# Thread: PLzzz Help!!!!!!!!!!!!!!-Skew?

1. ## PLzzz Help!!!!!!!!!!!!!!-Skew?

A random variable is uniformly distributed over[0,1].What is its skew?

Hi,

I've been moving in circles around this question for quite sometime now.

I would appreciate it someone would be kind enough to provide me with the full answer and some explanation please.

As you can see I've been provided with several leads and unfortunately I came up with some funny answer you can see in the post below by me with a similar heading.

I got sigma to be approximately 0.3 and cubed it to get 0.027

for E((x-1/2)^3=-0.5(using the equation above i.e integral)

final answer was -18.5 ?? :?: I 'm sure it is not correct

2. Originally Posted by KayPee
A random variable is uniformly distributed over[0,1].What is its skew?

Hi,

I've been moving in circles around this question for quite sometime now.

I would appreciate it someone would be kind enough to provide me with the full answer and some explanation please.

As you can see I've been provided with several leads and unfortunately I came up with some funny answer you can see in the post below by me with a similar heading.

I got sigma to be approximately 0.3 and cubed it to get 0.027

for E((x-1/2)^3=-0.5(using the equation above i.e integral)

final answer was -18.5 ?? :?: I 'm sure it is not correct
$\displaystyle E ((\frac{x-u^3}{o^3}))$$\displaystyle = \frac{E(x-u)^3)}{o^3} \displaystyle u=\frac{1}{2} \displaystyle o=\sqrt{\int 1 0 (x-1/2)^2 dx} (int 1 over 0) \displaystyle E((x-1/2^3)=\int 1 0 (x-1/2)^3 dx(int 1 over 0) Please excuse my lack of knowing the correct LaTex for the required symbols I hope you can interpret my work. 3. ## Thanks so much Originally Posted by TheMasterMind \displaystyle E ((\frac{x-u^3}{o^3}))$$\displaystyle = \frac{E(x-u)^3)}{o^3}$
$\displaystyle u=\frac{1}{2}$

$\displaystyle o=\sqrt{\int 1 0 (x-1/2)^2 dx}$ (int 1 over 0)

$\displaystyle E((x-1/2^3)=\int 1 0 (x-1/2)^3 dx$(int 1 over 0)

Please excuse my lack of knowing the correct LaTex for the required symbols I hope you can interpret my work.

Hello

never mind I'm also not good at using the latex .
I've got these steps above already.It is the final answer that I've struggled to reach.

4. Originally Posted by KayPee
Hello

never mind I'm also not good at using the latex .
I've got these steps above already.It is the final answer that I've struggled to reach.

5. To finish this off, the integral:

$\displaystyle \int_0^1 \left(x - \frac{1}{2}\right)^3 ~dx = \frac{1}{4}\left[\left(x - \frac{1}{2}\right)^4\right]^1_0$

$\displaystyle = \frac{1}{4}\left[\left(1 - \frac{1}{2}\right)^4 - \left(0 - \frac{1}{2}\right)^4\right]$

$\displaystyle = \frac{1}{4}\left[\left(\frac{1}{2}\right)^4 - \left(-\frac{1}{2}\right)^4\right]$

We know that for any even exponent b:

$\displaystyle (a)^b = (-a)^b$

So the inside of the brackets is zero, which makes the whole thing 0.