Exam 1 — Fall 2025: Fully Worked Solutions

The questions below reproduce the Fall 2025 Exam 1 in full accessible text. Each problem is followed by a complete worked solution. Point values reflect the actual exam.

Table 10 Point Summary
Section	Format	Points
Problem 1 — True/False	6 questions × 2 pts	12
Problem 2 — Multiple Choice	5 questions × 3 pts	15
Problem 3 — Free Response	4 parts	23
Problem 4 — Free Response	5 parts	27
Problem 5 — Free Response	4 parts	28
Total		105

Problem 1: True/False (12 points, 2 points each)

Indicate the correct answer by completely filling in the appropriate circle. If you indicate your answer by any other way, you may be marked incorrect.

Question 1.1 (2 pts)

Let $X$ and $Y$ be two discrete random variables with supports $x \in \{0, 1, 2\}$ and $y \in \{1, 2, 3\}$.

T or F: If $P(X=1,\; Y=1) = P(X=1)\cdot P(Y=1)$, then this implies that $X$ and $Y$ are independent random variables.

Question 1.2 (2 pts)

For three events $A$, $B$, and $C$ from the same sample space $\Omega$, it is known that $C \neq \emptyset$ and $P(A \cap B) > 0$.

T or F: If $P(A \cap B \cap C) = 0$, it must follow that $P(A \cap C) = P(B \cap C) = 0$.

Question 1.3 (2 pts)

Let $X$ be a continuous random variable with a PDF $f_X(x)$ defined over the symmetric interval $[-c, c]$ for some constant $c > 0$.

T or F: If the PDF is an even function (meaning $f_X(-x) = f_X(x)$ for all $x$), then this implies that the 50th percentile (median) of the distribution must be 0.

Question 1.4 (2 pts)

A censor reports a raw error measurement $X$ in millivolts. Historical data show a bias in these measurements with $E[X] = -2$ (millivolts) and $\text{SD}(X) = 5$ (millivolts).

T or F: For the transformed score $Y = 5X + 35$, it follows that $E[Y] = \text{SD}(Y)$.

Question 1.5 (2 pts)

In Normal distribution word problems, we distinguish “forward” and “backward” problems.

T or F: “Backward” problems solve for an $x$-value given a probability, while “forward” problems solve for a probability given an $x$-value.

Question 1.6 (2 pts)

In a class of 60 students, the instructor randomly selects 20 homework submissions without replacement to audit for possible AI policy violations. Each audited paper is labeled probable violation or no obvious violation. Let $X$ be the number of audited papers with probable violations.

T or F: Then $X$ is a Binomial random variable because the process satisfies the BINS conditions.

Problem 2: Multiple Choice (15 points, 3 points each)

Indicate the correct answer by completely filling in the appropriate circle. If you indicate your answer by any other way, you may be marked incorrect. For each question, there is only one correct option letter choice unless specified.

Question 2.1 (3 pts)

A researcher randomly selected 100 graduate students and surveyed their daily expenses on eating out. The collected data are visualized in the boxplot below.

A horizontal modified boxplot titled "Daily Expenses ($) - Uncorrected". The box spans approximately $3 to $15, with median near $7 and mean dot near $10. The lower whisker extends left to about $0. The upper whisker extends right to approximately $31. One explicit point (outlier) appears to the far right at approximately $33, representing the value 37.24.

The researcher received an email from one of the participants, stating that they had misreported the amount, and the corrected amount should be 27.24 instead of 37.24. After this correction, which of the two graphical components of the boxplot remain unchanged?

(A) Third Quartile, $Q_3$
(B) Sample Mean, $\bar{x}$
(C) Upper Whisker
(D) The number of explicit points
(E) The maximum value

Question 2.2 (3 pts)

At a large clinic, women’s heights are well modeled by a Normal distribution with mean 165 cm and standard deviation 7 cm. Identify the false statement from A–E.

(A) The mean, median, and mode are all 165 cm.
(B) The 25th and 75th percentiles are equidistant from 165 cm.
(C) Converting heights from centimeters to inches does not change anyone’s z-score or percentile.
(D) About 68% of women have heights within 7 cm of 165 cm.
(E) Every woman’s height lies within 3 standard deviations of 165 cm.

