UNIVERSITY OF HOUSTON
MATH 3339 Exam 1-2
Official Practice Exam | 2026/2027 Edition
75 Questions 90 Minutes 80% Passing Recertification
Table of Contents
Section 1: Descriptive Statistics & Probability (Questions 1-15)
Section 2: Discrete Probability Distributions (Questions 16-30)
Section 3: Continuous Probability Distributions (Questions 31-45)
Section 4: Sampling Distributions & Central Limit Theorem (Questions 46-60)
Section 5: Estimation & Confidence Intervals (Questions 61-75)
Answer Key (Final Page)
Instructions
This practice exam contains 75 multiple-choice questions divided into 5 sections.
You have 90 minutes to complete the entire exam. Pace yourself accordingly.
Each question has four answer choices (A, B, C, D). Select the single best answer.
A passing score of 80% requires at least 60 correct answers out of 75 questions.
Read each scenario carefully before selecting your response.
The correct answer and rationale are provided after each question for study purposes.
This exam reflects the 2026/2027 curriculum update for MATH 3339 at the University of Houston.
,Section 1 of 5: Descriptive Statistics & Probability | 2026/2027
Q1 Question 1 of 75
A researcher collects the daily high temperatures (in degrees Fahrenheit) recorded in Houston over
a 30-day period and obtains the values: 92, 95, 89, 91, 97, 94, 90, 93, 88, 96, 91, 94, 87, 95, 92, 93,
90, 96, 89, 94, 91, 95, 88, 93, 92, 94, 90, 97, 89, 91. A colleague asks the researcher to summarize
the central tendency of this dataset. Which measure is most appropriate for this symmetric,
unimodal distribution?
A. Median, because it is resistant to extreme values in any dataset
B. Mean, because the data are symmetric without outliers
C. Mode, because it identifies the most frequent value
D. Range, because it captures the full spread of the data
Correct Answer: B
Rationale:
For a symmetric, unimodal distribution without outliers, the mean is the most appropriate measure of
central tendency because it uses every data point and provides the balance point of the distribution. The
median is preferred when outliers are present, the mode only identifies frequency and ignores
magnitude, and the range is a measure of spread rather than central tendency.
Q2 Question 2 of 75
A quality control analyst at a manufacturing plant measures the diameters of 50 ball bearings and
finds the sample mean is 10.2 mm with a standard deviation of 0.3 mm. The analyst constructs a
boxplot and observes that the median, quartiles, and whiskers are nearly symmetric about the
center. Which statement best describes the shape of this distribution?
A. Right-skewed because the mean exceeds the median
B. Approximately symmetric because the median and mean are close
C. Left-skewed because the left whisker is longer
D. Bimodal because two clusters are present in the data
Correct Answer: B
MATH 3339 Exams 1-2 -- 2026/2027 | Passing Score: 80% | Page 2
, Rationale:
When the mean and median are approximately equal and the boxplot shows symmetry, the distribution
is approximately symmetric. Right-skewed distributions have a mean greater than the median with a
longer right whisker, left-skewed distributions have a longer left whisker with mean less than the median,
and bimodality would show two distinct peaks rather than a symmetric single peak.
Q3 Question 3 of 75
A statistics professor asks 200 students whether they prefer morning, afternoon, or evening classes,
and also records their GPA category (below 2.5, 2.5 to 3.0, above 3.0). The professor wants to
determine whether class-time preference is independent of GPA category. Which type of analysis is
most appropriate?
A. Chi-square test of independence using a contingency table
B. Two-sample t-test comparing mean GPAs across preferences
C. ANOVA comparing three group means for class time
D. Linear regression predicting GPA from class preference
Correct Answer: A
Rationale:
The chi-square test of independence evaluates whether two categorical variables are associated by
comparing observed frequencies in a contingency table to expected frequencies under independence. A
two-sample t-test requires quantitative data, ANOVA requires a quantitative response variable, and
linear regression requires a quantitative dependent variable, none of which apply to two categorical
variables.
Q4 Question 4 of 75
A data analyst computes the variance of a dataset containing the weights of 12 newborn infants and
obtains a value of 0.85 kg squared. A colleague points out that the analyst used the population
variance formula instead of the sample variance formula. What is the corrected sample variance?
