GELE Mathematics — Engineering Data Analysis (Probability and Statistics)Study Notes

Descriptive statistics allow us to summarize large datasets with a few meaningful numbers. In civil engineering, we constantly collect data: concrete compressive strength from test cylinders, steel yield stresses, traffic volumes, settlement measurements. We need ways to describe these datasets concisely. **Measures of Central Tendency (Location)** The mean (arithmetic average) is the most common measure: $$\bar{x} = \frac{\sum x}{n} = \frac{x_1 + x_2 + \cdots + x_n}{n}$$ For example, if we test 5 concrete cylinders and obtain compressive strengths (in MPa) of 28, 31, 29, 32, 30, the mean is: $$\bar{x} = \frac{28 + 31 + 29 + 32 + 30}{5} = \frac{150}{5} = 30 \text{ MPa}$$ The **median** is the middle value when data are arranged in order. For the concrete example (arranged: 28, 29, 30, 31, 32), the median is 30 MPa. The median is useful when outliers may skew the mean — if one cylinder fails catastrophically at 8 MPa, the mean becomes 24 MPa but the median remains 29 MPa, better representing typical strength. The **mode** is the most frequently occurring value. If we had strengths 28, 28, 30, 32, 35 MPa, the mode is 28 MPa. Less common in continuous engineering data but useful for categorical data (e.g., most common failure mode). **Measures of Spread (Dispersion)** Variability tells us how much individual measurements deviate from the average — critical for assessing consistency and setting safety factors. **Variance** measures average squared deviation from the mean: - **Population variance** (entire dataset): $$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$$ - **Sample variance** (sample from larger population): $$s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$$ We use $n-1$ for samples because a sample variance tends to underestimate the population variance; dividing by $n-1$ provides an unbiased estimator. In professional laboratory testing (ACI 318), when you test a sample of concrete cylinders, you use sample statistics. **Standard Deviation** is the square root of variance and has the same units as the data: $$\sigma = \sqrt{\sigma^2}, \quad s = \sqrt{s^2}$$ Standard deviation is more intuitive: it represents a typical deviation from the mean. **Worked Example 1.1: Concrete Strength Analysis** A structural engineer tests 6 concrete cylinders and obtains compressive strengths (MPa): 32, 35, 31, 36, 33, 35. Calculate the mean, median, and sample standard deviation. **Solution:** Step 1: Calculate the mean. $$\bar{x} = \frac{32 + 35 + 31 + 36 + 33 + 35}{6} = \frac{202}{6} = 33.67 \text{ MPa}$$ Step 2: Find the median. Arrange in order: 31, 32, 33, 35, 35, 36. With 6 values (even), the median is the average of the 3rd and 4th values: $$\text{Median} = \frac{33 + 35}{2} = 34 \text{ MPa}$$ Step 3: Calculate deviations from the mean and sum of squares. - $x_1 = 32$: $(32 - 33.67)^2 = 2.79$ - $x_2 = 35$: $(35 - 33.67)^2 = 1.78$ - $x_3 = 31$: $(31 - 33.67)^2 = 7.13$ - $x_4 = 36$: $(36 - 33.67)^2 = 5.44$ - $x_5 = 33$: $(33 - 33.67)^2 = 0.45$ - $x_6 = 35$: $(35 - 33.67)^2 = 1.78$ $$\sum (x_i - \bar{x})^2 = 2.79 + 1.78 + 7.13 + 5.44 + 0.45 + 1.78 = 19.37$$ Step 4: Calculate sample variance and standard deviation. $$s^2 = \frac{19.37}{6-1} = \frac{19.37}{5} = 3.87 \text{ (MPa)}^2$$ $$s = \sqrt{3.87} = 1.97 \text{ MPa}$$ **Answer:** Mean = 33.67 MPa, Median = 34 MPa, Sample Std Dev = 1.97 MPa. The low standard deviation indicates consistent concrete quality. **Worked Example 1.2: Coefficient of Variation** The coefficient of variation (CV) expresses standard deviation as a percentage of the mean, useful for comparing variability across different scales: $$CV = \frac{s}{\bar{x}} \times 100\%$$ For the concrete test above: $$CV = \frac{1.97}{33.67} \times 100\% = 5.85\%$$ A CV of ~5–6% is typical for well-controlled concrete testing (ACI 318-19). A CV > 10% might indicate problems with batching, curing, or testing procedures.

