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GELE MathematicsEngineering Data Analysis (Probability and Statistics)Revision Notes

Final-week revision notes for Engineering Data Analysis (Probability and Statistics). If you have already studied the full chapter, this page is your go-to refresher before sitting the GELE. Compact, high-yield, and aligned with what Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests in the Mathematics subtest.

Exam context

On the GELE 2026, the Mathematics subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Geodetic Engineering's pattern. Engineering Data Analysis (Probability and Statistics) lands at position 9th out of 10 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Mathematics on a typical GELE paper.

Engineering Data Analysis (Probability and Statistics) - Revision Notes

Engineering Data Analysis is a consistently tested topic in the PRC Civil Engineer Licensure Examination. It encompasses descriptive statistics (measures of central tendency and dispersion), probability theory, counting techniques, and probability distributions. A solid command of these topics allows engineers to quantify uncertainty, analyze experimental data, and make informed design decisions — skills directly applicable to quality control, structural reliability, and geotechnical risk assessment. This revision note consolidates all exam-critical formulas, definitions, solved board-style problems, common pitfalls, and exam strategies into one efficient study reference.

Sections

Formulas

Example

Data: 5, 7, 7, 10, 16 → x̄ = (5+7+7+10+16)/5 = 45/5 = 9.0

Formula

x̄ = (Σx) / n

Variables

x̄ = sample mean; Σx = sum of all data values; n = number of observations

Application

Compute the arithmetic average of any ungrouped data set.

Example

Data: 2, 4, 6, 8, 10; x̄ = 6 → deviations² = 16, 4, 0, 4, 16 → σ² = 40/5 = 8; σ = √8 ≈ 2.83

Formula

σ² = Σ(x − μ)² / N [population]; s² = Σ(x − x̄)² / (n − 1) [sample]

Variables

σ² = population variance; s² = sample variance; μ = population mean; x̄ = sample mean; N or n = count

Application

Quantify how spread out values are from the mean. Use sample formula when data represent a subset of a larger population.

Example

Data: 5, 7, 7, 10, 16; x̄ = 9 → Σ(x−x̄)² = 16+4+4+1+49 = 74 → s² = 74/4 = 18.5 → s ≈ 4.30

Formula

s = √[Σ(x − x̄)² / (n − 1)]

Variables

s = sample standard deviation

Application

Board exam problems often specify 'sample' — always check whether to use n or n − 1.

Example

Data (ordered): 5, 7, 7, 10, 16 → n = 5 (odd) → median = 3rd value = 7

Formula

Median: middle value of ordered data; for even n, average the two middle values

Variables

Position of median = (n+1)/2 when n is odd

Application

Used when data contain extreme outliers that would distort the mean.

Exam Tips

  • Always re-read the problem: 'population' → divide by n; 'sample' → divide by (n − 1).
  • For small odd-numbered data sets, the median is the ((n+1)/2)th value — memorize this.
  • When computing sample variance by hand, use the computational formula s² = [Σx² − (Σx)²/n] / (n−1) to reduce arithmetic errors.
  • Board exams frequently give data sets of 5 to 7 numbers — practice these by hand.
  • If asked for 'the most appropriate measure of central tendency' for skewed data, the answer is the median.

Key Points

  • The three measures of central tendency are the mean (arithmetic average), median (middle value when data are ordered), and mode (most frequently occurring value).
  • For a data set 2, 4, 6, 8, 10: mean = 6, median = 6, mode = none (all values appear once).
  • Population variance uses divisor n; sample variance uses divisor (n − 1) — the Bessel correction that accounts for estimating a population parameter from a sample.
  • Standard deviation (σ or s) is the square root of variance and has the same units as the original data, making it more interpretable than variance.
  • When data are grouped, use the class midpoint as the representative value for each class interval.
  • Coefficient of variation CV = (σ/μ) × 100% allows comparison of dispersion across data sets with different units or magnitudes.
  • The range = maximum − minimum is the simplest but least informative measure of spread.

