Program to accept numbers, then calculate the mean, median, and range
On top of Thomas Ward answer, here is a short and important comment:

More beautiful functions

Your functions are performing some mathematical computations and handling the formatting to show the result to the 2 users. These correspond to 2 concerns that could be separated.

Also, your functions rely on the list_one global variable. It would be clearer to have this provided to the functions using an argument.

You'd have something like:

def mean(lst):
    return sum(lst) / len(lst)

def range_(lst):
    return lst[-1] - lst[0]

def median(lst):
    length = len(lst)
    if length % 2 == 0:
        return sum(lst[length//2-1:length//2+1]) / 2
    else:
        return lst[length//2]


...
            print('Mean: ' + str(mean(list_one)))
            print('Median: ' + str(median(list_one)))
            print('Range: ' + str(range_(list_one)))


These are easier to understand and to reason about. They are also easier to test. This is an interesting exercise that you can try to practice.

Small improvements in median

The expression length//2 is used in multiple places which makes the code tedious to read and less efficient than it could be.
Using a temporary variable makes things clearer:

def median(lst):
    length = len(lst)
    mid = length//2
    if length % 2 == 0:
        return sum(lst[mid-1:mid+1]) / 2
    else:
        return lst[mid]


From here, I think it's pretty good already. Let's see what can be still be done for the sake of learning new techniques.

When you want to compute both the quotient and the remainder of the division, you can use "/" and "%" just like you did. Another option is to use the divmod builtin which returns both directly:

def median(lst):
    length = len(lst)
    mid, rem = divmod(length, 2)
    if rem == 0:
        return sum(lst[mid-1:mid+1]) / 2
    else:
        return lst[mid]


The operation sum(lst[mid-1:mid+1]) / 2 looks a lot like things we've just somewhere else: we compute the sum of a list and we divide by the number of elements which is happens to be 2.

We could reuse our mean function here.

def median(lst):
    length = len(lst)
    mid, rem = divmod(length, 2)
    if rem == 0:
        return mean(lst[mid-1:mid+1])
    else:
        return lst[mid]


To go further, we could use the ternary operator to have a single return value.
We could also use the fact that in a boolean context, non-zero integers are equivalent to True and zero is equivalent to false.
This would lead to the following code which is, from my point of view, not really better:

def median(lst):
    length = len(lst)
    mid, rem = divmod(length, 2)
    return lst[mid] if rem else mean(lst[mid-1:mid+1])