Returning nth lowest number without using built-in functions - Python

I am working on the following: Write a function that returns nth lowest number of a list (or iterable in general). Return the lowest if second argument not specified Note that if a list contains duplicates, they should be handled before determining nth lowest

I got the following to work fine:

numbers = [8,9, 1,300,5, 54, 54]
def nth_lowest(b, n='N/A'):
    nums = set(b)
    if n == 'N/A':
        return min(b)
    else:
        nums = sorted(nums)
        return nums[n-1]

print(nth_lowest(numbers))
print(nth_lowest('ananasgnasgzynrmas', 6))
print(nth_lowest(numbers, 4))

But I wanted to try to write it without using those built-in SET or MIN functions. Here is what I have:

numbers = [8,9, 1,300,5, 54, 54]
def nth_lowest(b, n='N/A'):
    nums = []
    new_list = []
    for i in b:  # instead of using set()
        if i not in nums:
            nums.append(i)
    nums = sorted(nums)
    if n == 'N/A':
        while b: # instead of using min()
            minimum = b[0]
            for x in b:
                if x < minimum:
                    minimum = x
            new_list.append(minimum)
            b.remove(minimum)
        return new_list
    else:
        return nums[n-1]

print(nth_lowest(numbers))
print(nth_lowest('ananasgnasgzynrmas', 6))
print(nth_lowest(numbers, 4))

but it is error'ing out on the following:

return nums[n-1]
IndexError: list index out of range

Any ideas?