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Units of years Now the lifetime Normal(2, 1) in of a bike wheel is than 1 year? (a) What is the will last more probability a whcel should be tested (6) What ' ...

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Units of years Now the lifetime Normal(2, 1) in of a bike wheel is than 1 year? (a) What is the will last more probability a whcel should be tested (6) What ' 's approximately the minimum numbestea tasted more than if we want percent tes sted a 0.9 probability that 75 one year? year when n Hint: Let .1, bc the number that last more than one are tested and use the normal approximna tion to the binomial

units of years Now the lifetime Normal(2, 1) in of a bike wheel is than 1 year? (a) What is the will last more probability a whcel should be tested (6) What ' 's approximately the minimum numbestea tasted more than if we want percent tes sted a 0.9 probability that 75 one year? year when n Hint: Let .1, bc the number that last more than one are tested and use the normal approximna tion to the binomial



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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 $98,200 .$

Into this lesson in this lesson, we have the likelihood of a number that is less than the mean or in a certain range. And that has given us the E to the power negative. Me and my standard deviation man is a mean squared divided by two types. Standard revision squared All of us on innovation I squared of two back. So the mean is giving us we're giving cowan 100,000 as the mean. Then 10,000 as the standard deviation. The pie has a regular value three A Constant Value 3.14159. No. Yeah. Yeah. So making the substitution in order to find the likelihood. So E. That would have Mhm squared Oliver too squared then All over. Okay, so this is equal to 0.000039. So this is Go in 12 Still zero Part 0. There were 39%. All right. So this is the likelihood. That's value is less than 98,200. All right. So times for a time this is the end of the lesson. Yeah.

Problem 52. A storage battery is normally distributed with the mean Of 4.5 years. I mean for unexpected lifetime 4.5 years. And standard division of oh boy. In five years we want to use a computer or a graphing utility to approximate the probability the probability For the given battery. We lost four, 4- five years. We can do not X as the storage battery lifetime. Then we want to compare the viability for X To be between four and five years. We can calculate it using the integration for F of X. The X From 4 to 5 for a normal distribution F of X is defined as undivided by segment To blow it by square root of two boy played by E. There's a lot of minus X minus mu. All squared divided by two sigma squared. I have entered this formula into Yes. Most to visit as a graphic utility and use it in calculating this integration. Let's get into this mess. The boat sigma Equals 4.5 years. We put the median equals 4.5 years. And this is the function And the integration is from four 25 from four to fight by getting this area. We can see that it's represented by this shaded area And this integration equals 1.68 city 0.68 City. This is for birdie for bored me is required To see if 10% of the batteries lost less than three years. Then we want to calculate the probability for ex To be less than three equals the integration from minus infinity 23 for F. Of X. D. X. The same type of eggs. Then we enter this moose and the change the integration limits a from minus infinity. Then we can take it as -1000 and be is strict. This is the area And it equals 0 .013, so equals 0.13 or its 1.3%. This means only 1.3% of the batteries lost less than three years. Then the answer is no Because 1.3% only we lost less than three years is the final answer for birth. And this is a for an answer of party.

We're gonna be using the function f of X equals e to the negative M Times X, where m is going to be 1/4 because it's a multiplication inverse of four X is going to be nine. And we're using this function because it says more than so. We're gonna substitute nine in for X and 1/4 in for em. We can simplify that a little bit too negative 9/4 which would give us about 105,000. It's


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