PMF Hypergeometric Distribution 2

Question 1

A lot of 20 items contains 4 defectives. A sample of 5 is drawn without replacement. What is E[X], the expected number of defectives?

Accepted Answer

1

Answer

E[X] = n × K/N = 5 × 4/20 = 5 × 0.2 = 1.

Question 2

What is the variance of a Hypergeometric distribution with parameters N, K, n?

Accepted Answer

n × (K/N) × (1 - K/N) × (N-n)/(N-1)

Answer

The Hypergeometric variance includes the finite population correction factor (N-n)/(N-1) compared to the Binomial variance.

Question 3

In quality control, 3 defective and 12 good items are in a batch of 15. Two items are drawn without replacement. What is P(both defective)?

Accepted Answer

Both A and C are correct

Answer

P(X=2) = C(3,2)×C(12,0)/C(15,2) = 3/105 ≈ 0.029; equivalently (3/15)(2/14) ≈ 0.029 — both give the same answer.

Question 4

The finite population correction factor in the Hypergeometric variance is (N-n)/(N-1). What happens to this factor as n approaches N?

Accepted Answer

It approaches 0, reducing the variance

Answer

As n → N, (N-n)/(N-1) → 0, meaning the variance shrinks to zero because when you sample the entire population, there is no randomness.

Question 5

Which term describes (N-n)/(N-1) in the Hypergeometric variance formula?

Accepted Answer

Finite population correction factor

Answer

The factor (N-n)/(N-1) is called the finite population correction factor, adjusting for sampling without replacement.

PMF - Probability Mass Function Practice Test