A. 0.85 x = 0.927 kg squared
B. 0.85 x = 0.779 kg squared
C. 0.85 x = 0.785 kg squared
D. 0.85 x = 0.935 kg squared
Correct Answer: A
MATH 3339 Exams 1-2 -- 2026/2027 | Passing Score: 80% | Page 3
, Rationale:
The population variance divides the sum of squared deviations by N, while the sample variance divides
by N minus 1. To convert, multiply the population variance by N divided by (N minus 1), giving 0.85 times
12 divided by 11 equals approximately 0.927. The other options apply incorrect conversion factors.
Q5 Question 5 of 75
A hospital records the blood type of every patient admitted during a one-month period: 90 patients
have type O, 45 have type A, 35 have type B, and 10 have type AB. If a single patient is selected at
random from these records, what is the probability that the patient has either type A or type B
blood?
A. 45/180 + 35/180 = 80/180 = 0.444
B. 45/180 x 35/180 = 1575/32400 = 0.049
C. (45/180) squared + (35/180) squared = 0.083
D. 1 - (90/180) = 0.500
Correct Answer: A
Rationale:
Since blood types A and B are mutually exclusive events (a patient cannot have both), the probability of
A or B equals the sum of their individual probabilities: 45/180 plus 35/180 equals 80/180 equals 0.444.
Multiplying the probabilities would apply to joint events, squaring individual probabilities has no statistical
meaning, and subtracting from 1 only removes type O but still includes type AB.
Q6 Question 6 of 75
A student randomly selects a card from a standard 52-card deck, records its suit, replaces it,
shuffles, and draws again. The student wants to find the probability of drawing a heart on the first
draw and a diamond on the second draw. Which probability rule applies?
A. Addition rule for mutually exclusive events
B. Multiplication rule for independent events
C. Bayes theorem for conditional probability
D. Complement rule for opposite outcomes
Correct Answer: B
MATH 3339 Exams 1-2 -- 2026/2027 | Passing Score: 80% | Page 4
MATH 3339 Exam 1-2
Official Practice Exam | 2026/2027 Edition
75 Questions 90 Minutes 80% Passing Recertification
Table of Contents
Section 1: Descriptive Statistics & Probability (Questions 1-15)
Section 2: Discrete Probability Distributions (Questions 16-30)
Section 3: Continuous Probability Distributions (Questions 31-45)
Section 4: Sampling Distributions & Central Limit Theorem (Questions 46-60)
Section 5: Estimation & Confidence Intervals (Questions 61-75)
Answer Key (Final Page)
Instructions
This practice exam contains 75 multiple-choice questions divided into 5 sections.
You have 90 minutes to complete the entire exam. Pace yourself accordingly.
Each question has four answer choices (A, B, C, D). Select the single best answer.
A passing score of 80% requires at least 60 correct answers out of 75 questions.
Read each scenario carefully before selecting your response.
The correct answer and rationale are provided after each question for study purposes.
This exam reflects the 2026/2027 curriculum update for MATH 3339 at the University of Houston.
,Section 1 of 5: Descriptive Statistics & Probability | 2026/2027
Q1 Question 1 of 75
A researcher collects the daily high temperatures (in degrees Fahrenheit) recorded in Houston over
a 30-day period and obtains the values: 92, 95, 89, 91, 97, 94, 90, 93, 88, 96, 91, 94, 87, 95, 92, 93,
90, 96, 89, 94, 91, 95, 88, 93, 92, 94, 90, 97, 89, 91. A colleague asks the researcher to summarize
the central tendency of this dataset. Which measure is most appropriate for this symmetric,
unimodal distribution?
A. Median, because it is resistant to extreme values in any dataset
B. Mean, because the data are symmetric without outliers
C. Mode, because it identifies the most frequent value
D. Range, because it captures the full spread of the data
Correct Answer: B
Rationale:
For a symmetric, unimodal distribution without outliers, the mean is the most appropriate measure of
central tendency because it uses every data point and provides the balance point of the distribution. The
median is preferred when outliers are present, the mode only identifies frequency and ignores
magnitude, and the range is a measure of spread rather than central tendency.