Question 2.3 (3 pts)

A clinical trial enrolls 10 women, each receiving one dose of a new drug. Based on a pre-measured genotype, 4 participants have a 30% chance of response (type $G_1$) and 6 have a 70% chance (type $G_2$). Responses are independent across participants, and each participant’s outcome is either response or no response. Let $X$ be the number who respond. Identify the correct statement from A–E.

(A) The number of trials is not a fixed constant.
(B) The trials are dependent.
(C) There are more than two possible outcomes on each trial.
(D) The probability of success is not the same for all trials.
(E) The trials are conducted without replacement from a small finite population.

Question 2.4 (3 pts)

A new type of biodegradable plastic is developed, and its degradation time $X$ (in years) is modeled by the probability density function

\[\begin{split}f_X(x) = \begin{cases} k\,x^2 & 0 \leq x \leq 3 \\ 0 & \text{otherwise} \end{cases}\end{split}\]

What is the expected lifetime of this plastic?

(A) 1 year
(B) 1.5 years
(C) 2.25 years
(D) 3 years
(E) 3.25 years
(F) 9 years

Question 2.5 (3 pts)

Let $X \sim \text{Binomial}(n=2,\; p)$ with unknown $p \in (0,1)$. Define a second random variable $Y$ conditionally on $X$ as follows:

$P(Y = 0 \mid X = 0) = 1$
$P(Y = 1 \mid X = 1) = P(Y = 2 \mid X = 1) = \dfrac{1}{2}$
$P(Y = y \mid X = 2) \sim \text{Poisson}(\lambda)$

Which expression equals $P(Y = 0)$?

(A) $(1-p)^2 + p^2 \cdot e^{-\lambda}$
(B) $(1-p)^2 \cdot e^{-\lambda} + p^2$
(C) $(1-p)^2 + 2p(1-p) + p^2 \cdot e^{-\lambda}$
(D) $(1-p) \cdot p \cdot e^{-\lambda}$

Free Response Questions 3–5

Show all work, clearly label your answers, and use four decimal places.

Problem 3 (23 points)

Problem 3 Setup

Assume men’s college basketball game lengths $X$ (minutes) are Normally distributed with mean 118 and standard deviation 10, where these values already account for fouls, timeouts, media breaks, and overtime.

\[X \sim N(\mu = 118,\;\sigma = 10).\]

Question 3a (6 pts)

Find the probability that a randomly selected game lasts within 1.5 standard deviations of the mean.

Question 3b (10 pts)

Compute the interquartile range (IQR) for the population of men’s college basketball game lengths.

Question 3c (4 pts)

A random sample of 10 game times is given below. Compute the population-level inner fences and determine if any of these points fall outside the 1.5 IQR rule.

\[88 \quad 89 \quad 108 \quad 111 \quad 115 \quad 115 \quad 116 \quad 124 \quad 130 \quad 134\]

Solution

The population-level inner fences use the population $Q_1$ and $Q_3$ from Question 3b, not the sample quartiles.

Inner fences:

\[\begin{split}\text{Inner lower fence} &= x_{0.25} - 1.5 \times IQR = 111.3 - 1.5 \times 13.4 = 111.3 - 20.1 = \boxed{91.2 \text{ min}}, \\[4pt] \text{Inner upper fence} &= x_{0.75} + 1.5 \times IQR = 124.7 + 1.5 \times 13.4 = 124.7 + 20.1 = \boxed{144.8 \text{ min}}.\end{split}\]

Check each observation:

Observation	Value (min)	Status
1	88	$88 < 91.2$ → outside lower fence ✗
2	89	$89 < 91.2$ → outside lower fence ✗
3–10	108–134	All between 91.2 and 144.8 → within fences ✓

Two points fall outside the inner lower fence: 88 and 89.

