Left of center?

[BMAD]

Allow for a random set of 32 numbers to be chosen from 0 to 100.

pair and average the 32 numbers in the order they appear ([1,2]; [3,4],...) to make 16 numbers

pair and average each of the 16 averaged numbers to make 8 new numbers

continue in this fashion until you have only 1 number.

What is the probability that this number will be be in the 40's?

Edited October 8, 2015 by BMAD

[k-man]

I'm getting the answer of about 47.32%

Here is how I'm getting it. 

Standard deviation (SD) of the entire population [0-100] is approximately 29.3. We take a sample of size n=32 from the population and calculate the mean. The standard error will be about SE=SD/sqrt(n)  ~ 5.18. This means that the mean of a sample will follow a normal distribution with the mean of 50 and standard deviation of ~5.18. Half of sample means will be above 50 and half will be below. To be in the 40s the sample mean needs to be above the 40 mark. Z score of the 40 mark is -1.93 and the p-value is 0.0268, which means that there is 2.68% chance that the mean of a sample will be below 40 resulting in the answer of .5 - .0268 = .4732

[jasen]

each number must be an unique number or not?

[BMAD]

each number must be an unique number or not?

not