.. _exam1_fall2026: STAT 350 — Exam 1 — Fall 2026 ================================ .. admonition:: Exam Information :class: info | **Course:** STAT 350 — Introduction to Statistics | **Semester:** Fall 2026 | **Version:** V1 | **Total Points:** 105 | **Time Allowed:** 60 minutes .. list-table:: :header-rows: 1 :widths: 50 25 25 * - Problem - Total Possible - Topic * - Problem 1 (True/False, 2 pts each) - 12 - Descriptive Statistics, Independence, Random Variables, Binomial, Normal * - Problem 2 (Multiple Choice, 3 pts each) - 15 - Descriptive Statistics, Poisson, LOTUS, Binomial, Exponential * - Problem 3 - 26 - PDFs and CDFs * - Problem 4 - 26 - Law of Partitions, Bayes' Theorem, Independence * - Problem 5 - 26 - Normal Distribution, Quartiles and Fences * - **Total** - **105** - ---- Problem 1 — True/False (12 points, 2 points each) ---------------------------------------------------- .. admonition:: Question 1.1 (2 pts) :class: note A dataset :math:`X` has an interquartile range (IQR) of :math:`\text{IQR}_X`. Each value in the dataset is multiplied by :math:`-2`, and then 10 is added, giving new values :math:`y = -2x + 10`. **True or False:** The IQR of the new values of the dataset :math:`Y` is :math:`-2 \cdot \text{IQR}_X + 10`. .. dropdown:: Solution :class-container: sd-border-success **Answer: FALSE** Adding 10 moves both quartiles up by 10, so it cancels in :math:`Q_3 - Q_1`. Multiplying by :math:`-2` reverses the order of the values, so the third quartile of :math:`Y` comes from the first quartile of :math:`X`, and the first quartile of :math:`Y` comes from the third quartile of :math:`X`: .. math:: Q_{3,Y} = -2\,Q_{1,X} + 10, \qquad Q_{1,Y} = -2\,Q_{3,X} + 10. Therefore .. math:: \text{IQR}_Y = Q_{3,Y} - Q_{1,Y} = 2\,(Q_{3,X} - Q_{1,X}) = 2 \cdot \text{IQR}_X. .. admonition:: Question 1.2 (2 pts) :class: note Let :math:`A` and :math:`B` be two non-empty events from the same sample space :math:`\Omega`, both with positive probabilities (:math:`P(A) > 0` and :math:`P(B) > 0`). **True or False:** If :math:`A` and :math:`B` are mutually exclusive (disjoint), then they cannot be independent events. .. dropdown:: Solution :class-container: sd-border-success **Answer: TRUE** Disjoint events have :math:`P(A \cap B) = 0`. Independence requires :math:`P(A \cap B) = P(A)P(B)`, and :math:`P(A)P(B) > 0` because both probabilities are positive. So :math:`P(A \cap B) \neq P(A)P(B)`, and the events are not independent. Equivalently, if :math:`A` occurs then :math:`B` cannot, so :math:`P(B \mid A) = 0 \neq P(B)`. .. admonition:: Question 1.3 (2 pts) :class: note Let :math:`X` be a discrete random variable with finite mean :math:`\mu` and standard deviation :math:`\sigma > 0`. Define a new random variable :math:`Y = 4 - 3X`. **True or False:** The standard deviation of :math:`Y` is equal to :math:`-3\sigma`. .. dropdown:: Solution :class-container: sd-border-success **Answer: FALSE** The constant 4 does not change the spread, and the multiplier enters the variance squared: .. math:: \text{Var}(Y) = (-3)^2\,\text{Var}(X) = 9\sigma^2, \qquad \sigma_Y = \sqrt{9\sigma^2} = 3\sigma. A standard deviation is never negative, while :math:`-3\sigma < 0` because :math:`\sigma > 0`. .. admonition:: Question 1.4 (2 pts) :class: note Let :math:`X \sim \text{Binomial}(n, p)` have :math:`n \geq 1` independent trials and success probability :math:`p \in [0, 1]`. **True or False:** The variance :math:`\text{Var}(X)` can never exceed :math:`\dfrac{n}{4}`. .. dropdown:: Solution :class-container: sd-border-success **Answer: TRUE** The variance is :math:`\text{Var}(X) = np(1-p)`. The product :math:`p(1-p)` is largest at :math:`p = \tfrac{1}{2}`: .. math:: p(1-p) = \tfrac{1}{4} - \left(p - \tfrac{1}{2}\right)^2 \leq \tfrac{1}{4}. Therefore :math:`\text{Var}(X) = np(1-p) \leq \dfrac{n}{4}`, with equality only when :math:`p = \tfrac{1}{2}`. .. admonition:: Question 1.5 (2 pts) :class: note Let :math:`X` be a continuous random variable with probability density function :math:`f(x)`. **True or False:** The density must satisfy :math:`f(x) \leq 1` for all :math:`x \in \mathbb{R}`. .. dropdown:: Solution :class-container: sd-border-success **Answer: FALSE** A density must be nonnegative with total area 1 under its curve; its height can exceed 1. For example, :math:`X \sim \text{Uniform}(a = 0, b = 0.5)` has .. math:: f(x) = \begin{cases} 2 & 0 \leq x \leq 0.5 \\ 0 & \text{otherwise} \end{cases} and its total area is :math:`2 \times 0.5 = 1`. Values of :math:`f(x)` are not probabilities; probabilities are areas under :math:`f`. .. admonition:: Question 1.6 (2 pts) :class: note Suppose :math:`X \sim N(\mu, \sigma^2)`. **True or False:** The probability that :math:`X` falls within :math:`1.28` standard deviations of the mean, :math:`P(\mu - 1.28\sigma \leq X \leq \mu + 1.28\sigma)`, is :math:`0.80`. .. dropdown:: Solution :class-container: sd-border-success **Answer: TRUE** Standardize with :math:`Z = (X - \mu)/\sigma` and use the standard normal table: .. math:: P(\mu - 1.28\sigma \leq X \leq \mu + 1.28\sigma) &= P(-1.28 \leq Z \leq 1.28) \\ &= \Phi(1.28) - \Phi(-1.28) \\ &= 0.8997 - 0.1003 = 0.7994 \approx 0.80. ---- Problem 2 — Multiple Choice (15 points, 3 points each) --------------------------------------------------------- .. admonition:: Question 2.1 (3 pts) :class: note A survey of 150 recent Purdue engineering graduates revealed that their starting annual salaries had a sample mean of $86,500 and a sample median of $78,000. Based on these summary statistics, which of the following statements is most likely true regarding the distribution of starting salaries? (A) The distribution is symmetric and mound-shaped because engineering salaries are standardized. (B) The distribution is right-skewed (positively skewed) because a relatively small number of exceptionally high salaries pull the sample mean above the sample median. (C) The distribution is left-skewed (negatively skewed) because the majority of graduates earn less than the mean. (D) The sample standard deviation must be equal to zero. (E) The interquartile range (IQR) is guaranteed to be larger than the full range of the data. .. dropdown:: Solution :class-container: sd-border-success **Answer: (B)** - **(A) FALSE.** In a symmetric distribution the sample mean and sample median are close; here the sample mean is $8,500 above the sample median. - **(B) TRUE.** The sample mean is sensitive to extreme values and the sample median is not. A few very high salaries pull the sample mean above the sample median, which is the pattern of a right-skewed distribution. - **(C) FALSE.** In a left-skewed distribution, low values pull the sample mean below the sample median. Here the sample mean is above the sample median. - **(D) FALSE.** A standard deviation of zero means every salary is the same, so the sample mean and sample median would be equal. - **(E) FALSE.** The IQR can never exceed the range, because :math:`Q_1 \geq \min` and :math:`Q_3 \leq \max`. .. admonition:: Question 2.2 (3 pts) :class: note During peak lunch hours at the Purdue Memorial Union, mobile food orders arrive according to a Poisson process at an average rate of :math:`0.4` orders per minute. What is the probability that exactly :math:`2` orders arrive in a randomly selected :math:`5`-minute interval? (A) 0.0544 (B) 0.1465 (C) 0.2707 (D) 0.4000 (E) 0.6767 .. dropdown:: Solution :class-container: sd-border-success **Answer: (C)** A 5-minute interval has mean :math:`\lambda = 0.4 \times 5 = 2` orders, so the number of orders :math:`X` in the interval satisfies :math:`X \sim \text{Poisson}(\lambda = 2)`: .. math:: P(X = 2) = \frac{e^{-2}\,2^2}{2!