Statistics question involving the beta distribution?

The mean is alpha / (alpha + beta) according to this reference.

(Be careful here, because personally I really don't remember ....)

Then the mean is 2 / (2+2) = 1/2

http://en.wikipedia.org/wiki/Beta_distribution

Therefore the expected value over 52 weeks is 52 * 1/2 = $26? Am I getting that right?

The variance is 1/20 = a b / [(a+b)^2 (a+b+1)]

I think this is where you use the Central Limit Theorem

to say you can approximate it by a normal.

Normalized to std. dev. 1 and mean 0, your random variables are yi = sqrt(20)(xi - 1/2)

Then 1/sqrt(52) sum(yi) ~ N(0,1)

The number you want to look at compared to the N(0,1) curve is when the sum of xi is equal to 60

then the sum of yi is 30 sqrt(20)

so the number to compare is 30 sqrt(20)/sqrt(52)

I think that might be right.

The cost it is less than $60 must be 1.00000 or slightly less.

The pattern length is the bigger, the greater advantageous. yet you are going to be able to found out it from SD, and F. in the event that they're fantastic, your pattern length isn't a difficulty. T-attempt will inform you that your hypothesis is widespread or no longer with an significant of ninety 5% and up. with the aid of fact it is going to inform you that the parameter isn't far remote from 0.if so your R*2 would be additionally lifeless. shop your 10 greenbacks.it incredibly is my excitement to respond to a question like this.