MATH 110 MODULE 8 EXAM – INTRODUCTION TO STATISTICS – PORTAGE
LEARNING COMPLETE PRACTICE EXAM QUESTIONS AND ANSWERS | VERIFIED
SOLUTIONS | UPDATED 2026/2027 STUDY GUIDE
Examiner/Administrator: Portage Learning
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MATH 110 – INTRODUCTION TO STATISTICS – MODULE 8 FINAL ASSESSMENT
2026/2027 EDITION
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COMPLETE PRACTICE EXAM
30 MULTIPLE-CHOICE QUESTIONS
EXACT OFFICIAL COUNT: 30 QUESTIONS
PASSING SCORE: 70%
TESTING TIME: 90 MINUTES
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PORTAGE LEARNING || ALIGNED WITH CURRENT INTRODUCTION TO STATISTICS
CURRICULUM BLUEPRINTS || DESCRIPTIVE & INFERENTIAL STATISTICS ||
PROFESSIONAL EXAM PREPARATION GUIDE || 100% ORIGINAL VERIFIED SOLUTIONS ||
COMPREHENSIVE STATISTICAL ANALYSIS PRACTICE || PREPARED FOR ONLINE
ACADEMIC ASSESSMENT USE || UPDATED 2026/2027 EXAMINATION FORMAT
━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Questions 1–10 → Probability & Discrete Random Variables
Q1. A manufacturing company reports that 12% of its products are defective. A quality
inspector randomly selects 8 products for testing. What is the probability that exactly 2
products are defective?
A. 0.1977
B. 0.2741
C. 0.3012
D. 0.1489
Correct Answer: 🔴 A. 0.1977
,Explanation: 🔹 This problem follows a binomial probability model because there are a
fixed number of trials (8), each trial has two outcomes (defective or not defective), and
the probability remains constant at 0.12. Using the binomial formula:
P (X = 2) = (82)(0.12)2 (0.88)6
Evaluating the expression gives approximately 0.1977. Option B and C are too large
given the low defect rate, while D underestimates the probability. The exact computation
confirms Option A.
Q2. A random variable has the following probability distribution:
X 1 2 3 4
P(X) 0.2 0.3 0.1 0.4
What is the expected value of X?
A. 2.1
B. 2.7
C. 3.0
D. 2.4
Correct Answer: 🔴 B. 2.7
Explanation: 🔹 The expected value of a discrete random variable is computed by
multiplying each value by its probability and summing the products:
E(X) = ∑ xP (x) = 1(0.2) + 2(0.3) + 3(0.1) + 4(0.4)
This equals 0.2 + 0.6 + 0.3 + 1.6 = 2.7. Option B is correct. The other values result from
arithmetic mistakes or omission of one term.
Q3. A statistics professor states that student quiz scores are normally distributed with a
mean of 78 and a standard deviation of 6. What percentage of students score between
72 and 84?
,A. 34%
B. 68%
C. 95%
D. 99.7%
Correct Answer: 🔴 B. 68%
Explanation: 🔹 The interval from 72 to 84 is one standard deviation below and above
the mean:
x−μ
z= σ
x 1.2
μ 0.0
σ 1.0
x−μ
z= σ
≈ 1.2
Φ(z) ≈ 88.5%
For x = 72, z = −1; for x = 84, z = +1. According to the Empirical Rule for normal
distributions, approximately 68% of data falls within one standard deviation of the mean.
Options C and D correspond to two and three standard deviations respectively.
Q4. A card is drawn from a standard deck of 52 cards. What is the probability of
drawing either a king or a heart?
A. 4/52
B. 13/52
C. 16/52
D. 17/52
Correct Answer: 🔴 C. 16/52
Explanation: 🔹 There are 4 kings and 13 hearts, but the king of hearts is counted twice,
so the overlap must be subtracted:
, P (K ∪ H) = P (K) + P (H) −
P (K ∩ H)
P (A) 0.55
A 0.20 B
P (B) 0.45
P (A ∩ B)0.20
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) ≈ 0.80
Thus, favorable outcomes = 4 + 13 − 1 = 16. Therefore, the probability is 16/52. Option
D incorrectly double-counts the king of hearts.
Q5. An insurance company estimates that 5% of policyholders file claims annually. If 20
policyholders are selected, what is the probability that none file a claim?
A. 0.3585
B. 0.6415
C. 0.1216
D. 0.2500
Correct Answer: 🔴 A. 0.3585
Explanation: 🔹 This is another binomial setting with n = 20 and p = 0.05. The
probability of zero claims is:
P (X = 0) = (20
0
)(0.05)0 (0.95)20
This simplifies to (0.95)^20 ≈ 0.3585. Option B is the complement probability of at least
one claim.
Q6. Which condition is necessary for a binomial experiment?