Probability is the mathematical language of uncertainty. In civil engineering, we use probability to answer questions like: What is the probability that a steel member reaches its design yield stress? What is the probability that annual rainfall exceeds 2000 mm? What is the probability that a foundation will settle more than 25 mm? **Basic Probability Definitions** For an event A: $$P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of equally likely outcomes}}$$ Probability ranges from 0 (impossible) to 1 (certain): $0 \le P(A) \le 1$. Example: When rolling a fair die, the probability of rolling a 4 is: $$P(\text{rolling a 4}) = \frac{1}{6} \approx 0.1667$$ In structural engineering, we often work with the complement — the probability that something does NOT happen: $$P(\text{not } A) = 1 - P(A)$$ For instance, if a structure has a 95% probability of meeting its service load, the probability it does NOT meet service load is: $$P(\text{failure}) = 1 - 0.95 = 0.05 = 5\%$$ **Addition Rule: Probability of OR (Union)** When two events can occur, the probability that event A OR event B occurs is: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ The overlap $P(A \cap B)$ is subtracted to avoid double-counting. If A and B are **mutually exclusive** (cannot both happen), then $P(A \cap B) = 0$, so: $$P(A \cup B) = P(A) + P(B) \quad \text{(mutually exclusive)}$$ **Multiplication Rule: Probability of AND (Intersection)** For independent events (occurrence of one does not affect the other): $$P(A \cap B) = P(A) \cdot P(B)$$ For dependent events, we use conditional probability: $$P(A \cap B) = P(A) \cdot P(B|A)$$ where $P(B|A)$ is the probability of B given that A has occurred. **Worked Example 2.1: Structural Reliability** A bridge has three independent cable-stayed support systems. Each system has a 98% probability of functioning correctly. What is the probability that all three systems function (assume this is required for safe operation)? **Solution:** Since the systems are independent and we need ALL three to function (AND operation): $$P(\text{all 3 work}) = P(\text{System 1 works}) \times P(\text{System 2 works}) \times P(\text{System 3 works})$$ $$P(\text{all 3 work}) = 0.98 \times 0.98 \times 0.98 = (0.98)^3 = 0.9412 = 94.12\%$$ **Answer:** The probability that all three systems function is 94.12%. **Physical Interpretation:** Even though each system is 98% reliable, the combined system reliability drops to 94% because failure of any single system causes overall failure. This is why redundancy is critical in safety-critical structures. **Worked Example 2.2: Card Probability (Union with Overlap)** A card is drawn from a standard 52-card deck. Find the probability that it is either a King OR a Heart (or both). **Solution:** Let A = "King" and B = "Heart." - There are 4 Kings in the deck: $P(A) = 4/52$ - There are 13 Hearts in the deck: $P(B) = 13/52$ - There is 1 King of Hearts (overlap): $P(A \cap B) = 1/52$ Using the addition rule: $$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \approx 0.3077$$ **Answer:** Probability ≈ 30.77% or 4/13. **Common Mistake:** If you forgot to subtract the overlap, you would get (4+13)/52 = 17/52 ≈ 32.7%, which is incorrect because you counted the King of Hearts twice. **Worked Example 2.3: Conditional Probability in Quality Control** A concrete batch fails strength testing 5% of the time. If a batch fails the first test, the probability it will also fail a second independent test is 5%. What is the probability that a batch fails both tests? **Solution:** Let A = "fails first test" and B = "fails second test." Since the tests are independent: $$P(A \cap B) = P(A) \times P(B) = 0.05 \times 0.05 = 0.0025 = 0.25\%$$ **Answer:** Only 0.25% chance of failing both tests. This is why retesting is valuable — a single failure may be a fluke, but two failures in a row is strong evidence of a genuine problem with the batch.

