Hypergeometric Mean and Variance

Problem

For sampling draws items without replacement from a population of pop items containing successes successes, return [mean, variance] of the hypergeometric count. mean = draws * K / N and variance = draws * (K / N) * (1 - K / N) * (pop - draws) / (pop - 1), where K / N = successes / pop. Return [0.0, 0.0] for invalid counts and variance 0.0 when pop == 1.

Hint

This is the binomial variance reduced by the finite population correction factor.

Test cases

case 1

in: [50,5,10]

expected: [1,0.7346938775510204]

case 2

in: [10,3,4]

expected: [1.2,0.56]

case 3

in: [52,4,5]

expected: [0.38461538461538464,0.32718412808910546]

case 4

in: [5,5,5]

expected: [5,0]

case 5

in: [10,3,11]

expected: [0,0]

Starter codePython

def hypergeometric_mean_variance(pop, successes, draws):
    # Your code here
    pass

5 test cases · run your code against them in the editor