Page 1 of 49
WGU PUBH 6002 QUIZ 2 – 2026
BIOSTATISTICS PUBLIC HEALTH COMPLETE
(115) CURRENT TESTING QUESTIONS AND
CORRECT ANSWERS WITH DAETAILED
RATIONALES.
PUBH
Prepare for the WGU PubH 6002 Quiz 2 – Biostatistics Public Health with
focused review materials covering probability distributions, sampling
distributions, standard error, z-scores, the Central Limit Theorem,
measures of association, p-values, and Type I/II errors. Designed to build
confidence in analyzing data, interpreting results, and making evidence-
based public health decisions. Suitable for WGU public health students
preparing for Quiz 3 in biostatistics and related examinations.
MULTIPLE CHOICE.
SECTION 1: PROBABILITY & PROBABILITY DISTRIBUTIONS
(Questions 1–25)
1. What is the probability that both the father and the mother are
current smokers if the probability that the mother is a current
smoker is 0.4 and the probability that the father is a current
smoker is 0.5, assuming independence?
A. 0.10
B. 0.20
C. 0.30
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D. 0.50
E. 0.90
Correct answer: B. 0.20
Rationale: If the smoking habits of parents are independent, the
probability that both events occur is the product of their individual
probabilities: P(Mother AND Father) = P(Mother) × P(Father) = 0.4 ×
0.5 = 0.20.
2. What is the probability that the father is a current smoker
given that the mother is not a current smoker?
A. 0.3
B. 0.4
C. 0.5
D. 0.6
E. 0.7
Correct answer: C. 0.5
Rationale: If the smoking habits of parents are independent, the
probability that the father is a smoker is unaffected by the mother's
smoking status. Therefore, P(Father | Mother not) = P(Father) = 0.5.
3. Can two events be both mutually exclusive and independent?
A. Yes, always
B. Yes, but only if probabilities are equal
C. No, never
D. Yes, only if one event has probability 0
E. Cannot be determined
Correct answer: C. No, never
Rationale: If A and B are nontrivial events, they cannot be both
mutually exclusive and independent. Being mutually exclusive
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(P(A∩B)=0) is a lack of independence unless one event has a
probability of zero.
4. The formula for the probability of the union of two events A
and B is:
A. P(A) + P(B)
B. P(A) × P(B)
C. P(A) + P(B) − P(A∩B)
D. P(A) + P(B) − P(A∪B)
E. P(A) × P(B) + P(A∩B)
Correct answer: C. P(A) + P(B) − P(A∩B)
Rationale: The union of events A and B includes outcomes in A, B,
or both. To avoid double-counting the intersection, subtract P(A∩B).
This is the general addition rule.
5. What is the probability that a visit to a primary care
physician's office results in lab work or referral to a specialist if
P(lab)=0.40, P(referral)=0.30, and P(both)=0.05?
A. 0.12
B. 0.35
C. 0.65
D. 0.70
E. 0.75
Correct answer: C. 0.65
Rationale: P(lab ∪ referral) = P(lab) + P(referral) − P(lab ∩ referral) =
0.40 + 0.30 − 0.05 = 0.65.
6. The probability of the intersection of two events A and B is
expressed as:
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A. P(A) + P(B)
B. P(A) × P(B)
C. P(A|B) × P(B)
D. P(A) − P(B)
E. P(A∪B) − P(A)
Correct answer: C. P(A|B) × P(B)
Rationale: The multiplication rule states that P(A∩B) = P(A|B) × P(B).
For independent events, this reduces to P(A) × P(B).
7. The normal distribution is characterized by all of the following
EXCEPT:
A. It is symmetric
B. It is bell-shaped
C. The mean equals the median
D. It is bounded at zero
E. It is defined by its mean and standard deviation
Correct answer: D. It is bounded at zero
Rationale: The normal distribution is unbounded, meaning it
extends from −∞ to +∞. It is symmetric, bell-shaped, has mean =
median = mode, and is fully defined by its mean and standard
deviation.
