• Dec 16th 2009, 08:42 PM
mokash34
I've been asked by a friend to help with a probability assignment... I don't know how to solve it so I figured I would post it here... thanks for the help!

A fair die is tossed 40 times. Use the given equation (in the attachment) to determine the probability of the total number of dots being between 130 and 150.

Repeat using the central limit theorem.

• Dec 16th 2009, 10:01 PM
matheagle
1 The CLT is more accurate than the attachment (Cheyshev's)
2 Between 130, 150 is vague. You need to know if you're including the endpoints.

This is the sum of 40 i.i.d. random variables with mean 3.5.
As I guessed the mean of the sum is 40 times 3.5 which is.....140.
• Dec 16th 2009, 10:22 PM
mokash34
okay sorry, i typed it word for word but i see now that it is vague. let us assume that 130 and 150 are included. what would the probability be using the given equation (which he is required to use for the first part)

and then, using Central limit theorem, find the probability...

thanks again!!!
• Dec 16th 2009, 10:24 PM
matheagle
$\displaystyle \sum_{i=1}^{40}X_i\approx N(\mu ,\sigma^2)$ via the CLT.

Where $\displaystyle \mu$ is the mean of the sum =(40)(3.5)=140.

Likewise $\displaystyle \sigma^2$ is the variance of the sum, which is 40 times the variance of one X.

So compute V(X) for a random variable with 1/6 probability on 1,2,3,4,5,6.
• Dec 16th 2009, 10:40 PM
mokash34
I just saw the rest of your post... which answers most of my questions... i'll post in a minute with another question... i'm sure of it haha
i've attached what the given textbook has for the CLT...
• Dec 16th 2009, 10:53 PM
mokash34
Var(X)=E[X^2]-E[X]^2 .... right? or is this a different equation
• Dec 16th 2009, 10:55 PM
matheagle
Quote:

Originally Posted by mokash34
Var(X)=E[X^2]-E[X]^2 .... right? or is this a different equation

and $\displaystyle M_n$ is just the sample mean, $\displaystyle \bar X$

I thought $\displaystyle M_n$ was the sum of the 40 random variables.

It's the sum divided by 40.
• Dec 16th 2009, 11:00 PM
mokash34
hmm... i think i'm in over my head... i don't seem to understand.

this is one of three problems and i though it looked the easiest...

so there will be 2 answers... one from CLT and one from the equation in my first post...

is the CLT the equation you gave me? or is it the second picture i posted?

are the answers going to be different?

i understand if you feel like this is a waste of your time, thanks nonetheless
• Dec 17th 2009, 07:51 AM
matheagle
Did you calculate the variance of one of these tosses?
The CLT is much more accurate than chebyshev's.
That's the point of this.
Now it's not a binomial, in that case we add and subtract .5.
We are approximating a discrete rv (number of spots) with a continuous rv (normal)
so we should throw in some correction factor, but I doubt people will do that.
That's why you should know if you're including the endpoints.
• Dec 17th 2009, 07:54 AM
matheagle
Quote:

Originally Posted by mokash34
Var(X)=E[X^2]-E[X]^2 .... right? or is this a different equation

DO this for the distribution above.