Definitions

Term

Mean (Arithmetic)

Definition

The sum of all data values divided by the number of observations.

Importance

Most common measure of center; used in variance and standard deviation calculations.

Term

Median

Definition

The middle value of an ordered data set; resistant to outliers.

Importance

Preferred when data are skewed (e.g., income distributions, construction cost data).

Term

Mode

Definition

The value(s) that appear most frequently. A data set may be unimodal, bimodal, or have no mode.

Importance

Useful for categorical data; identifies the most common result.

Term

Standard Deviation

Definition

The square root of the variance; measures average distance of data points from the mean.

Importance

Essential for normal distribution problems and quality control in engineering.

Term

Variance

Definition

The average of the squared deviations from the mean.

Importance

Foundation for inferential statistics and distribution analysis.

Section Title

1. Descriptive Statistics — Measures of Central Tendency and Dispersion

Common Mistakes

  • Using divisor n instead of (n − 1) when the problem states 'sample standard deviation' — this is a very frequent board exam trap.
  • Forgetting to order the data before finding the median.
  • Confusing 'no mode' with a mode of zero.
  • Computing variance but reporting it as standard deviation without taking the square root.
  • Not using class midpoints when computing the mean of grouped frequency data.

Formulas

Example

Rolling a 6-sided die: P(even) = 3/6 = 1/2 = 0.50

Formula

P(A) = favorable outcomes / total outcomes

Variables

Classical probability — assumes equally likely outcomes

Application

Dice, cards, coins, and random selection problems.

Example

P(King or Heart): P(K) = 4/52, P(H) = 13/52, P(K∩H) = 1/52 → P = 4/52 + 13/52 − 1/52 = 16/52 ≈ 0.308

Formula

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Variables

∪ = union (or); ∩ = intersection (and)

Application

Finding P(A or B) for any two events. If mutually exclusive, P(A ∩ B) = 0.

Example

P(all 4 independent units work) = (0.95)⁴ = 0.8145 ≈ 81.45%

Formula

P(A ∩ B) = P(A) × P(B) [independent events]

Variables

Valid only when events A and B are statistically independent

Application

Series/parallel reliability systems, repeated independent trials.

Example

P(at least one defective in 3 items, each 10% defective) = 1 − (0.90)³ = 1 − 0.729 = 0.271

Formula

P(A') = 1 − P(A)

Variables

A' = complement of A

Application

Extremely useful for 'at least one' problems — compute P(none) then subtract from 1.

Example

P(drawing 2nd ace | 1st was ace, no replacement) = 3/51 ≈ 0.0588

Formula

P(A | B) = P(A ∩ B) / P(B)

Variables

P(A|B) = conditional probability of A given B has occurred

Application

Problems involving dependent events, Bayes-type reasoning.

Exam Tips

  • Key words: 'OR' → addition rule; 'AND' → multiplication rule; 'given that' → conditional probability.
  • For reliability problems with independent components: series system P(works) = product of all; parallel system P(fails) = product of individual failure probabilities.
  • When drawing without replacement, probabilities change with each draw — use conditional probability.
  • Card problems: 52 cards total; 4 suits of 13 each; 4 of each face value — memorize the deck structure.
  • For dice: two dice → 36 total outcomes; enumerate favorable outcomes systematically.

Key Points

  • Probability is a number between 0 (impossible) and 1 (certain); P(sample space) = 1.
  • The classical definition: P(A) = (number of favorable outcomes) / (total number of equally likely outcomes).
  • The complement rule is one of the most powerful tools: P(not A) = 1 − P(A).
  • Two events are mutually exclusive if they cannot occur simultaneously: P(A ∩ B) = 0.
  • Two events are independent if the occurrence of one does not affect the other: P(A ∩ B) = P(A) × P(B).
  • Conditional probability P(A|B) = P(A ∩ B) / P(B) — the probability of A given that B has occurred.
  • Mutually exclusive ≠ independent. If A and B are mutually exclusive and both have P > 0, they are NOT independent.
  • For 'at least one' problems, use complement: P(at least one) = 1 − P(none).

