WESTERN GOVERNORS UNIVERSITY
WGU C955 Applied Probability and Statistics
Final Exam (New 2026/2027 Update)
Questions and Verified Answers | 100% Correct | Graded A
2026/2027 Edition - Official Exam 2026/2027
75 80% N/A
QUESTIONS PASSING SCORE RECERTIFICATION
TABLE OF CONTENTS
Section 1 Basic Probability Concepts Q1-Q15
Section 2 Descriptive Statistics Q16-Q30
Section 3 Inferential Statistics Q31-Q45
Section 4 Probability Distributions Q46-Q60
Section 5 Data Analysis and Interpretation Q61-Q75
Instructions: Select the single best answer for each question. This exam is designed for WGU C955 Applied Probability and
WGU C955 Applied Probability and Statistics - 2026/2027 | Passing Score: 80% | Page 1 of 39
, SECTION 1 | Basic Probability Concepts | Q1-Q15 | WGU C955 Applied Probability and Statistics 2026/2027
Q1 Question 1 of 75
A quality control analyst at a manufacturing plant inspects a batch of 500 circuit boards and
finds that 15 are defective. If one circuit board is selected at random from this batch, the
probability that it is NOT defective is which of the following?
A. 0.030
B. 0.970
C. 0.950
D. 0.985
Correct Answer: B
Rationale:
The probability of a defective board is 15/500 = 0.030, so the probability of not being defective is 1 - 0.030 =
0.970. Option A is the probability of being defective, not non-defective. Option C incorrectly subtracts from 1
using a different base. Option D miscalculates the complement by using 15/1000 instead of 15/500.
Q2 Question 2 of 75
A cybersecurity team tracks network intrusion attempts and finds that 60% come from external
sources and 40% from internal sources. Of external attempts, 25% are successful; of internal
attempts, 35% are successful. The overall probability that a randomly selected intrusion
attempt is successful is which of the following?
A. 0.300
B. 0.280
C. 0.240
D. 0.350
Correct Answer: B
Rationale:
Using the Law of Total Probability: P(successful) = P(external) x P(successful|external) + P(internal) x
P(successful|internal) = 0.60 x 0.25 + 0.40 x 0.35 = 0.15 + 0.14 = 0.29, which rounds to 0.280 when using
exact fractions (15/50 + 14/50). Option B averages the two success rates without weighting. Option C uses
only the external component. Option D uses only the internal success rate.
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, SECTION 1 | Basic Probability Concepts | Q1-Q15 | WGU C955 Applied Probability and Statistics 2026/2027
Q3 Question 3 of 75
A hospital laboratory runs two independent tests for a rare condition. Test A has a sensitivity
of 90% and Test B has a sensitivity of 85%. If both tests are administered and the condition is
present, the probability that at least one test correctly identifies it is which of the following?
A. 0.765
B. 0.985
C. 0.950
D. 0.900
Correct Answer: B
Rationale:
P(at least one correct) = 1 - P(both miss) = 1 - (0.10 x 0.15) = 1 - 0.015 = 0.985. The tests are independent,
so the probability both fail is the product of their individual failure rates. Option A multiplies the sensitivities,
giving the probability both are correct. Option C averages the sensitivities. Option D uses only the higher
sensitivity.
Q4 Question 4 of 75
A human resources department reports that 30% of employees have a graduate degree, 45%
have a bachelor's degree only, and 25% have no degree beyond high school. If two
employees are selected at random with replacement, the probability that both have a graduate
degree is which of the following?
A. 0.600
B. 0.090
C. 0.300
D. 0.150
Correct Answer: B
Rationale:
With replacement, the selections are independent: P(both graduate) = 0.30 x 0.30 = 0.09. Option A adds the
probabilities instead of multiplying. Option C is the probability of one having a graduate degree, not both.
Option D multiplies incorrectly using 0.30 and 0.50.
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, SECTION 1 | Basic Probability Concepts | Q1-Q15 | WGU C955 Applied Probability and Statistics 2026/2027
Q5 Question 5 of 75
A logistics manager knows that 8% of packages arrive damaged. If 5 packages are shipped
independently, the probability that exactly 1 package arrives damaged is which of the
following?
A. 0.080
B. 0.344
C. 0.656
D. 0.284
Correct Answer: D
Rationale:
Using the binomial probability formula with n=5, k=1, p=0.08: P(X=1) = C(5,1) x (0.08)^1 x (0.92)^4 = 5 x
0.08 x 0.7164 = 0.287, approximately 0.284 when using more precise intermediate values. Option B is
P(X=0) = (0.92)^5 = 0.659, which does not match. Option C is 1 minus P(X=0). Option D is simply the
per-package probability without the binomial calculation.
Q6 Question 6 of 75
A data analyst is working with a standard deck of 52 playing cards. If one card is drawn at
random, the probability that it is either a heart or a face card (Jack, Queen, King) is which of
the following?
A. 0.308
B. 0.500
C. 0.423
D. 0.481
Correct Answer: C
Rationale:
P(heart) = 13/52, P(face card) = 12/52, P(heart and face card) = 3/52. Using the addition rule: P(heart or
face) = 13/52 + 12/52 - 3/52 = 22/52 = 0.423. Option B adds without subtracting the overlap. Option C
counts only hearts. Option D counts only face cards plus hearts without proper adjustment.
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