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Use the following data. The lifetimes of a certain type of automobile tire have been found to be distributed normally with a mean lifetime of $100,000 \mathrm{km}$ and a standard deviation of $10,000 \mathrm{km}$ Answer the following questions.In a sample of 100 of these tires, find the likelihood that the mean lifetime for the sample is less than 98,200 .

Intro Stats / AP Statistics

Chapter 22

Introduction to Statistics

Section 4

Normal Distributions

Descriptive Statistics

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Lectures

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07:46

The mean lifetime of a cer…

02:42

Lifetime of a tire Assume …

03:44

A particular brand of tire…

04:31

A recent study of the life…

07:26

Stop the car! A car compan…

00:56

A natural gas utility is c…

05:41

The mean number of miles d…

03:45

For Exercises $5-8,$ use t…

04:35

Find the mean, variance, a…

04:25

01:57

A car manufacturer claims …

05:31

The average miles driven a…

06:04

Carbon Monoxide Deaths A s…

00:46

A normal distribution has …

01:16

00:47

Use the following informat…

01:12

00:36

00:43

Find the probabilities fo…

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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