Probability that the range includes the mean in a sample of $n=4$ from a normal distribution? If we select one random sample with 4 elements from a normal distribution, and we denote the minimum value among the sample with $a$, and denote the maximum value among the sample with $b$, what is the probability that distance between $a$ and $b$ (i.e. the range) includes the true population mean?
Could anyone help me how to solve this problem from the 2012 local-math-contest?
The short solution of this question: $7/8$.  
 A: I would solve the inverse: what is the probability that the range does not include the mean, then subtract that from 1.
So that is the sum of the 2 cases: a is greater than the mean; b is less than the mean.  Calculate the probability of these 2 cases (mutually exclusive) using the distribution of the min(max) from n=4, and subtract the 2 values (should be the same) from 1.
Instead of using the formula for order statistics you can also use the fact that the median and the mean are the same for the normal and therefore there is a 50% chance of each point being above(below) the mean, so what is the probability of all 4 values being greater(less) than the mean?
A: Here is an illustration of what @GregSnow suggested:

We start at the top and go down. At the top, there are two possibilities for the first value: 


*

*The first value is below the median.

*The first value is above the median.


Because the median and mean are the same in a normal distribution, there is a 50% chance that the first value lies below and a 50% chance that it lies above the median.
For the second value, we have again two possibilities: either it lies above or below the median. Again, the chances are 50/50.
We are only interested in the paths where either all 4 values fall below or above the median. These paths are marked with green arrows in the graphic. The other paths are omitted for clarity.
As we go down, we multiply the probabilities. At the bottom, we add the probabilities.
Can you solve it from here?
