Find the Speedcuber's times (Codewars)

'''

Speedcubing is the hobby involving solving a variety of twisty puzzles, the most famous being the 3x3x3 puzzle or Rubik's Cube, as quickly as possible.

In the majority of Speedcubing competitions, a Cuber solves a scrambled cube 5 times, and their result is found by taking the average of the middle 3 solves (ie. the slowest and fastest times are disregarded, and an average is taken of the remaining times).

In some competitions, the unlikely event of a tie situation is resolved by comparing Cuber's fastest times.

Write a function cube_times(times) that, given an array of floats times returns an array / tuple with the Cuber's result (to 2 decimal places) AND their fastest solve.

For example:

cube_times([9.5, 7.6, 11.4, 10.5, 8.1]) = (9.37, 7.6)
# Because (9.5 + 10.5 + 8.1) / 3 = 9.37 and 7.6 was the fastest solve.
Note: times will always be a valid array of 5 positive floats (representing seconds)

'''


def cube_times(times):
    max = times[0]
    min = times[1]
    for i in range(1, len(times)):
        if times[i] > max:
            max = times[i]
        if times[i] < min:
            min = times[i]

    # I've found the min and max elements of the list
    mediumElements = []
    for i in range(0, len(times)):
        if times[i] != min and times[i] != max:
            mediumElements.append(times[i])

    # My list with medium elements is ready
    sum = 0
    for i in range(0, len(mediumElements)):
        sum += mediumElements[i]
    average = sum / 3
    # Rounding average in 2 digits
    average = round(100 * average) / 100

    # New list = returned list
    returnedList = []
    returnedList.append(average)
    returnedList.append(min)

    # Convert the list to a tuple
    returnedTuple = tuple(returnedList)

    return returnedTuple


# MAIN FUNCTION
print()
print("cube_times([10.5, 9.2, 8.4, 11.3, 12.6]) = " + str(cube_times([10.5, 9.2, 8.4, 11.3, 12.6])))