Q2 Question 2 of 75
A quality control analyst at a manufacturing plant measures the diameters of 50 ball bearings and
finds the sample mean is 10.2 mm with a standard deviation of 0.3 mm. The analyst constructs a
boxplot and observes that the median, quartiles, and whiskers are nearly symmetric about the
center. Which statement best describes the shape of this distribution?
A. Right-skewed because the mean exceeds the median
B. Approximately symmetric because the median and mean are close
C. Left-skewed because the left whisker is longer
D. Bimodal because two clusters are present in the data
Correct Answer: B
MATH 3339 Exams 1-2 -- 2026/2027 | Passing Score: 80% | Page 2
, Rationale:
When the mean and median are approximately equal and the boxplot shows symmetry, the distribution
is approximately symmetric. Right-skewed distributions have a mean greater than the median with a
longer right whisker, left-skewed distributions have a longer left whisker with mean less than the median,
and bimodality would show two distinct peaks rather than a symmetric single peak.
Q3 Question 3 of 75
A statistics professor asks 200 students whether they prefer morning, afternoon, or evening classes,
and also records their GPA category (below 2.5, 2.5 to 3.0, above 3.0). The professor wants to
determine whether class-time preference is independent of GPA category. Which type of analysis is
most appropriate?
A. Chi-square test of independence using a contingency table
B. Two-sample t-test comparing mean GPAs across preferences
C. ANOVA comparing three group means for class time
D. Linear regression predicting GPA from class preference
Correct Answer: A
Rationale:
The chi-square test of independence evaluates whether two categorical variables are associated by
comparing observed frequencies in a contingency table to expected frequencies under independence. A
two-sample t-test requires quantitative data, ANOVA requires a quantitative response variable, and
linear regression requires a quantitative dependent variable, none of which apply to two categorical
variables.
Q4 Question 4 of 75
A data analyst computes the variance of a dataset containing the weights of 12 newborn infants and
obtains a value of 0.85 kg squared. A colleague points out that the analyst used the population
variance formula instead of the sample variance formula. What is the corrected sample variance?
A. 0.85 x = 0.927 kg squared
B. 0.85 x = 0.779 kg squared
C. 0.85 x = 0.785 kg squared
D. 0.85 x = 0.935 kg squared
Correct Answer: A
MATH 3339 Exams 1-2 -- 2026/2027 | Passing Score: 80% | Page 3
, Rationale:
The population variance divides the sum of squared deviations by N, while the sample variance divides
by N minus 1. To convert, multiply the population variance by N divided by (N minus 1), giving 0.85 times
12 divided by 11 equals approximately 0.927. The other options apply incorrect conversion factors.
Q5 Question 5 of 75
A hospital records the blood type of every patient admitted during a one-month period: 90 patients
have type O, 45 have type A, 35 have type B, and 10 have type AB. If a single patient is selected at
random from these records, what is the probability that the patient has either type A or type B
blood?
A. 45/180 + 35/180 = 80/180 = 0.444
B. 45/180 x 35/180 = 1575/32400 = 0.049
C. (45/180) squared + (35/180) squared = 0.083
D. 1 - (90/180) = 0.500
Correct Answer: A
Rationale:
Since blood types A and B are mutually exclusive events (a patient cannot have both), the probability of
A or B equals the sum of their individual probabilities: 45/180 plus 35/180 equals 80/180 equals 0.444.
Multiplying the probabilities would apply to joint events, squaring individual probabilities has no statistical
meaning, and subtracting from 1 only removes type O but still includes type AB.
Q6 Question 6 of 75
A student randomly selects a card from a standard 52-card deck, records its suit, replaces it,
shuffles, and draws again. The student wants to find the probability of drawing a heart on the first
draw and a diamond on the second draw. Which probability rule applies?
A. Addition rule for mutually exclusive events
B. Multiplication rule for independent events
C. Bayes theorem for conditional probability
D. Complement rule for opposite outcomes
Correct Answer: B
MATH 3339 Exams 1-2 -- 2026/2027 | Passing Score: 80% | Page 4