Question 3d (3 pts)

Consider three distribution models used in this course: $\text{Normal}(\mu, \sigma^2)$, $\text{Exponential}(\lambda)$, and $\text{Uniform}(a, b)$. The interquartile range ($IQR = Q_3 - Q_1$) measures the spread of the middle 50% of the distribution. In the statements below, “does not depend on the mean” means the IQR cannot be determined from the mean ($E[X]$) alone; “constant multiple of the mean” means $IQR = k \cdot E[X]$ for some constant $k$ that does not vary. (If needed use scratch space on pg 2 of exam.)

Which statement about the $IQR$ and the mean ($E[X]$) is incorrect?

(A) For an Exponential distribution, the IQR is a constant multiple of the mean.
(B) For a Normal distribution, the IQR does not depend on the mean.
(C) For a Uniform distribution on $[0, b]$, the IQR equals the mean.
(D) For a Uniform distribution on $[a, b]$, the IQR is determined by the mean alone.

Problem 4 (27 points)

Problem 4 Setup

Heekyung is training her cat, Meredith, to give a high-five. From her extensive experience, Meredith’s response depends on whether a treat is offered. Heekyung offers a treat 70% of the time. Meredith’s behavior on an attempt is exactly one of the following: high-five, ignore, or nag.

If a treat is offered:

Meredith gives a high-five with probability 0.8.
Meredith ignores Heekyung with probability 0.2.

If a treat is not offered:

Meredith gives a high-five with probability 0.35.
Meredith ignores Heekyung with probability 0.6.
Otherwise, Meredith Nags Heekyung.

Heekyung will restart the training next month if the probability that no treat was offered, given that a high-five occurred, is greater than one-half.

Question 4a (3 pts)

Are the two events {Treat is offered} and {Give a high-five} independent? State yes or no and provide mathematical justification.

Question 4b (2 pts)

For each branch A, B, and C in the tree diagram below, write the probability statement and find its probability.

A probability tree diagram. The root splits into Treat (probability 0.7) and Did not give a Treat (probability 0.3). From Treat, two branches lead to HighFive (probability 0.8) and branch A leading to Ignore. From Did not give a Treat, three branches lead to HighFive (probability 0.35), branch B leading to Ignore, and branch C leading to Nag.

Question 4c (4 pts)

Determine the probability that Meredith Nags Heekyung.

Question 4d (8 pts)

Determine the probability that Meredith gives a high-five.

Question 4e (10 pts)

Based on your results, compute the probability that no treat was offered, given that a high-five occurred. According to these calculations and Heekyung’s initial no retraining requirements stated at the start of this question, should Heekyung restart the training next month?

Problem 5 (28 points)

Problem 5 Setup

Purdue analyzed how long IT tickets take to resolve. Routine requests are handled in a quick “triage window” and are about equally likely to finish at any time during the first 10 minutes. If a ticket survives past 10 minutes, the remaining time decays exponentially. Let $T$ be the resolution time (minutes) with PDF:

\[\begin{split}f_T(t) = \begin{cases} k & 0 \leq t \leq 10 \\[4pt] k\,e^{-(t-10)} & t > 10 \\[4pt] 0 & \text{otherwise} \end{cases}\end{split}\]

where $k > 0$ is to be determined.

Question 5a (10 pts)

Determine $k$ so that $f_T(t)$ is a valid probability density function.

The cumulative distribution function for the resolution time $T$ is given below, defined up to the normalizing constant $k$ determined in part (a):

\[\begin{split}F_T(t) = \begin{cases} 0 & t < 0 \\[4pt] k \cdot t & 0 \leq t \leq 10 \\[4pt] 1 - k\,e^{-(t-10)} & t > 10 \end{cases}\end{split}\]

Question 5b (6 pts)

What is the probability that a ticket is finished between 5 and 15 minutes? For your convenience, the CDF of the resolution time $T$ is given above, defined up to the normalizing constant $k$ determined in part (a).

Question 5c (6 pts)

Given that a ticket has been in the system for at least 5 minutes, determine the probability that the total time the ticket remains unresolved is at least 15 minutes.

Question 5d (6 pts)

Only 5% of tickets take longer than $t^*$ to solve. Determine $t^*$.