PMF Hypergeometric Distribution 3

Question 1

For a Hypergeometric distribution, what is the support of X (the number of successes)?

Accepted Answer

max(0, n+K-N) ≤ X ≤ min(n, K)

Answer

X ranges from max(0, n+K-N) to min(n, K) because you cannot draw more successes than available (K) or than sampled (n), and must draw enough to fill the sample.

Question 2

A hypergeometric scenario has N=10, K=6, n=4. What is the maximum possible value of X?

Accepted Answer

4

Answer

The maximum is min(n, K) = min(4, 6) = 4 since you can draw at most all 4 sampled items as successes.

Question 3

How does the Hypergeometric distribution compare to the Binomial in terms of variance?

Accepted Answer

Hypergeometric has smaller variance due to the finite population correction

Answer

The Hypergeometric variance is always less than or equal to the Binomial variance because the finite population correction factor (N-n)/(N-1) ≤ 1.

Question 4

In a committee of 10 (6 men, 4 women), 3 are randomly selected. What distribution models the number of women selected?

Accepted Answer

Hypergeometric (N=10, K=4, n=3)

Answer

Selecting from a finite group without replacement uses the Hypergeometric distribution with N=10 total, K=4 women, n=3 selected.

Question 5

For the committee example (N=10, K=4, n=3), what is P(X=2)?

Accepted Answer

C(4,2)×C(6,1)/C(10,3) = 36/120 = 0.3

Answer

P(X=2) = C(4,2)×C(6,1)/C(10,3) = 6×6/120 = 36/120 = 0.3.

PMF - Probability Mass Function Practice Test

PMF - Probability Mass Function Practice Test

PMF Hypergeometric Distribution 3