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The useful life of a battery is thought tobe normallydistributed with σ = 3.85months and μ = 50.4months. For how many months should the battery beguaranteed i...

Question

The useful life of a battery is thought tobe normallydistributed with σ = 3.85months and μ = 50.4months. For how many months should the battery beguaranteed if the manufacturer wishesonly 1.5% of batteries to be subject toreplacement under guarantee? Enter your answer to oneplace of decimal. Months will be understood and should notbe entered.

The useful life of a battery is thought to be normally distributed with σ = 3.85 months and μ = 50.4 months. For how many months should the battery be guaranteed if the manufacturer wishes only 1.5% of batteries to be subject to replacement under guarantee? Enter your answer to one place of decimal. Months will be understood and should not be entered.



Answers

A storage battery has an expected lifetime of 4.5 years with a standard deviation of 0.5 year. Assume that the useful lives of these batteries are normally distributed. (a) Use a computer or graphing utility and Simpson's Rule (with $n=12$ ) to approximate the probability that a given battery will last for 4 to 5 years. (b) Will $10 \%$ of the batteries last less than 3 years?

So 87 a. Defined the random variable so the random, variable X and this comes right from the problem would be the useful life, all the car battery. So useful life of a car Battery B is ex continuous or discreet, so that would be continuous months. Time in general is continuous. See, just wants the distribution. So it is an exponential distribution, and they give you the parameter. It is 0.25 the on average. How long would you expect one car battery the last? So it's asking for the mean So I know Emma the parameters point tooth 025 And that should equal one over the mean. So if you multiply by mu and divide by 0.25 that gives us the mean equals one over 10.25 or 40 months. If e. On average, how long would you expect nine batteries to last that they're used one after another? What each battery should last an average of 40 months times denying batteries, and that gives us 360 months find the probability of car battery lasts more than 36 months, So more than so, we could just use e to the negative 0.25 times 36. And that is 0.4066 G 70% of batteries last at least how long. So this is talking about the top 70%. So I know that 0.7 equals e to the negative 0.0 to 5 times X. So if I take the Ln of those sides, that would give me the natural log of 0.7. Equals negative 0.25 x divided by negative 0.25 and I would have the natural log of 0.7 over negative 0.25 for the value of X, and that will give us 14 0.27 months, so that would represent the number that would cut off the top 70%.

On this question. We're told that the mean lifetime risk watches 20 minutes to me that there's a standard deviation five months, that this judicious means t is equal to X minus you. We asked how many months to get you made if the manufacturer is not want to change from 10% of the watchers. So we know that the bottom 10% I have 10% of data values below. It's cut off. Therefore, I can determine disease score value for 10% probability of June 1. Using our G table, we'll find the G is equal to negative 14 to 8. And you can just plug that in Derek's into from thanks. Does he look to you? Plus C. And this is just saying that disease score is a measure of the number. Standard deviations of the value is moved from the mean just the basic definition for so now, if you plug it with in 25 plus negative 1.2 eight times five, which will give us results of 18.6 that h imprint six months

Well, since the problem just tells you to go use a graphing calculator, that's exactly what I'm going to do. And perps did that backwards. And in this program, you can actually go to the functions under miscellaneous. And you can actually just find the answer right here and plugging in the values 48 to 60 of the function. Now, we could work this out if anybody's interested. But again, I'm following the directions. So 139 No, please. Um, T minus 48. Okay, Weird. Jeez, um, just double check that you typed it incorrectly, and this is your answer as a decimal. But the way the problem set up is that they want you to Oh, I'm breaking this at times. You know, if you take that value multiplied by 100 that your percent 47.72%. That's what we're looking for anyway. No need for the graph. All you need is the calculator

48. What kind of distribution? We're gonna use an exponential and we for exponential use em. In this case, we know our exponent RM equals 1 10 that's made


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