} = 2e^{-2} = 0.2707. .. admonition:: Question 2.3 (3 pts) :class: note Let :math:`X` be a discrete random variable representing the number of defective components found in an electronic assembly pack, with probability mass function: .. math:: P(X = 0) = 0.60, \quad P(X = 1) = 0.30, \quad P(X = 2) = 0.10 The repair cost (in dollars) for a pack is modeled by the function :math:`C(X) = 50 + 20X^2`. What is the expected repair cost :math:`E[C(X)]`? (A) $50.00 (B) $57.00 (C) $60.00 (D) $64.00 (E) $78.00 .. dropdown:: Solution :class-container: sd-border-success **Answer: (D)** By linearity of expectation, :math:`E[C(X)] = 50 + 20\,E[X^2]`. By LOTUS, .. math:: E[X^2] = 0^2(0.60) + 1^2(0.30) + 2^2(0.10) = 0 + 0.30 + 0.40 = 0.70. Therefore :math:`E[C(X)] = 50 + 20(0.70) = 64`, an expected repair cost of $64.00. .. admonition:: Question 2.4 (3 pts) :class: note An inspector examines a small shipment of :math:`N = 25` laptops. Unbeknownst to the inspector, the shipment contains exactly :math:`4` defective units. The inspector randomly selects :math:`n = 5` laptops without replacement and counts the number of defective units :math:`X` among the :math:`5` sampled. Which of the following explains why :math:`X` does NOT strictly follow a Binomial distribution? (A) The draws are dependent because each selection changes which laptops remain in the shipment. (B) A draw cannot have binary outcomes because defect status is learned only after inspection. (C) The number of trials is not fixed because the defect count is unknown before inspection. (D) The expected defect count cannot be computed before the inspector checks the selected laptops. (E) The variance must exceed the sample size because the inspector samples without replacement. .. dropdown:: Solution :class-container: sd-border-success **Answer: (A)** - **(A) TRUE.** Each laptop drawn changes what remains. The second laptop is defective with probability :math:`3/24` if the first was defective and :math:`4/24` if it was not, so the draws are dependent and the **Independent** criterion of BInS fails. - **(B) FALSE.** Each laptop is either defective or not, so every draw has two outcomes. When the status is learned does not matter. - **(C) FALSE.** The number of draws is fixed in advance at :math:`n = 5`. - **(D) FALSE.** An expected value is computed from the distribution of :math:`X` before any laptop is inspected, and it is not one of the BInS criteria. - **(E) FALSE.** Nothing requires the variance of :math:`X` to exceed the sample size, and the variance is not one of the BInS criteria. .. admonition:: Question 2.5 (3 pts) :class: note The delivery time :math:`T` (in minutes) for autonomous delivery robots on the Purdue campus follows an Exponential distribution with rate :math:`\lambda = 0.10` per minute. The dining service promises a delivery fee refund if an order takes longer than :math:`c` minutes. If they wish to issue refunds for only the top :math:`2.5\%` slowest deliveries, what should the threshold :math:`c` be? (A) 27.73 minutes (B) 29.96 minutes (C) 36.89 minutes (D) 39.12 minutes (E) 46.05 minutes .. dropdown:: Solution :class-container: sd-border-success **Answer: (C)** For :math:`T \sim \text{Exp}(\lambda = 0.10)`, :math:`P(T > t) = e^{-0.10t}` for :math:`t \geq 0`. The threshold :math:`c` satisfies :math:`P(T > c) = 0.025`: .. math:: e^{-0.10c} = 0.025 \quad\Rightarrow\quad c = \frac{-\ln(0.025)}{0.10} = \frac{3.6889}{0.10} = 36.89 \text{ minutes}. ---- Problem 3 (26 points) — HPC Simulation Runtime ------------------------------------------------- .. admonition:: Problem 3 Setup :class: important A High-Performance Computing (HPC) cluster at Purdue processes large-scale scientific simulations. Every