8. A binomial distribution is appropriate when:
A. The outcome is continuous
B. There are two possible outcomes and trials are independent
C. The sample size is small
D. The population is infinite
E. The outcome is ordinal
WGU PUBH 6002 QUIZ 2 – 2026
BIOSTATISTICS PUBLIC HEALTH COMPLETE
(115) CURRENT TESTING QUESTIONS AND
CORRECT ANSWERS WITH DAETAILED
RATIONALES.
PUBH
Prepare for the WGU PubH 6002 Quiz 2 – Biostatistics Public Health with
focused review materials covering probability distributions, sampling
distributions, standard error, z-scores, the Central Limit Theorem,
measures of association, p-values, and Type I/II errors. Designed to build
confidence in analyzing data, interpreting results, and making evidence-
based public health decisions. Suitable for WGU public health students
preparing for Quiz 3 in biostatistics and related examinations.
MULTIPLE CHOICE.
SECTION 1: PROBABILITY & PROBABILITY DISTRIBUTIONS
(Questions 1–25)
1. What is the probability that both the father and the mother are
current smokers if the probability that the mother is a current
smoker is 0.4 and the probability that the father is a current
smoker is 0.5, assuming independence?
A. 0.10
B. 0.20
C. 0.30
, Page 2 of 49
D. 0.50
E. 0.90
Correct answer: B. 0.20
Rationale: If the smoking habits of parents are independent, the
probability that both events occur is the product of their individual
probabilities: P(Mother AND Father) = P(Mother) × P(Father) = 0.4 ×
0.5 = 0.20.
2. What is the probability that the father is a current smoker
given that the mother is not a current smoker?
A. 0.3
B. 0.4
C. 0.5
D. 0.6
E. 0.7
Correct answer: C. 0.5
Rationale: If the smoking habits of parents are independent, the
probability that the father is a smoker is unaffected by the mother's
smoking status. Therefore, P(Father | Mother not) = P(Father) = 0.5.
3. Can two events be both mutually exclusive and independent?
A. Yes, always
B. Yes, but only if probabilities are equal
C. No, never
D. Yes, only if one event has probability 0
E. Cannot be determined
Correct answer: C. No, never
Rationale: If A and B are nontrivial events, they cannot be both
mutually exclusive and independent. Being mutually exclusive
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(P(A∩B)=0) is a lack of independence unless one event has a
probability of zero.
4. The formula for the probability of the union of two events A
and B is:
A. P(A) + P(B)
B. P(A) × P(B)
C. P(A) + P(B) − P(A∩B)
D. P(A) + P(B) − P(A∪B)
E. P(A) × P(B) + P(A∩B)
Correct answer: C. P(A) + P(B) − P(A∩B)
Rationale: The union of events A and B includes outcomes in A, B,
or both. To avoid double-counting the intersection, subtract P(A∩B).
This is the general addition rule.
5. What is the probability that a visit to a primary care
physician's office results in lab work or referral to a specialist if
P(lab)=0.40, P(referral)=0.30, and P(both)=0.05?
A. 0.12
B. 0.35
C. 0.65
D. 0.70
E. 0.75
Correct answer: C. 0.65
Rationale: P(lab ∪ referral) = P(lab) + P(referral) − P(lab ∩ referral) =
0.40 + 0.30 − 0.05 = 0.65.
6. The probability of the intersection of two events A and B is
expressed as:
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A. P(A) + P(B)
B. P(A) × P(B)
C. P(A|B) × P(B)
D. P(A) − P(B)
E. P(A∪B) − P(A)
Correct answer: C. P(A|B) × P(B)
Rationale: The multiplication rule states that P(A∩B) = P(A|B) × P(B).
For independent events, this reduces to P(A) × P(B).
7. The normal distribution is characterized by all of the following
EXCEPT:
A. It is symmetric
B. It is bell-shaped
C. The mean equals the median
D. It is bounded at zero
E. It is defined by its mean and standard deviation
Correct answer: D. It is bounded at zero
Rationale: The normal distribution is unbounded, meaning it
extends from −∞ to +∞. It is symmetric, bell-shaped, has mean =
median = mode, and is fully defined by its mean and standard
deviation.
8. A binomial distribution is appropriate when:
A. The outcome is continuous
B. There are two possible outcomes and trials are independent
C. The sample size is small
D. The population is infinite
E. The outcome is ordinal