Definitions

Term

Mutually Exclusive Events

Definition

Two events that cannot both occur in the same trial. P(A ∩ B) = 0.

Importance

Simplifies the addition rule: P(A ∪ B) = P(A) + P(B).

Term

Independent Events

Definition

Events where the occurrence of one does not change the probability of the other.

Importance

Allows multiplication rule: P(A ∩ B) = P(A) × P(B). Critical for reliability problems.

Term

Complement

Definition

The event consisting of all outcomes NOT in A.

Importance

Simplifies calculation of 'at least one' and 'not' type problems — always easier to compute P(none).

Term

Conditional Probability

Definition

The probability of event A occurring given that event B has already occurred.

Importance

Foundation for Bayes' Theorem and dependent event problems.

Section Title

2. Probability Theory — Rules and Calculations

Common Mistakes

  • Treating mutually exclusive events as independent — these are completely different concepts.
  • Forgetting to subtract P(A ∩ B) in the addition rule when events overlap (e.g., king AND heart).
  • Using the multiplication rule P(A)P(B) for dependent events (e.g., drawing without replacement).
  • Adding probabilities instead of multiplying for 'A AND B' with independent events.
  • Not recognizing that 'at least one' problems are most efficiently solved using the complement.

Formulas

Example

How many 3-digit codes from digits 1–9 (no repeat)? P(9,3) = 9!/(9−3)! = 9!/6! = 9×8×7 = 504

Formula

P(n, r) = n! / (n − r)!

Variables

n = total objects; r = objects selected; order MATTERS

Application

Arranging r items chosen from n distinct items. Used for passwords, race positions, officer elections.

Example

Choose 2 from 5 people for a committee: C(5,2) = 5!/(2!3!) = 120/12 = 10 ways

Formula

C(n, r) = n! / [r!(n − r)!]

Variables

n = total objects; r = objects selected; order DOES NOT matter

Application

Selecting r items from n without regard to arrangement. Used for committees, lotto, card hands.

Example

Arrangements of MISSISSIPPI: 11!/(4!4!2!) = 39,916,800 / 1152 = 34,650

Formula

Permutations of n objects with repetition: n! / (n₁! × n₂! × … × nₖ!)

Variables

n₁, n₂, … = number of identical objects of each type

Application

Arranging words with repeated letters (e.g., ENGINEER).

Example

5 people around a round table: (5−1)! = 4! = 24 ways

Formula

Circular permutation = (n − 1)!

Variables

n = number of objects arranged in a circle

Application

Seating arrangements around a round table.

Exam Tips

  • Quick check: 'arrange/rank/order/schedule/assign positions' → Permutation; 'choose/select/committee/team/group' → Combination.
  • C(n, r) = C(n, n−r): choosing 3 to include is the same as choosing (n−3) to exclude — use this to simplify large calculations.
  • P(n, n) = n! — arranging ALL n objects in a line.
  • For board exam speed: cancel factorials algebraically before multiplying — e.g., 9!/6! = 9×8×7 directly.
  • Lotto 6/42 problems: C(42,6) = 5,245,786 — know how to compute this rapidly.

Key Points

  • The fundamental principle of counting: if event A can occur in m ways and event B in n ways, together they occur in m × n ways.
  • Permutations count arrangements where ORDER MATTERS — think of ranking, positioning, assigning roles.
  • Combinations count selections where ORDER DOES NOT MATTER — think of committees, teams, groups.
  • Factorial: n! = n × (n−1) × (n−2) × … × 2 × 1; by definition, 0! = 1.
  • Distinguishing permutation vs. combination is the single most important skill in this topic area.
  • Permutations of identical objects: n! / (n₁! n₂! … nₖ!) accounts for repeated elements.
  • Circular permutations: (n−1)! — one element is fixed to remove rotational equivalence.