submitted task runs a mandatory one-hour CPU calibration stage before its simulation can finish or reach a checkpoint. Let :math:`X` denote the total active CPU runtime (in hours), including that calibration stage, of a randomly submitted simulation task before completion or checkpointing. Thus, :math:`X` cannot be less than one hour. The probability density function (PDF) of :math:`X` is modeled by the following piecewise function: .. math:: f(x) = \begin{cases} cx, & 1 \leq x \leq 3 \\ 3c, & 3 < x \leq 5 \\ 0, & \text{otherwise} \end{cases} .. admonition:: Question 3a (8 pts) :class: note Determine the value of the constant :math:`c` such that :math:`f(x)` is a valid probability density function. Show all integration steps clearly. .. dropdown:: Solution :class-container: sd-border-success The total area under :math:`f` must be 1: .. math:: 1 = \int_1^3 cx\,dx + \int_3^5 3c\,dx = c\left[\frac{x^2}{2}\right]_1^3 + 3c\,\big[x\big]_3^5 = c\left(\frac{9}{2} - \frac{1}{2}\right) + 3c(5 - 3) = 4c + 6c = 10c. Thus :math:`c = \dfrac{1}{10} = 0.1000`. .. admonition:: Question 3b (6 pts) :class: note The intervals of the Cumulative Distribution Function :math:`F(x)` are shown below. Find the expression for each labeled blank (i)–(iv) to complete :math:`F(x)` for all real numbers :math:`x \in \mathbb{R}`. .. math:: F(x) = \begin{cases} \boxed{\text{(i)}} & x < 1 \\ \boxed{\text{(ii)}} & 1 \leq x \leq 3 \\ \boxed{\text{(iii)}} & 3 < x \leq 5 \\ \boxed{\text{(iv)}} & x > 5 \end{cases} .. dropdown:: Solution :class-container: sd-border-success No probability lies below 1, and all of it lies at or below 5, so the outer blanks are **(i)** :math:`0` and **(iv)** :math:`1`. With :math:`c = \tfrac{1}{10}`, the two inner cases are: .. math:: \text{(ii)}\quad F(x) = \int_1^x \frac{t}{10}\,dt = \frac{x^2 - 1}{20}, .. math:: \text{(iii)}\quad F(x) = F(3) + \int_3^x \frac{3}{10}\,dt = \frac{2}{5} + \frac{3(x - 3)}{10} = \frac{3x - 5}{10}. The complete CDF is .. math:: F(x) = \begin{cases} 0 & x < 1 \\[4pt] \dfrac{x^2 - 1}{20} & 1 \leq x \leq 3 \\[8pt] \dfrac{3x - 5}{10} & 3 < x \leq 5 \\[8pt] 1 & x > 5 \end{cases} .. admonition:: Question 3c (4 pts) :class: note What is the probability that a randomly chosen simulation task takes between :math:`2` and :math:`4` hours to finish? .. dropdown:: Solution :class-container: sd-border-success Using the two inner cases of :math:`F(x)`, :math:`F(4) = \tfrac{7}{10}` and :math:`F(2) = \tfrac{3}{20}`. Hence, .. math:: P(2 \leq X \leq 4) = F(4) - F(2) = \frac{7}{10} - \frac{3}{20} = \frac{11}{20} = 0.5500. .. admonition:: Question 3d (4 pts) :class: note Given that a simulation task has already run for at least :math:`2` hours, what is the conditional probability that it takes more than :math:`4` hours to finish? .. dropdown:: Solution :class-container: sd-border-success Since :math:`\{X > 4\} \subseteq \{X \geq 2\}`, the intersection of the two events is :math:`\{X > 4\}`: .. math:: P(X > 4 \mid X \geq 2) = \frac{P(X > 4)}{P(X \geq 2)} = \frac{1 - F(4)}{1 - F(2)} = \frac{0.3000}{0.8500} = \frac{6}{17} \approx 0.3529. .. admonition:: Question 3e (4 pts) :class: note Find the median execution duration :math:`\tilde{\mu}` (in hours) of a simulation task. .. dropdown:: Solution :class-container: sd-border-success Since :math:`F(3) = 0.4000 < 0.50`, the median lies in :math:`3 < x \leq 5`. Solve .. math:: F(\tilde{\mu}) = \frac{3\tilde{\mu} - 5}{10} = 0.50 \quad\Rightarrow\quad \tilde{\mu} = \frac{10}{3} \approx 3.3333 \text{ hours}. ---- Problem 4 (26 points) — Autonomous Shuttle Alerts ---------------------------------------------------- .. admonition:: Problem 4 Setup :class: important Purdue's autonomous electric shuttle fleet navigates three campus routes: Route A (Engineering Quad), Route B (Stadium Drive), and