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QUESTION 4(n D)s? W = where s2 is the sample variance.Standard normal distributionStudent t distribution with degrees of freedom (n-1)Chi-square distribution with d...

Question

QUESTION 4(n D)s? W = where s2 is the sample variance.Standard normal distributionStudent t distribution with degrees of freedom (n-1)Chi-square distribution with degrees of freedom (n-1)Normal distribution with mean of / and variance of 0=

QUESTION 4 (n D)s? W = where s2 is the sample variance. Standard normal distribution Student t distribution with degrees of freedom (n-1) Chi-square distribution with degrees of freedom (n-1) Normal distribution with mean of / and variance of 0=



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The variance for each of Student's $t$ -distributions is equal to $df/(df -2 )$. Find the standard deviation for a Student's $t$ -distribution with each of the following degrees of freedom: a. 10 b. 20 c. 30 In summary: d. Explain how this verifies Property 5 of the $t$ -distributions listed on page 414.

Slowly got mu that is just me, you are average and we got a standard deviation of sigma Where you're looking for Q one, Q 2 and Q three, or to express them just as they are. To find out what our boundaries are. We can just use the simple formula of X is equal to mu plus Z sigma. That gets us our boundary lines between which numbers as we're doing with the problems with measuring spiders and hard things. Yeah. So we can find out that at the 25% mark, RZ score is negative 0.67. At 50%, we're going straight down the average. Which is that new? The mu isn't the average isn't any standard deviations away. So that turns out to just be zero for that Z score. And this is the score over here Is positive 6.67. So our X. or maybe we'll just call it Q one. It's a better way to call it Q one is equal to mu plaza not or else sigma, so many greek letters and this is going to equal what that Z is supposed to be. And we said that over here, Z score is negative .67. This 10 and this one's 100.67 So this means that out here we are 2/3 of a standard deviation away from the mean. Because our sigma is taken to be normally distributed for a normal standard normal distribution, we take that the zero is the average for the mean and the standard deviation is one. That's why 0.67 is our Z score for one standard deviation away. So I'll go ahead and replace the Z right here that I undo something. Yes, I did. Okay, I don't know what I ended here but we are looking at for fred quartile plus a negative 0.67 sigma. Q two is simply going to be the average. So that's pretty cool the mean and the median of the same and Q three is going to be equal to mu plus 30.67 sigma. So that does it for the first part. Now we're looking to express the cave percentile PK in terms of new sigma and Kay so the cave KF percentile that's weird things sake if KF percentile Um this would be the 50 if we're looking for something like you know the 75th%ile actually that would be over there. But let's say this 60th We would be looking at all the values that are 60% of the observations were looking at this part right here and the way that we saw that would just be to calculate the Z. Score at that corresponding um the score so P. K. Is going to be um you plus the Z score of K. So that's just going to be Z of K sigma. Yeah, maybe to specify that Zk a little bit more, we can say that ZK is equal to the probability at Z is less than okay. All we need to find a Z score for K. Okay? But then our Z score is going to be given by x minus mu divided by sigma. So really this is as simple as we can get it because we just rearranged it. We multiplied this sigma here and then added the average over there and that's how we got the X. So really all we were doing here with our Z score is letting X the the T. P fifth percentile. So this is pretty much as simple as we can get it.

Here we are given this information and we are asked to use it to find an unbiased estimator for Sigma in the form some constant times s Now, if we rearrange the equation to solve for Sigma, we get the following now for a given end. This is in the form of C times Expected value of s now just to recall for a given parameter or estimator for a parameter it is unbiased If it's expected, value equals the true parameter. So we have see time's expected value of s is Equity Sigma. Since he is a constant, this is also expected to This is also equal to the expected value See time's s which means that see Time's s is an unbiased estimator for Sigma And then we're given And it was 20. What is C equal to you in this case, so see would be equal to the following and using our rules for the for gamma functions when you get the falling and this comes out to 259 0.39

This question wants us to find the mean variance and standard deviation of a binomial distribution. We're told that the parameters of this binomial distribution are that and the number of trials is equal to 50 and p. The probability of success is equal to 0.4. Let's start with the mean which we call Mu the Greek letter you view we know for a binomial distribution is equal toe end times p the number of trials times the number of successes. In this case, that's gonna be 50 times 0.4. And when we do that multiplication, we find that Mu is equal to 20. Now the variance is looks similar, but in this time is gonna be n times p times Cute. Now we don't have a Q listed up here as one of our parameters, so we need to figure what what that is. Q. Is the probability of failure, and it's always equal to one minus p so we can do this. Quick. Calculation one minus 0.4 is, of course, 0.6, so Q is 0.6. That's the probability that an individual trial fails, and when we multiply all of these together we'll have Sigma squared is equal to 50 times 0.4 times 0.6, which is equal to 12. And that is our variance. We have a variance of 12 and then we want to find our standard deviation under standard deviation is just the square root of our variants. So it's gonna be the square root of 12 which we can plug into a calculator and find is about equal to 3.46 So these air your final answers. We have a mean of 20 variance of 12 and a standard deviation of about 3.46

This question asks us to find the mean variance and standard deviation of a binomial distribution. We're told that the parameters of this distribution are that we have 124 trials and a probability of success equal to 0.26 So and is 1 24 and P is 0 26. Let's start with the mean, which we right, as the Greek letter mu mean, is going to be equal to end times p or the number of trials, times of probability of success. In this instance, it will be 1 24 times 0.26 which is equal to 32.24 So this is the mean of our binomial distribution. Now let's take a look at the variance we write. Variance as Sigma squared in various is equal to end times P. Times Q. Now we don't have AH value for Q Given, but it's something that we can find. Q is just the probability of failure. So the opposite of P, which means Q will always be equal to one minus p. If we subtract 0.26 from one will find that Q is equal to zero point 74 So any times p times Q in this instance is going to be 1 24 times 0.26 times 0.74 And when we plug this into a calculator will get that our variance is 23.8576 And finally, now we want to find our standard deviation are standard deviation is just the square root of our variants. That's all we need to do is so we're gonna take the square root of that number. We just found 23.8576 and we'll find that our standard deviation is about equal to four point eight eight. So these are your final answers. We found them mean the variants and the standard deviation of this distribution, and that was that.


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