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You may need to use the appropriate appendixtable or technology to answer this question.X has a normal distribution with the given meanand a standard deviation. Fin...

Question

You may need to use the appropriate appendixtable or technology to answer this question.X has a normal distribution with the given meanand a standard deviation. Find the indicated probability. (Roundyour answer to four decimal places.)𝜇 = 101, 𝜎 = 15, find P(113 ≤ X ≤ 128)

You may need to use the appropriate appendix table or technology to answer this question. X has a normal distribution with the given mean and a standard deviation. Find the indicated probability. (Round your answer to four decimal places.) 𝜇 = 101, 𝜎 = 15, find P(113 ≤ X ≤ 128)



Answers

Assume that $x$ has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities.$$P(x \geq 120) ; \mu=100 ; \sigma=15$$

We have a normal distribution with mean 100 standard deviation 15. We want to find the probability that x is greater than or equal to 120 HD on this distribution. We're going to approach this problem. Not in terms of X, but in terms of Z scores, which make it easier to understand area and probability for normal distributions, remember that is the score is equal to the value X minus mu, divided by the standard deviation. So in this particular problem we can convert are bound of X being greater than or equal to 1 22. A Z score as follows, Z is equal to 120 minus $100.15 or 1.33 Now we can express this problems in terms of Z scores as follows, the probability Z is greater than equal to 1.33 or one minus the probability that Z is less than or equal to 1.33 Putting it into the notation on the right allows us to use a Z. Look up table, which identifies probabilities associated With particular Z scores, namely probabilities that are less than that Z score. Using as you look up table, we obtain 1 -0.9082 or our final solution 0.0912.

In this question, we are given a probability function. So we are told P X equals five minus X over 10 for X equals 123 and four. And with the information we need to find the mean understand a deviation to the first thing it's important to do is write ah, probability distribution table. So ah, Pierre X and then, ah, expertise of one to three and four. So that start both in the out. So what? X is one? We've got five minus wants. That's 4/10. I'm gonna keep everything. I would 10 because it might make things a bit simpler later. So when X is to put five minus 263 over 10 excess three, we have to overturn. On what? X is four. We have one overtime, and a quick check is just to make sure that does some toe one. So four plus 37 plus two is nine plus one is tense. That does some toe. What? So no, we want ah x p x. So now we're times ing overall your p x by X So for over 10 times one is still for overtime. 3/10 times two is not six out, two of 10 times three is also six out. 10 on four times. 1/10 is four out of 10. Now, if we add all those up, we get 20/10 which is to on this is all mean, which is the sum of X p X. So we now need to find X squared p x. No, but her total just to show that some up could be summed up warm there. Uh, so now winnings times are value of p x from the first column first row by our X value squared. So one squared is still one on that times four of 10. Still four after two squared is full. Full times three out of 10 is 12 after three squared is knowing nine times two out of 10 is 18 Onda Force read of 16 16 times. One overturn is 16 overturn. So this is why I kept all over 10 because now we do 12 plus 18 past 16 plus four and we get 50 over 10 which is five. So there's not your numbers and we know that our variance is the expectation off X squared, which we've just found here, which is five Onda. We minus expectation of backs, which we knows too. Squares. So the minus sing two squared, which is four. So I've got five minus four, which is one on a stunt. Deviation is the root of the variants, but Route one is still one. So that is Austin a deviation.

So this question were given exa normally distributed random variable with mean zero and standard diminishment .75 and we're asked to find the probability is indicated, so he's driven to transform X into our standard normal random variable x minus mean over standard deviation. We're going to apply this transformation to X. So part day we basically have probably TZ is between -5.36 95.0933. Now we know, call it easy, less than 5.0933 is one. Probably easy, less than -5.360. So the probability here is one. Our theme for asked for the probability that looks is more than 4.11, which is the same as probably to Z. more than 5.48, That is 1- authorities, Z And minus the probability of c less than 5.4 ft. That here is zero Less than frequent, courageous one. So the answer here is zero.

Question 20 says, find the standard deviation. The standard deviation can be found by taking the square root of the Somme of the X values minus the mean squared multiplied by the expected probability. So in question 20 either given us a probability distribution with additional information, we have the X times, the probability of their X. And then we also have 1/4 column in which they have already given us each value of X subtracted by the means already been squared, multiplied with probability. So we're gonna take those values and add them together, never gonna take the square root of their values. So we're giving 1.16 is the euro. I have a date zero 0.144 and 1.352 Some of those four values would be 3.4, and we take the square root to get our center deviation to be 1.8


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