Many probability problems require counting outcomes. How many ways can you arrange design alternatives? How many ways can a committee select members? Counting principles provide systematic methods. **Fundamental Counting Principle** If an event has $m$ possible outcomes and another independent event has $n$ possible outcomes, the total number of combined outcomes is $m \times n$. Example: A structural design requires choosing one of 4 steel types AND one of 3 connection types. Total possible designs = $4 \times 3 = 12$. **Permutations: Order Matters** A **permutation** is an arrangement of objects where order is important. The number of permutations of $n$ objects taken $r$ at a time is: $$P(n,r) = \frac{n!}{(n-r)!}$$ where $n! = n \times (n-1) \times (n-2) \times \cdots \times 1$ (read "n factorial"). Example: In how many ways can 3 workers be assigned to 3 different concrete test stations? - First station: 3 choices - Second station: 2 remaining choices - Third station: 1 remaining choice - Total: $3 \times 2 \times 1 = 3! = 6$ ways Using the formula: $P(3,3) = \frac{3!}{(3-3)!} = \frac{6}{1} = 6$. **Combinations: Order Does Not Matter** A **combination** is a selection of objects where order does not matter. The number of combinations of $n$ objects taken $r$ at a time is: $$C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$ Example: From 5 available design proposals, a committee must choose 2 to develop further. Order does not matter (choosing proposals A and B is the same as choosing B and A). The number of ways is: $$C(5,2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \times 3!} = \frac{120}{2 \times 6} = \frac{120}{12} = 10$$ If order DID matter (e.g., first choice receives funding, second receives partial funding), we would use: $$P(5,2) = \frac{5!}{(5-2)!} = \frac{120}{6} = 20$$ Key Insight: Combinations $C(n,r)$ are always smaller than permutations $P(n,r)$ because $P(n,r) = C(n,r) \times r!$ (we multiply by the number of arrangements of the $r$ selected items). **Worked Example 3.1: Arrangement of Structural Elements** A structural engineer must arrange 4 different load-bearing columns in a specific sequence in a warehouse design. How many different sequential arrangements are possible? **Solution:** This is a permutation problem because the order (sequence) matters — each column position affects load distribution. $$P(4,4) = \frac{4!}{(4-4)!} = \frac{4!}{0!} = 4! = 4 \times 3 \times 2 \times 1 = 24$$ **Answer:** There are 24 possible sequential arrangements. **Physical Interpretation:** Changing the column sequence changes load paths and deflection patterns, so position matters significantly. **Worked Example 3.2: Committee Selection** A construction safety committee is formed by selecting 3 supervisors from 8 available candidates. How many different committees are possible? **Solution:** This is a combination problem because the order of selection does not matter — each committee member has the same role (we are not assigning ranks). $$C(8,3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$$ **Answer:** There are 56 possible committees. **Check:** If order mattered (assigning chairperson, vice-chairperson, secretary), we would get: $$P(8,3) = \frac{8!}{5!} = 8 \times 7 \times 6 = 336$$ Indeed, $336 = 56 \times 6 = C(8,3) \times 3!$ ✓ **Worked Example 3.3: Probability Using Combinations** A batch of 20 steel samples contains 3 with slightly lower yield strength (but still acceptable). An inspector randomly selects 5 samples to test. What is the probability that exactly 1 of the 5 selected samples has lower yield strength? **Solution:** Total ways to select 5 from 20: $$C(20,5) = \frac{20!}{5! \times 15!} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = \frac{1,860,480}{120} = 15,504$$ Ways to select exactly 1 lower-strength sample (from 3 available) and 4 normal samples (from 17 available): $$C(3,1) \times C(17,4) = 3 \times \frac{17!}{4! \times 13!} = 3 \times \frac{17 \times 16 \times 15 \times 14}{4 \times 3 \times 2 \times 1} = 3 \times 2,380 = 7,140$$ Probability: $$P(\text{exactly 1 lower}) = \frac{7,140}{15,504} \approx 0.461 = 46.1\%$$ **Answer:** Approximately 46.1% chance that exactly 1 of the 5 sampled bars is lower-strength.