Definitions

Term

Permutation

Definition

An ordered arrangement of objects. Changing the order creates a different permutation.

Importance

Use when sequence or position matters: passwords, rankings, scheduling.

Term

Combination

Definition

A selection of objects without regard to order. Rearranging the same objects gives the same combination.

Importance

Use when only membership in a group matters: committees, lottery numbers, card hands.

Term

Factorial (n!)

Definition

The product of all positive integers from 1 to n. Defined as 0! = 1.

Importance

Building block of all permutation and combination calculations.

Section Title

3. Counting Techniques — Permutations and Combinations

Common Mistakes

  • Using permutation formula when combination is required (and vice versa) — always ask 'does order matter?'
  • Forgetting that 0! = 1, which is needed when r = n in permutation problems.
  • Not dividing by repeated-element factorials when arranging objects that are not all distinct.
  • Confusing circular permutation (n−1)! with linear permutation n!.
  • In lotto-type problems, automatically assuming order does not matter — verify from context.

Formulas

Example

P(exactly 3 heads in 5 coin flips): C(5,3)(0.5)³(0.5)² = 10 × 0.125 × 0.25 = 0.3125

Formula

P(X = x) = C(n,x) · pˣ · (1−p)^(n−x)

Variables

n = number of trials; x = number of successes; p = probability of success per trial

Application

Finding the probability of exactly x successes in n independent Bernoulli trials.

Example

n = 20 trials, p = 0.3 → μ = 6; σ² = 20(0.3)(0.7) = 4.2; σ = √4.2 ≈ 2.05

Formula

μ = np; σ² = np(1−p); σ = √[np(1−p)]

Variables

Mean and standard deviation of the Binomial distribution

Application

Quickly find central value and spread without computing the full distribution.

Example

Compressive strength: μ = 28 MPa, σ = 3 MPa. P(X < 25) → z = (25−28)/3 = −1.00 → from z-table: 0.1587

Formula

z = (x − μ) / σ

Variables

z = standard normal score (z-score); x = observed value; μ = mean; σ = std dev

Application

Convert any normally distributed variable to the standard normal for table lookup.

Example

Average 2 cracks per 10 m of weld. P(exactly 3 cracks in 10 m): P = e^(−2)(2³)/3! = (0.1353)(8)/6 ≈ 0.1804

Formula

P(X = x) = (e^(−λ) · λˣ) / x!

Variables

λ = average rate of occurrence (mean); e ≈ 2.71828; x = number of events

Application

Modeling arrivals, defects per unit length, calls per hour, etc.

Exam Tips

  • Binomial checklist: (1) fixed n, (2) two outcomes, (3) constant p, (4) independent. All four must be true.
  • Normal distribution: ALWAYS sketch the bell curve and shade the region before computing z-scores.
  • Symmetry of normal: P(Z < −a) = P(Z > a) = 1 − P(Z < a). This halves your table lookups.
  • Poisson: if the problem gives an average rate (events per unit) and asks for exact count probability → Poisson.
  • 68-95-99.7 rule: if asked about probability within 1, 2, or 3 standard deviations — use this rule directly without table.
  • Binomial mean np is the expected value — also the answer to 'how many are expected to succeed?'

Key Points

  • A probability distribution describes all possible outcomes of a random variable and their associated probabilities.
  • The Binomial distribution applies when: (1) fixed number of trials n, (2) each trial has exactly two outcomes (success/failure), (3) P(success) = p is constant, (4) trials are independent.
  • The Normal (Gaussian) distribution is symmetric and bell-shaped, fully described by its mean μ and standard deviation σ.
  • The Empirical Rule (68-95-99.7 Rule): approximately 68% of data fall within μ ± σ, 95% within μ ± 2σ, and 99.7% within μ ± 3σ.
  • The Standard Normal Distribution has μ = 0 and σ = 1; standardize any normal variable using z = (x − μ)/σ.
  • The Poisson distribution models the number of rare, random events occurring in a fixed interval of time or space.
  • For the binomial distribution: mean μ = np; variance σ² = np(1−p); std dev σ = √[np(1−p)].
  • For the Poisson distribution: mean = λ; variance = λ (mean equals variance — a key identifying feature).

