Hello

Here's a question:

Let X have a continuous uniform distribution on the interval [a,b]. Find the density function for this dristribution and then verify that $\displaystyle \mu = \frac{(a + b)}{2}$ and $\displaystyle \sigma^2 = \frac{(b - a)^2}{12}$ .

ok

So here's what I "think" I should do ...

Let f(x) = 1/(b-a) = 1 assuming that a=0 and b=1

For a <= x <= b

I know I need to integrate, but I'm not sure how to set up the problem

$\displaystyle \mu = E(x) = \int xf(x)dx$ where f(x) = 1

then

$\displaystyle \int xdx = \frac{1}{2}x^2$ which equals 1/2 when pluging in [a.b]

Verifying ...

$\displaystyle \mu = \frac {(a+b)}{2} $ you get $\displaystyle \mu = \frac {(0+1)}{2} = \frac {1}{2}$

and then we have $\displaystyle \sigma^2$

$\displaystyle \sigma^2 = E((x-\mu)^2) = \int (x-\mu)^2f(x)dx$

so ...

$\displaystyle \int (x-\frac{1}{2})^2dx = \int x^2-x+\frac{1}{4}dx = \frac{x^3}{3}-\frac{x^2}{2}+\frac{x}{4}$

When you plug in the interval [0,1] you get $\displaystyle \frac{1}{12}$

verifying $\displaystyle \mu^2$ You also get $\displaystyle \frac{1}{12}$

Is this correct? wow, that was long, sorry, I just tried to do it myself without any help at first. Be nice to me, it's my first post and it's also my birthday! WOOHOO

Thanks for all the help