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Suppose X is a continuous uniform random variable over theinterval (a,b) where a =2 and b=5.Find E[X^2]...

Question

Suppose X is a continuous uniform random variable over theinterval (a,b) where a =2 and b=5.Find E[X^2]

Suppose X is a continuous uniform random variable over the interval (a,b) where a =2 and b=5.Find E[X^2]



Answers

A continuous random variable $X$ is said to have a uniform distribution on the interval $[a, b]$ if the PDF has the $$f(x)=\left\{\begin{array}{ll}\frac{1}{b-a}, & \text { if } a \leq x \leq b \\0, & \text { otherwise }\end{array}\right.$$ (a) Find the probability that the value of $X$ is closer to $a$ than it is to $b$ (b) Find the expected value of $X$ (c) Find the CDF of $X$

The key concept that we will be focusing on in this particular problem is graphing a uniform distribution function. So in this problem we are given A continuous random variable X. That's uniformly distributed on the interval from 5 to 12. And what we want to do is you want to graph the density function for X. And so Given that were given the X's uniformly distributed from on this interval from 5 to 12, we know that a uniform distribution, the probability density function of uniformly distributed random variable is like this, where we take one over, B minus A for an interval from A to B. And so in this particular pro um problem, since we have this uniform distribution and this interval here, we can just plug this in. And so we have that for this, our random variable, we have Uniform distribution density function that's equal to 17 For the interval of 5- 12. And so we can graph this. So we can Make this 1/7 over here and we can make a horizontal line. And here we have our probability density function for our random variable X, Where we just have a horizontal straight line at 1/7 from the interval 5 to 12, and we have solved our problem.

The key concept that we will be focusing on in this particular problem is the graphing of a uniformly distributed random variable. And so we are in this problem we are given this continuous random variable X. That's uniformly distributed on the interval from negative three 23 And what we want to do in this problem is we want to graph F of X or the density function for this random variable X. And so given the X's and uniformly distributed, we have the probability density function of a uniformly distributed random variable is such that You have one over b minus A. Where we have that this is on an interval from A to be. And so given that we have a uniformly distributed Random variable from this interval from -323, we can find a probability density function. So we have that 1/3 minus negative three for the interval Of -3, 2, 3. And so we can simplify this and we find that we have 16 for the interval -32, 3. And so we have that this is our Probability density function on the interval from -32, 3. And so now we can graph it over here and so we can Mark This and we could say that this represents 16 over here. And yeah, we can just draw A line from negative three 23 Mhm. Mhm. Yeah. And 16 over here. And so now we have our probability density function graft. And so we have that the probably ability density function is a horizontal line at 1/6 from the from negative X equals negative 32 X equals three. And so we have graft our random variable and we are done.

This question discusses proving the parameters for uniform distribution. Now we know that a uniform distribution has all of the values Given by one over B -A. Now we want to find A. You and in general when you want to find mu we can actually take the expectation of some value here, which we're going to call X. And that will allow us to find me right, it's expected values an average. Um So here we're actually going to take the integral from A to B. Of our distribution, which is you which is given by this value here. So this is going to be one over b minus A. And this is going to be a function of X, let's say D X. Now this is all a constant because our only variable here is X. So this just becomes X squared over to Times Our one Over B -A. Yeah, here in front. Now we can plug in for X here A. And B. And this gives us you plug in the B. And we plug in the A. And this will give us our values for a meal here. Now we can do the same thing but we're going to find prove the variance and the variance which I'm gonna general here by V is given by this following formula here, which is the expectation of X squared minus the expectation of X, that is quantity squared. Now we are resolved for the expectation here, right? This is the expectation of the value here. Now we can actually do the integral again plugging in from A to B. But this time we're actually going to go from one, this is just the value of the distribution here and this time we're gonna plug in an X squared here and integrate this out. Once you integrate this out, that will allow you to solve, you have now solved for this term here and then once you take this term and you square, that allows you to plug in for this value here. Once you plug in both of those values, that allows you to solve, and you should get the given answer for the variance of the uniform distribution Yeah.

You know this problem, we are told that we have a uniform distribution on the interval 15 To 5.5. And there are a few things that we would like to find now, in general, if we have a uniform distribution on the interval A to B, we know that are mean Is equal to 1/2 of a plus b. Okay, We know that our variants is able to 1 12 of b minus a squared. Yeah, I mean our standard deviation is always the square two variants. That doesn't just apply for a uniform distribution, that's for anything. So, here or mean is one half of 1.5, It was 5.5, Which is 3.5 are variants is 1/12 Times 5, 5 -1, 5 squared, Which is 4/3 are variances 4/3 And then standard deviation is always the Square two variants. And So that's the square to 4/3, which is two or 3/3. And so that is our standard deviation L and B On B. We would like to find the probability that X is less than 2.5. Mhm. In order to find this probability this would be equal to the integral from 1.5 to 2.5 because we know that can't be less than than 1.5 because that's our lower bound on the interval uh 1/4, because that was the length of the interval for our uniform distribution, So this is 1/4 of acts evaluated From X is 1, 5, 2.5, Which is 1 4th times to 5 -1.5. Yeah. Which is 1/4. And so the probability that actually is less than 25. There's 1/4. Yeah. Now, lastly we would like to find the cumulative distribution function and we know that for a uniform distribution the cumulative distribution function is april zero. Mhm. The facts is less than night. And so this means f of X is equal to zero. If X is less than 1.5 before we take care of them, I'll take care of the bottom round. It's one if X is greater than B. This means here it's one of X is greater than 5.5. Yeah. And that affects us between A and B. It's x minus A over b minus a. This is A A is less than or equal to X less than full of meat. And this will be X -1.5 over four. All right. If 1.5 is less than or equal to X Unless they're able to 5.5. And so there's a cumulative distribution function


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