A probability distribution describes all possible outcomes of an uncertain event and their probabilities. In civil engineering, distributions help us model material properties, environmental data, and component failures. **Binomial Distribution: Fixed Trials, Two Outcomes** The binomial distribution applies when: - There are $n$ independent trials (tests, attempts). - Each trial has exactly two outcomes: success (probability $p$) or failure (probability $1-p$). - The probability $p$ is constant across all trials. - We want to find the probability of exactly $x$ successes. The probability mass function is: $$P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$$ where $\binom{n}{x} = C(n,x) = \dfrac{n!}{x!(n-x)!}$ is the number of ways to choose $x$ successes from $n$ trials. The mean (expected value) of a binomial distribution is: $$\mu = np$$ The standard deviation is: $$\sigma = \sqrt{np(1-p)}$$ **Example Context:** A contractor uses a concrete mix that has an 85% probability of meeting strength specifications. If 10 batches are made, what is the probability that exactly 8 batches meet specifications? Here, $n = 10$, $x = 8$, $p = 0.85$. **Worked Example 4.1: Binomial Probability** A structural steel component supplier reports a defect rate of 2% (98% acceptance). A contractor orders 5 components. What is the probability that exactly 1 component is defective? **Solution:** This is a binomial problem with: - $n = 5$ trials (5 components ordered) - $x = 1$ successes desired (exactly 1 defect) - $p = 0.02$ (probability of defect) - $1 - p = 0.98$ (probability of no defect) $$P(X = 1) = \binom{5}{1} (0.02)^1 (0.98)^4$$ Step 1: Calculate the binomial coefficient. $$\binom{5}{1} = \frac{5!}{1! \times 4!} = 5$$ Step 2: Calculate the probability terms. $$(0.02)^1 = 0.02$$ $$(0.98)^4 = 0.9224$$ Step 3: Multiply. $$P(X = 1) = 5 \times 0.02 \times 0.9224 = 0.0922 = 9.22\%$$ **Answer:** The probability that exactly 1 component (out of 5) is defective is approximately 9.22%. **Interpretation:** There is about a 1-in-11 chance of finding exactly one defect, which is reasonable given the low 2% defect rate. **Normal (Gaussian) Distribution: Continuous Data** The normal distribution is the most important distribution in statistics. Many engineering properties approximately follow a normal distribution: concrete strength, steel yield stress, building settlements, traffic volumes. The normal distribution is symmetric, bell-shaped, and completely described by two parameters: - **Mean** $\mu$ (center of the distribution) - **Standard deviation** $\sigma$ (spread of the distribution) Probability is found using the **standard normal variable** (Z-score): $$z = \frac{x - \mu}{\sigma}$$ where $x$ is a measurement, $\mu$ is the population mean, and $\sigma$ is the population standard deviation. This $z$ value is then looked up in a standard normal table to find the probability. **The 68-95-99.7 Rule (Empirical Rule):** - About 68% of data falls within $\mu \pm 1\sigma$ - About 95% of data falls within $\mu \pm 2\sigma$ - About 99.7% of data falls within $\mu \pm 3\sigma$ This rule is very useful for quick estimates without tables. **Worked Example 4.2: Concrete Strength Probability** Concrete test data from