A. The trials must be dependent
B. The probability of success changes each trial
LEARNING COMPLETE PRACTICE EXAM QUESTIONS AND ANSWERS | VERIFIED
SOLUTIONS | UPDATED 2026/2027 STUDY GUIDE
Examiner/Administrator: Portage Learning
━━━━━━━━━━━━━━━━━━━━━━━━━━━━
MATH 110 – INTRODUCTION TO STATISTICS – MODULE 8 FINAL ASSESSMENT
2026/2027 EDITION
━━━━━━━━━━━━━━━━━━━━━━━━━━━━
COMPLETE PRACTICE EXAM
30 MULTIPLE-CHOICE QUESTIONS
EXACT OFFICIAL COUNT: 30 QUESTIONS
PASSING SCORE: 70%
TESTING TIME: 90 MINUTES
━━━━━━━━━━━━━━━━━━━━━━━━━━━━
PORTAGE LEARNING || ALIGNED WITH CURRENT INTRODUCTION TO STATISTICS
CURRICULUM BLUEPRINTS || DESCRIPTIVE & INFERENTIAL STATISTICS ||
PROFESSIONAL EXAM PREPARATION GUIDE || 100% ORIGINAL VERIFIED SOLUTIONS ||
COMPREHENSIVE STATISTICAL ANALYSIS PRACTICE || PREPARED FOR ONLINE
ACADEMIC ASSESSMENT USE || UPDATED 2026/2027 EXAMINATION FORMAT
━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Questions 1–10 → Probability & Discrete Random Variables
Q1. A manufacturing company reports that 12% of its products are defective. A quality
inspector randomly selects 8 products for testing. What is the probability that exactly 2
products are defective?
A. 0.1977
B. 0.2741
C. 0.3012
D. 0.1489
Correct Answer: 🔴 A. 0.1977
,Explanation: 🔹 This problem follows a binomial probability model because there are a
fixed number of trials (8), each trial has two outcomes (defective or not defective), and
the probability remains constant at 0.12. Using the binomial formula:
P (X = 2) = (82)(0.12)2 (0.88)6
Evaluating the expression gives approximately 0.1977. Option B and C are too large
given the low defect rate, while D underestimates the probability. The exact computation
confirms Option A.
Q2. A random variable has the following probability distribution:
X 1 2 3 4
P(X) 0.2 0.3 0.1 0.4
What is the expected value of X?
A. 2.1
B. 2.7
C. 3.0
D. 2.4
Correct Answer: 🔴 B. 2.7
Explanation: 🔹 The expected value of a discrete random variable is computed by
multiplying each value by its probability and summing the products:
E(X) = ∑ xP (x) = 1(0.2) + 2(0.3) + 3(0.1) + 4(0.4)
This equals 0.2 + 0.6 + 0.3 + 1.6 = 2.7. Option B is correct. The other values result from
arithmetic mistakes or omission of one term.
Q3. A statistics professor states that student quiz scores are normally distributed with a
mean of 78 and a standard deviation of 6. What percentage of students score between
72 and 84?
,A. 34%
B. 68%
C. 95%
D. 99.7%
Correct Answer: 🔴 B. 68%
Explanation: 🔹 The interval from 72 to 84 is one standard deviation below and above
the mean:
x−μ
z= σ
x 1.2
μ 0.0
σ 1.0
x−μ
z= σ
≈ 1.2
Φ(z) ≈ 88.5%
For x = 72, z = −1; for x = 84, z = +1. According to the Empirical Rule for normal
distributions, approximately 68% of data falls within one standard deviation of the mean.
Options C and D correspond to two and three standard deviations respectively.
Q4. A card is drawn from a standard deck of 52 cards. What is the probability of
drawing either a king or a heart?
A. 4/52
B. 13/52
C. 16/52
D. 17/52
Correct Answer: 🔴 C. 16/52
Explanation: 🔹 There are 4 kings and 13 hearts, but the king of hearts is counted twice,
so the overlap must be subtracted:
, P (K ∪ H) = P (K) + P (H) −
P (K ∩ H)
P (A) 0.55
A 0.20 B
P (B) 0.45
P (A ∩ B)0.20
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) ≈ 0.80
Thus, favorable outcomes = 4 + 13 − 1 = 16. Therefore, the probability is 16/52. Option
D incorrectly double-counts the king of hearts.
Q5. An insurance company estimates that 5% of policyholders file claims annually. If 20
policyholders are selected, what is the probability that none file a claim?
A. 0.3585
B. 0.6415
C. 0.1216
D. 0.2500
Correct Answer: 🔴 A. 0.3585
Explanation: 🔹 This is another binomial setting with n = 20 and p = 0.05. The
probability of zero claims is:
P (X = 0) = (20
0
)(0.05)0 (0.95)20
This simplifies to (0.95)^20 ≈ 0.3585. Option B is the complement probability of at least
one claim.
Q6. Which condition is necessary for a binomial experiment?
A. The trials must be dependent
B. The probability of success changes each trial