Route C (Discovery Park). Shuttles are assigned randomly to routes based on campus demand: - Route A: assigned :math:`50\%` of the time, :math:`P(A) = 0.50` - Route B: assigned :math:`30\%` of the time, :math:`P(B) = 0.30` - Route C: assigned :math:`20\%` of the time, :math:`P(C) = 0.20` During operation, high pedestrian density or sun glare may trigger a safety sensor alert (:math:`S`). Telemetry data shows: - On Route A: :math:`P(S \mid A) = 0.08` - On Route B: :math:`P(S \mid B) = 0.15` - On Route C: :math:`P(S \mid C) = 0.25` .. admonition:: Question 4a (4 pts) :class: note Calculate the joint probabilities for the three alert scenarios: :math:`P(A \cap S)`, :math:`P(B \cap S)`, and :math:`P(C \cap S)`. .. dropdown:: Solution :class-container: sd-border-success Apply :math:`P(R \cap S) = P(R)\,P(S \mid R)` for each route :math:`R`: .. math:: P(A \cap S) &= (0.50)(0.08) = 0.0400 \\ P(B \cap S) &= (0.30)(0.15) = 0.0450 \\ P(C \cap S) &= (0.20)(0.25) = 0.0500 .. admonition:: Question 4b (6 pts) :class: note Calculate the overall probability :math:`P(S)` that a randomly selected shuttle trip experiences a sensor alert. .. dropdown:: Solution :class-container: sd-border-success The three routes partition the sample space, so by the Law of Partitions, .. math:: P(S) = P(A \cap S) + P(B \cap S) + P(C \cap S) = 0.0400 + 0.0450 + 0.0500 = 0.1350. .. admonition:: Question 4c (8 pts) :class: note Suppose a shuttle trip is known to have experienced a sensor alert (:math:`S`). Compute the probability that this shuttle was traveling on Route A. .. dropdown:: Solution :class-container: sd-border-success By Bayes' Theorem, .. math:: P(A \mid S) = \frac{P(A \cap S)}{P(S)} = \frac{0.0400}{0.1350} = \frac{8}{27} \approx 0.2963. .. admonition:: Question 4d (5 pts) :class: note Are the event :math:`\{\text{Shuttle is on Route A}\}` and the event :math:`\{\text{Sensor alert occurs}\}` independent? State YES or NO and provide clear mathematical justification. .. dropdown:: Solution :class-container: sd-border-success **NO.** Since :math:`P(S \mid A) = 0.0800 \neq P(S) = 0.1350`, the two events are dependent. Equivalently, :math:`P(A \cap S) = 0.0400`, while :math:`P(A)\,P(S) = (0.50)(0.1350) = 0.0675`. .. admonition:: Question 4e (3 pts) :class: note Suppose :math:`4` independent shuttle trips on Route C are monitored. What is the probability that at least one of these :math:`4` trips experiences a sensor alert? .. dropdown:: Solution :class-container: sd-border-success Let :math:`Y` be the number of the 4 trips with an alert. Each trip has an alert or not, the trips are independent, :math:`n = 4` is fixed, and each trip has the same alert probability :math:`P(S \mid C) = 0.25`, so :math:`Y \sim \text{Binomial}(n = 4, p = 0.25)`. Use the complement: .. math:: P(Y \geq 1) = 1 - P(Y = 0) = 1 - (0.75)^4 = 1 - 0.3164 = 0.6836. ---- Problem 5 (26 points) — Silicon Wafer Thickness -------------------------------------------------- .. admonition:: Problem 5 Setup :class: important High-precision silicon wafer substrates manufactured at the Birck Nanotechnology Center have a thickness :math:`X` (in micrometers) that follows a Normal distribution with unknown mean :math:`\mu` and unknown standard deviation :math:`\sigma`. Quality assurance testing on a large production run establishes the following two specifications: 1. Exactly :math:`10\%` of the manufactured wafers have a thickness strictly less than :math:`47.812` micrometers: :math:`P(X < 47.812) = 0.10`. 