a project shows that compressive strength approximately follows a normal distribution with mean $\mu = 35$ MPa and standard deviation $\sigma = 3$ MPa. What is the probability that a randomly selected test cylinder has strength between 32 and 38 MPa? **Solution:** Step 1: Convert the strength values to Z-scores. For $x = 32$ MPa: $$z_1 = \frac{32 - 35}{3} = \frac{-3}{3} = -1$$ For $x = 38$ MPa: $$z_2 = \frac{38 - 35}{3} = \frac{3}{3} = 1$$ Step 2: Use the 68-95-99.7 rule or a standard normal table. We need $P(-1 \le Z \le 1)$. According to the empirical rule, about 68% of the data lies within $\mu \pm 1\sigma$, which means between $z = -1$ and $z = +1$. **Answer:** The probability is approximately 0.68 or 68% that a cylinder has strength between 32 and 38 MPa. **Physical Interpretation:** If you test many cylinders, you expect roughly 2 out of 3 to fall in this strength range. **Worked Example 4.3: Setting Acceptance Criteria** A structural design requires concrete with a minimum strength of 28 MPa. Data from past projects shows strengths follow a normal distribution with $\mu = 35$ MPa and $\sigma = 2.5$ MPa. What percentage of batches are expected to fail (fall below 28 MPa)? **Solution:** Step 1: Calculate the Z-score for the minimum strength. $$z = \frac{28 - 35}{2.5} = \frac{-7}{2.5} = -2.8$$ Step 2: Find $P(Z \le -2.8)$ using a standard normal table (or software). From a standard normal table, $P(Z \le -2.8) \approx 0.0026$ or about 0.26%. **Answer:** Only about 0.26% of batches are expected to fall below 28 MPa — a very low failure rate. This indicates the design has a strong safety margin. **Practical Note:** In quality assurance (ACI 318), if observed failures exceed expected theoretical failures, this signals a process problem (e.g., poor curing, contaminated materials). **Poisson Distribution: Rare Events Over Time/Space** The Poisson distribution models the number of rare events occurring over a fixed interval (time or space). Examples: - Number of potholes in a 10 km road section - Number of cracks in a 100 m³ concrete pour - Number of traffic accidents at an intersection per month The Poisson distribution is characterized by a single parameter $\lambda$ (lambda), the average rate of occurrence. The probability of exactly $x$ events is: $$P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$$ where $e \approx 2.718$ (Euler's number), $\lambda$ is the average number of events in the interval, and $x!$ is factorial. **Worked Example 4.4: Pothole Distribution** Historical data shows an average of 3.5 potholes per 10 km road section. What is the probability of exactly 5 potholes in a randomly selected 10 km section? **Solution:** $$P(X = 5) = \frac{e^{-3.5} \times 3.5^5}{5!}$$ Step 1: Calculate $e^{-3.5} \approx 0.0302$. Step 2: Calculate $3.5^5 = 525.22$. Step 3: Calculate $5! = 120$. Step 4: Compute the probability. $$P(X = 5) = \frac{0.0302 \times 525.22}{120} = \frac{15.86}{120} \approx 0.132 = 13.2\%$$ **Answer:** There is approximately a 13.2% probability of finding exactly 5 potholes in a 10 km section. **Expected Value:** For a Poisson distribution, the mean is $E[X] = \lambda = 3.5$, so on average you expect 3.5 potholes per 10 km. The variance is also $\lambda = 3.5$.

