PMF Hypergeometric Distribution 2 ā Questions and Answers
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?
- 1 (Correct answer)
- 0.8
- 2
- 4
Correct answer: 1
E[X] = n Ć K/N = 5 Ć 4/20 = 5 Ć 0.2 = 1.
Question 2: What is the variance of a Hypergeometric distribution with parameters N, K, n?
- n Ć (K/N) Ć (1 - K/N) Ć (N-n)/(N-1) (Correct answer)
- n Ć p Ć (1-p)
- n Ć (K/N) Ć (1 - K/N)
- K(N-K)n / NĀ²
Correct answer: n Ć (K/N) Ć (1 - K/N) Ć (N-n)/(N-1)
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)?
- C(3,2)ĆC(12,0)/C(15,2) = 3/105 ā 0.029
- C(3,2)/C(15,2) ā 0.057
- (3/15) Ć (2/14) ā 0.029
- Both A and C are correct (Correct answer)
Correct answer: Both A and C are correct
P(X=2) = C(3,2)ĆC(12,0)/C(15,2) = 3/105 ā 0.029; equivalently (3/15)(2/14) ā 0.029 ā 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?
- It approaches 0, reducing the variance (Correct answer)
- It approaches 1, giving the Binomial variance
- It becomes negative
- It approaches N
Correct answer: It approaches 0, reducing the variance
As n ā N, (N-n)/(N-1) ā 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?
- Finite population correction factor (Correct answer)
- Normalizing constant
- Skewness correction
- Continuity correction
Correct answer: Finite population correction factor
The factor (N-n)/(N-1) is called the finite population correction factor, adjusting for sampling without replacement.
Question 6: A card is drawn from a deck of 52 without replacement for 5 draws. Which distribution models the number of aces drawn?
- Hypergeometric (N=52, K=4, n=5) (Correct answer)
- Binomial (n=5, p=4/52)
- Poisson (Ī» = 5Ć4/52)
- Geometric (p=4/52)
Correct answer: Hypergeometric (N=52, K=4, n=5)
Drawing cards without replacement from a finite deck requires the Hypergeometric distribution with N=52, K=4 aces, and n=5 draws.
PMF Hypergeometric Distribution 2
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?