2. Exactly :math:`5\%` of the manufactured wafers have a thickness strictly greater than :math:`53.944` micrometers: :math:`P(X > 53.944) = 0.05`. Use a standard normal table with :math:`z` values rounded to two decimal places. If two table entries are equally close to a target cumulative area, use the smaller :math:`z` value. .. admonition:: Question 5a (8 pts) :class: note Calculate the population mean :math:`\mu` and standard deviation :math:`\sigma`. .. dropdown:: Solution :class-container: sd-border-success From the standard normal table, :math:`\Phi(-1.28) = 0.1003` is the entry closest to 0.10, so :math:`z = -1.28`. Specification 2 gives a cumulative area of :math:`P(X \leq 53.944) = 0.95`; :math:`\Phi(1.64) = 0.9495` and :math:`\Phi(1.65) = 0.9505` are equally close, and the stated tie rule selects :math:`z = 1.64`. Thus, .. math:: 47.812 = \mu - 1.28\sigma, \qquad 53.944 = \mu + 1.64\sigma. Subtracting the first equation from the second gives :math:`6.132 = 2.92\sigma`, so .. math:: \sigma = \frac{53.944 - 47.812}{1.64 + 1.28} = \frac{6.132}{2.92} = 2.1 \text{ micrometers}, .. math:: \mu = 47.812 + 1.28(2.1) = 50.5 \text{ micrometers}. .. admonition:: Fallback values for 5(b)–5(d) :class: important If you cannot determine :math:`\mu` and :math:`\sigma` in 5(a), you may use :math:`\mu = 50` micrometers and :math:`\sigma = 2.0` micrometers for parts 5(b)–5(d). **These are not the correct answers to 5(a)**; they are provided only so you can continue with the rest of the problem. If you use them, say so clearly in your work. You will not receive credit for 5(a) by using these values. .. admonition:: Question 5b (6 pts) :class: note Find the probability that a randomly chosen wafer has a thickness between :math:`47` micrometers and :math:`53` micrometers. .. dropdown:: Solution :class-container: sd-border-success With :math:`\mu = 50.5` and :math:`\sigma = 2.1`, the endpoints standardize to .. math:: z = \frac{47 - 50.5}{2.1} = -1.6667 \approx -1.67, \qquad z = \frac{53 - 50.5}{2.1} = 1.1905 \approx 1.19. The table gives :math:`\Phi(-1.67) = 0.0475` and :math:`\Phi(1.19) = 0.8830`, so .. math:: P(47 \leq X \leq 53) = \Phi(1.19) - \Phi(-1.67) = 0.8830 - 0.0475 = 0.8355. **With the fallback values,** the endpoints standardize to :math:`z = -1.50` and :math:`z = 1.50`, and :math:`P(47 \leq X \leq 53) = 0.9332 - 0.0668 = 0.8664`. .. admonition:: Question 5c (6 pts) :class: note Determine the first quartile :math:`Q_1`, the third quartile :math:`Q_3`, and the population Interquartile Range (IQR) for wafer thickness. .. dropdown:: Solution :class-container: sd-border-success The standard normal table gives :math:`\Phi(-0.67) = 0.2514` and :math:`\Phi(0.67) = 0.7486`, the entries closest to 0.25 and 0.75, so use quartile :math:`z` values :math:`\pm 0.67`: .. math:: Q_1 &\approx \mu - 0.67\sigma = 50.5 - 0.67(2.1) = 49.093 \text{ micrometers} \\ Q_3 &\approx \mu + 0.67\sigma = 50.5 + 0.67(2.1) = 51.907 \text{ micrometers} \\ \text{IQR} &= Q_3 - Q_1 = 2.814 \text{ micrometers} **With the fallback values,** :math:`Q_1 \approx 48.66` micrometers, :math:`Q_3 \approx 51.34` micrometers, and :math:`\text{IQR} \approx 2.68` micrometers. .. admonition:: Question 5d (6 pts) :class: note Calculate the population lower and upper inner fences. Using these fences as a quality-control rule, would a manufactured wafer with a thickness of :math:`44.20` micrometers be flagged as a potential outlier requiring further inspection? Explain in a complete sentence. .. dropdown:: Solution :class-container: sd-border-success .. math:: \text{Lower inner fence} &= Q_1 - 1.5\,\text{IQR} = 49.093 - 1.5(2.814) = 44.872 \text{ micrometers} \\ \text{Upper inner fence} &= Q_3 + 1.5\,\text{IQR} = 51.907 + 1.5(2.814) = 56.128 \text{ micrometers} Because :math:`44.20` micrometers is below the lower inner fence of :math:`44.872` micrometers, the wafer would be flagged as a potential outlier. **With the fallback values,** the fences are :math:`44.64` and :math:`55.36` micrometers. Because :math:`44.20 < 44.64`, the wafer would again be flagged as a potential outlier.