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For a < x <b elsewhere1) a) Show that the uniform distribution, given by p(x) =is normalized.b) If a = 3 and b = 8, find the variance of the distribution:...

Question

For a < x <b elsewhere1) a) Show that the uniform distribution, given by p(x) =is normalized.b) If a = 3 and b = 8, find the variance of the distribution:

for a < x <b elsewhere 1) a) Show that the uniform distribution, given by p(x) = is normalized. b) If a = 3 and b = 8, find the variance of the distribution:



Answers

Suppose $X$ has a continuous uniform distribution over the interval [-1,1] . (a) Determine the mean, variance, and standard deviation of $X$. (b) Determine the value for $x$ such that $P(-x<X<x)=0.90$. (c) Determine the cumulative distribution function.

Okay. So is this a problem? A is negative one, B is one, This is the domain size and we got this. It's from negative 1 to 1 is a uniform distribution. So first we have the expected value of X. It's gonna be a plus B over to And the result is zero. The various SB minus a square or two or 12. So it's to the power to over town which is for all over town. And there is always one of the three. So the standard evasion is just the square of the variance and it's Square ribs 3/3. So this is the first problem. Okay. Problem be we just used so I suppose we have a this negative variable X. That is more than our variable X. And the barber actually squared. And that's then we can calculate its probability is just one or two X minus negative X. Which exactly X. So if it's probably is .9 that X should also be .9 the problem. See, we can write down its competitive distribution function which is zero when X is more than active one, it's one of the two X plus one. If negative one is more than X more than one and it's one if X is square or you go than one. So that's a

Well here we are, given a random variable X. That has a discreet uniform distribution On the into jersey fixed between one and 3 and first for us to determine the mean and the variance. So the mean and the various formulas are given here. So the meanest P plus K over two, which is basically wants to to experience, is the minus a plus one squared minus 1/12. So three squared is nine, 9 -1 is eight. 8/12, yes .667. So we have our mean and variance.

For this exercise, we have a random variable X. That has a discreet uniform distribution on the integers one through three. So in other words, the probability of X taking on either of the three values 12 or three is one third for each. We are asked in the question to determine the mean and the variance of X. The mean or expected value of X. When it is a uniform discrete variable is A plus B, divided by two where a is the smallest integer, and B is the largest integer, And that gives us a mean of two. Now to solve the variants, we use this formula, so we have three minus one plus one, All squared -1 all over 12. And this comes out to you to over three. And so this random variable with a discreet uniform distribution has a mean of two and a variance of 2/3.

In this exercise, we are told that X has a continuous uniform distribution over the range 1.5 To 5.5. And for part A we are asked to find the mean variance and standard deviation of X. Now, for a uniform distribution, the mean or expected value is A plus B over to where A is the bottom of the range and be at the top of the range. So this is 1.5 plus 5.5, divided by two, Which is 3.5. The variance is given by b minus a squared over 12, And this is 1.25 Or that's more like 1.33. And the standard deviation is the square root of the variance, and this is Approximately 1.155. Next for part B, we are asked for the probability that X is less than 2.5. Now, first note that the density function Is going to be one divided by B -A. So in our case it's one quarter And that exists on the range 1.5 to 5.5. And so this probability Is simply 2.5 minus the bottom of the range times the height or times one quarter, And that comes out to one quarter. And then for part C, we are asked to find the cumulative distribution function. Now, for a continuous uniform distribution, the CDF is always given by this form, so it's zero for x is less than a, it's x minus a, divided by B minus A. For ex at least as big as a, but less than B, and it's one for access at least as big as B. So for this particular example, Going to be zero For X is less than 1.5, It's going to be x -1.5, divided by four for ex Is at least as big as 1.5, But smaller than 5.5, and then it's one for X Greater than or equal to 5.5.


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