Given a continuous uniform distribution, show that (a)...

Transcript

00:01 Hi, i'm david and i'm here to help you answer your question.

00:04 Now let me bring up your question here.

00:06 In this question we're going to discuss about the continuous uniform distribution.

00:11 Let me remind you that if we have the x followed by the uniform from a to b and then the density of the x, it will equal to 1 over p minus a and for the x between the interval from a to b.

00:31 B.

00:33 And then in this question we want to find the mean and the standard and the variance on the random variable x.

00:39 Now for the part a, when you find the mule which is equal to a of the x.

00:46 By formula, the mean of the continuous random variable to the integral x, fx, the x from minus infinity to infinity.

00:56 Now because x between a and b, so this integral from a to b, x, and now fx equals.

01:03 2 1 over b minus a the x we have one of a b minus a constant coming outside untidy derivative of the x equal to the x square over 2 and this is an evaluate from a to p now if we compute it we should have in saugabe the b square minus a square then we get equal to on the top b square equal to b minus a square equal to b minus a times from the b plus a over the b minus a.

01:38 And then at the end we should get equal to we can cancel this out.

01:43 And then we don't forget we have a 2 in front here.

01:47 So we should get equal to the b plus a over 2.

01:51 Or we can write this down to a plus b over 2.

01:55 And this exactly the one we have to prove here for the a.

01:59 For the b, once you show that the sigma square equal to the b minus a, over 12.

02:05 To find the variance, we need to find the e of the x square first...

Question

Given a continuous uniform distribution, show that (a) $\mu=\frac{A+B}{2}$ and (b) $\sigma^{2}=\frac{(B-A)^{2}}{12}$.

Question

Given a continuous uniform distribution, show that (a) $\mu=\frac{A+B}{2}$ and (b) $\sigma^{2}=\frac{(B-A)^{2}}{12}$.

Please give Ace some feedback