000. We bought it 70 because X is in thousands of miles equals integration From 70 to the maximum value of X. There is no maximum value of X. Then we bought infinity for f of X. Dx equals the same integration. We substitute from 70 to infinity. We can substitute by X equals infinity. Then we take the limit the name it for the integration. When X approaches infinity my interest Mr Bush to buy x equals 70, gives -1.4 Equals it was about 1.4 -1.4. And the limit here gives zero. Then it's it's about negative 1.4 which equals all point to 466

Okay, So we're gonna look at how to evaluate our standard deviation and are mean in a normal distribution on. Then look at different percentages based off of that to see how much of our data would lie within the state or deviation that were given. So, for example, for this problem, let's look at a certain life span of a tire. So if you buy a tire from a dealership before you have to get it replaced, it might have a certain amount of miles that you can have on it. So let's just say the lifetime of this particular one is 30,000 miles, right? So we want to see that. So that would be our mean, which also come here in, like this. You and then our senior deviation, which is given to us, is 25 100 miles. So standard deviation. So typically, the average lifespan of this Tyree 30,000. But the standard deviation how far away from that? So they're giving this interview aviation of 2500. So within 2500 of the mean would be the standard deviation. And that can also you're in as such, Justin TV Asia. Okay, so We want to look at the percentage of tires that would have a mileage or the lifetime That would be between 20 2500 miles and 37,500 miles. So what percentage is going to be between those mileage numbers, huh? So to evaluate this, we would have to take and find dizzy that you and how we do that is we take X minus mu or are mean, if you will, over our standard deviation. Huh? That would be the formula you would use. And this is on Lee used for normal distribution. So if you remember normal distribution, that bell shaped curve where you mean, um, you would be right here, and then your standard deviations would go out this way. 12 and three, Negative. 12 and three. Standard deviations. Right. So that's a normal bell curve. So we're trying to find how many would fall within the 559,500 of the standard deviation, which they gave us of 2500 so soft we would simply plug in. The 1st 1 they're giving us is the 22,500 which would be your axe, minus your mean, which was given, which is the 30,000 and divide by that standard deviation which again given to you. I think that hunter. Now, when you solve for that, you should get negative 3.0. So that would be your first Z that we would find negative three. Okay, that would be like yours, E. But we also have to find between 37,500. So the same exact thing we plug. So the acts that they're giving us 37 1005 100 this time, Myers, the same given of your me over the standard deviation which you get given to you. And when she saw for this, you should get positive. Three. Okay, so that would be your second Z. Okay, so what does that mean? So now that we have positive three and a negative three as your Z's on a normal curve, we would look to see what that would mean. Now, if you remember from the empirical rule within three, standard deviations of your mean is that 99.7%. And that's just if you know it off your top of your head. So 123 standard deviations out a negative three. So your data that falls in between here. So how maney mileage is would fall within. That would be 99.7% for the empirical rule. Now, let's say you weren't sure how to find up, or you just couldn't remember off the top of your head in your book. Or you can simply go online and search for what's called a Z table, and what you do is you would find your values for three and negative three and you'll see what it looks like on a curve. And it would basically give you the decimals so it would be 0.49 Eat seven if you really want to get granular with it on the point for 987 So that would be for each side. So if we were to draw this so that 0.4987 would just be one side, so maybe the negative and then the other side that would come together. So that's why you have to add them so you would get the 99.74%. If you want to be that precise Okay, so that's how you would find it in two different ways. So to go back to the problem, the percentage of having those mileage is of that tire between 559,500 would be 99.7% of the time would fall within that. Okay, Now, let's say for if it's not a normal distribution so it wouldn't look like that bell curve and there was no additional information that you had about it being normally distributed, you could use a different theory. And that would state that, um, 89% of your data would be within three standard deviations of the mean so practically 89% 89% If it's not a normal distribution, okay. And that's within three standard deviations. So again, that was the easy that we had solved for, and you can use it off of a different the're, um, heavy chose right here. Okay, here. So again, that's if it's not. If you didn't get any additional information, that rule would just state within three standard deviations of your mean approximately 89% under that theory would fall within that. So that percentage between 559,500

Welcome to new Madrid. In the current problem we are given to observe the uh, light time of tires with the normal distribution. Then the current question they asked that the manufacturer guarantees that if the tires do not last at least 75,000 kilometers, Okay, that is if X does not exceed 75,000, Yes, Okay, they will replace, Okay, they will be replaced and then this guarantee, so we have to find them what percent of the tires may be needed to do that. Okay, so that means what what person, if it is exceeds this, then it is fine. But if it does not accept then they have to reduce. So we need to know this percentage. Right? So let's go and check. So it will be probability of x minus mu by sigma, less than 75,000 minus mu by sigma. Which will be probability of said less than now 75 minus 100. So it is minus 25 1000 by 10,000. Right, So that will be probability of zero, less than minus 2.5. Now think of a standard normal distribution, this goes from minus three to plus three, correct? So minus 2.5 is pretty much a rear value. Right? So there should be very thin chance of this. Let us verify. Let us bring the normal table over here and check. So now if you see 2.5 over here, give this 1 to 0.99 38 correct? So this value will be because the probability of said greater than 2.5. Right? That will be one minus probability of then or less than 2.5 mm That is we will have two and this will be force of six and then zero. You know, So that means 0.62 Or in terms of percentage it will be a 0.62% of the total sample Should be uh, X cd. Yeah. Uh, sorry. Should be should not shouldn't be getting there. That is out of this entire Okay, only 0.62 That is 0.62 It's merely 1%. So 1% of the data, mainly the replacement by the manufacturer. So 0.62% of users of tires will required uh huh replacement under the guarantee. So I hope you could understand this. Let me know if you have any questions.


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