In real engineering projects, we rarely have complete data for an entire population. Instead, we collect a sample and use it to estimate population parameters. Confidence intervals and hypothesis testing are tools for quantifying uncertainty in these estimates. **Confidence Intervals: Range of Plausible Values** A confidence interval is a range of values that likely contains the true population parameter (e.g., true mean strength) with a specified confidence level. For a sample mean, when the population standard deviation $\sigma$ is known (or estimated by sample $s$), the confidence interval is: $$\bar{x} \pm z^* \frac{s}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean - $z^*$ is the critical Z-value (depends on confidence level) - $s$ is the sample standard deviation - $n$ is the sample size - $s/\sqrt{n}$ is the standard error Common confidence levels and their $z^*$ values: - 90% confidence: $z^* = 1.645$ - 95% confidence: $z^* = 1.96$ - 99% confidence: $z^* = 2.576$ **Interpretation:** A 95% confidence interval means: if we repeated our sampling 100 times and calculated an interval each time, about 95 of those intervals would contain the true population mean (and 5 would not). **Worked Example 5.1: Concrete Strength Confidence Interval** An engineer tests 25 concrete cylinders and obtains a mean strength $\bar{x} = 33.2$ MPa with sample standard deviation $s = 2.1$ MPa. Calculate a 95% confidence interval for the true mean strength. **Solution:** Step 1: Identify the parameters. - $\bar{x} = 33.2$ MPa - $s = 2.1$ MPa - $n = 25$ - Confidence level = 95%, so $z^* = 1.96$ Step 2: Calculate the standard error. $$\text{SE} = \frac{s}{\sqrt{n}} = \frac{2.1}{\sqrt{25}} = \frac{2.1}{5} = 0.42 \text{ MPa}$$ Step 3: Calculate the margin of error. $$\text{ME} = z^* \times \text{SE} = 1.96 \times 0.42 = 0.82 \text{ MPa}$$ Step 4: Construct the interval. $$\text{CI} = \bar{x} \pm \text{ME} = 33.2 \pm 0.82 = [32.38, 34.02] \text{ MPa}$$ **Answer:** We are 95% confident that the true mean concrete strength lies between 32.38 and 34.02 MPa. **Practical Use:** If the specification requires a minimum strength of 32 MPa, the lower bound (32.38 MPa) exceeds this, suggesting the batch meets requirements with high confidence. **Hypothesis Testing: Is a Claim Supported by Data?** Hypothesis testing is used to determine whether data support or contradict a claim about a population parameter. Basic structure: 1. **Null Hypothesis ($H_0$):** A claim about the population (e.g., "the true mean strength equals 35 MPa"). 2. **Alternative Hypothesis ($H_a$ or $H_1$):** What we believe if the null is false (e.g., "the true mean strength is NOT 35 MPa"). 3. **Test Statistic:** A value calculated from the sample (e.g., Z-score). 4. **P-value:** The probability of observing the sample data (or more extreme) if the null hypothesis is true. 5. **Decision:** If the p-value is very small (typically < 0.05), reject the null hypothesis. **Worked Example 5.2: Testing a Design Specification** A specification requires steel beams to have a mean yield stress of at least 250 MPa. A sample of 20 beams shows $\bar{x} = 248$ MPa with $s = 5$ MPa. Is there evidence at the 5% significance level that the true mean falls below 250 MPa? **Solution:** Step 1: Set up the hypotheses. - $H_0$: $\mu = 250$ MPa (meets specification) - $H_a$: $\mu < 250$ MPa (fails specification, one-tailed test) Step 2: Calculate the test statistic (t-score, since population $\sigma$ is unknown). $$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{248 - 250}{5/\sqrt{20}} = \frac{-2}{1.118} = -1.789$$ Step 3: Compare to the critical t-value. With $df = n-1 = 19$ degrees of freedom and $\alpha = 0.05$ (one-tailed), the critical value is approximately $t_{crit} = -1.729$. Since $t = -1.789 < -1.729$, the test statistic falls in the rejection region. **Answer:** We reject the null hypothesis at the 5% significance level. There is statistically significant evidence that the true mean yield stress is below 250 MPa. This suggests a problem with the batch — perhaps improper heat treatment or alloy composition — and the engineer should investigate. **Type I and Type II Errors:** - **Type I Error:** Rejecting a true null hypothesis (false positive). Probability = $\alpha$ (significance level). If we set $\alpha = 0.05$, we accept a 5% chance of Type I error. - **Type II Error:** Failing to reject a false null hypothesis (false negative). Probability = $\beta$. Related to statistical power = $1 - \beta$. In safety-critical applications (e.g., bridge structures), engineers often choose very low $\alpha$ (e.g., 0.01) to minimize the risk of incorrectly accepting unsafe materials.