Anyone have a weird affection for Statistics?

If so, I have a homework problem I could use some help with.

It goes like this...

Let X-bar be the mean of a random sample of size n=10 from a distribution with p.d.f. f(x) = 6x(1-x), for 0 < x < 1. Find the mean and variance of X-bar.

The answers are 1/2 and 1/200. How?

Thanks for any help!

berkeley you.
That was just for making me remember statistics. I really hate statistics.

Are you sure it is 1/200, and not 1/20? I keep getting .05 for the variance, but I was never too great at this.

In any case, this is a pretty easy one. For a continuous variable, just integrate x*f(x) for the mean.

Int(x(f(x))) from 0 to 1. I like to make it as easy as possible for me, and break it into the base parts. So instead of x(6x(1-x)), I get x(6x-6x^2), which then gets me 6x^2-6x^3. Integrate this from 0 to 1, and you get 2x^3-1.5x^4, which from 0 to 1 is 2-1.5=.5

The variance is calculated using the formula: INT (x^2)*f(x) dx over domain - (mean)^2.

So we do Int(x^2(6x-6x^2))=Int(6x^3-6x^4) from 0 to 1, which gives us 3/2-6/5, which gives us 3/10. Subtract mean^2, we get 3/10-1/4=1/20.

Is this easy enough to read? If need be, I can make a google document that is easier to follow. Not sure on the variance thing though, I hope that was a typo on your end or in the book.

Also, I assume you know the calculus--do you know the calculus?

Last question, are you allowed a calculator, and if so, what kind?

mtn wrote: So instead of x(6x(1-x))

The pdf is f(x) = 6x(1-x). How are you getting x(6x(1-x))?

according to the book, mu = 1/2 and I am not totally sure why.

I approached it as 6x - 6x^2. Integrating each of those I got 6(x^2/2) - 6(x^3/3) - mu^2

I reduced that to 3x^2 - 2x^3 - 1/4 Evaluating from 0 to 1 I got: 3 - 2 - 1/4 = .75 I thought you divide that by 10 (because of the sample size) which equals .075. That is the deviation but the variance is deviation squared = .005625.

To answer your other question, I am very rusty on calculus. We are allowed any type of calculator.

Thanks for your help!

I am not even sure if I am in the right ballpark.

CamaroKeith wrote:

mtn wrote: So instead of x(6x(1-x))

The pdf is f(x) = 6x(1-x). How are you getting x(6x(1-x))?

Bolded text added to this post for clarity.

mtn wrote: [To find the mean of a pdf] for a continuous variable , just integrate x*f(x) for the mean.

I'm going to repeat that to make sure you got it, since I usually needed to be told things over and over:
To find the mean of a pdf for a continuous variable, just integrate X*f(X) over the domain.

Also, if you are allowed any calculator and are "very rusty on calculus", I highly recommend that you get a TI-89. It will do the integration for you. It is going to save you a ton of time on the exams.

EDIT: TI-83 will also do integration for you, but it is a little more cumbersome.

A couple questions:

1. Are you sure the variance is 1/200, and not 1/20?
2. Do you want me to put this into a google document where it will be easier to read and follow?

My dad was a statistical engineer. That stuff may as well be swahili to me.

Lesley wrote: My dad was a statistical engineer. That stuff may as well be swahili to me.

"In mathematics you don't understand things. You just get used to them."
~Johann von Neumann

I just finished my BS in Math with a minor in Economics, and this quote is entirely accurate.

Statistically speaking, 9 out of 10 people enjoy gang rape. Hope that helps.

87% of statistics are made up on the spot.

Statistics are over rated

mtn wrote:

Lesley wrote: My dad was a statistical engineer. That stuff may as well be swahili to me.

"In mathematics you don't understand things. You just get used to them." ~Johann von Neumann I just finished my BS in Math with a minor in Economics, and this quote is entirely accurate.

How true!

I just punch the numbers into Minitab for my Design of Experiments.

there are liars

damn liars

and statisticians

mtn: Thanks for your help