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Suppose that we are attempting to locate & target in the three-dimentional space, and that the three coordinate errors (in meters) of the point chosen are indep...

Question

Suppose that we are attempting to locate & target in the three-dimentional space, and that the three coordinate errors (in meters) of the point chosen are independent normal random variables with mean 0 and stardard deviation 2. Find the probability that the distance between the point and the target exceed 6 meters.

Suppose that we are attempting to locate & target in the three-dimentional space, and that the three coordinate errors (in meters) of the point chosen are independent normal random variables with mean 0 and stardard deviation 2. Find the probability that the distance between the point and the target exceed 6 meters.



Answers

For any normal distribution, find the probability that the random variable lies within two standard deviations of the mean.

Okay, So what we're looking for is the probability that a random number lies within two standard deviations of the mean and refining it for any normal distribution. So I'll just draw out a normal distribution just so we can look at it. And that's kind of what it's gonna look like. And this will be the mean and we want two standard deviations to left into the right. Okay, so there's multiple ways of doing it. This the first way is to know that in any normal distribution 90 approximately 95% uh, the numbers are gonna fall within two standard deviations. So since we're looking for the probability, that would just translate to 0.95 But that's not super specific. So if you want on answer that goes to more decimal points. You want to be a little more specific with it. You can use the Z score, and so Z score equals the, um, your value. So we'll just call that X minus the man over the standard deviation. So we know that the value is to ST's, which were those away from the mean. So no matter what, it's just gonna be two standard deviations, and we can just call the means zero, because it's for any distribution. And so the standard deviation just cancels, leaving you with a Z score of just two. And if you have something called table A, you can look it up online, and it will show you all of the, um, corresponding um proportions for Z score of two so you can use that and you'll get around. You will get around 9545 and that is in decimal form because it's a probability, and so that is your more exact answer.

All right, We have normal distribution acts with mean four and standard deviation to. We want to find the probability that X lies between three and six. We're going to solve this problem in terms of Z scores, which makes the standard normal distribution easier to understand to start off with. Let's remember that a Z score is by definition x minus the mean, divided by the standard deviation. We can convert the two bounds for this probability to Z scores. Using this formula first for Z one, we have three minus 4/2 equals negative one half, and secondly we have Z equals six minus 4/2 equals one. So now we can restate this problem as the probability that the Z score falls between negative one half and one, or the probability that Z is less than equal to one minus the probability Z is less than negative 10.5. This second restating of the probabilities. In terms of Z scores allows us to use a Z table look up in order to identify this problem. So using a Z table, we find these quantities equal 2.8413 minus 0.3085 which gives us our final probability of 0.53 to eight.

All right, We have normal distribution acts with mean four and standard deviation to. We want to find the probability that X lies between three and six. We're going to solve this problem in terms of Z scores, which makes the standard normal distribution easier to understand to start off with. Let's remember that a Z score is by definition x minus the mean, divided by the standard deviation. We can convert the two bounds for this probability to Z scores. Using this formula first for Z one, we have three minus 4/2 equals negative one half, and secondly we have Z equals six minus 4/2 equals one. So now we can restate this problem as the probability that the Z score falls between negative one half and one, or the probability that Z is less than equal to one minus the probability Z is less than negative 10.5. This second restating of the probabilities. In terms of Z scores allows us to use a Z table look up in order to identify this problem. So using a Z table, we find these quantities equal 2.8413 minus 0.3085 which gives us our final probability of 0.53 to eight.


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