Definitions

Term

Binomial Distribution

Definition

A discrete probability distribution for the number of successes in n independent trials, each with success probability p.

Importance

Most common discrete distribution in board exams; applies to quality control and reliability scenarios.

Term

Normal Distribution

Definition

A continuous, symmetric, bell-shaped probability distribution defined by mean μ and standard deviation σ.

Importance

Foundation of statistical inference; many engineering measurements are approximately normal.

Term

Z-Score (Standard Normal Score)

Definition

The number of standard deviations a value x is from the mean: z = (x − μ)/σ.

Importance

Allows the use of the standard normal table (z-table) for any normal distribution problem.

Term

Poisson Distribution

Definition

A discrete distribution for the count of rare, independent events in a fixed interval, characterized by rate λ.

Importance

Used for defect counts, accident rates, and queue modeling — mean equals variance.

Term

68-95-99.7 Empirical Rule

Definition

For a normal distribution: 68.27% of data within ±1σ, 95.45% within ±2σ, 99.73% within ±3σ.

Importance

Quick approximation without a z-table; very commonly tested in board exams.

Section Title

4. Probability Distributions — Binomial, Normal, and Poisson

Common Mistakes

  • Applying the binomial formula when trials are not independent (e.g., drawing without replacement from a small population — use hypergeometric instead).
  • Forgetting to check all four conditions of the binomial distribution before applying its formula.
  • Using the wrong tail of the z-table: P(X < a) vs P(X > a) vs P(a < X < b) — sketch the curve first.
  • Confusing P(X = x) with P(X ≤ x) — binomial point probability vs. cumulative probability.
  • Misidentifying a Poisson problem as binomial because the problem mentions 'probability' — look for rare events over an interval.
  • Not converting negative z-scores correctly: P(Z < −1) = P(Z > 1) by symmetry = 1 − P(Z < 1).

Connections

  • Descriptive statistics (mean, std dev) are the input parameters for Normal distribution problems — you must be able to compute μ and σ before applying z-scores.
  • The Binomial distribution is the parent of both the Normal approximation (when n is large: np > 5 and n(1−p) > 5) and the Poisson approximation (when n is large and p is small, λ = np).
  • Combinations C(n,r) appear directly inside the Binomial probability formula — mastery of counting is prerequisite to distribution problems.
  • Probability rules (addition, multiplication, complement) are applied within distribution problems — e.g., P(X ≥ 3) = 1 − P(X ≤ 2) uses the complement rule within a binomial context.
  • Standard deviation from descriptive statistics becomes σ in the z-score formula for Normal distribution — the two topics are mathematically linked.
  • The complement rule P(not A) = 1 − P(A) is the bridge between simple probability and 'at least one' type problems, which appear in both basic probability and binomial distribution sections.
  • Quality control in structural engineering (NSCP 2015 concrete mix design) relies on Normal distribution to set acceptance criteria — connecting probability theory to code applications.
  • Conditional probability underpins Bayes' Theorem, which connects to reliability engineering — the probability that a structure has failed given that an inspection signal was observed.

Exam Strategy

Approach Engineering Data Analysis problems systematically: (1) READ the problem twice and classify it — is it descriptive statistics, pure probability, counting, or a named distribution? (2) For descriptive statistics, immediately identify whether 'population' (use n) or 'sample' (use n−1) applies — this is the most common source of error. (3) For probability, map out key words: 'or' → addition rule, 'and' → multiplication rule, 'given that' → conditional, 'at least one' → complement. (4) For counting, ask yourself immediately: 'Does order matter?' — YES → Permutation, NO → Combination. (5) For distributions: check all four binomial conditions; for normal problems, ALWAYS sketch the bell curve and shade the region before computing z; for Poisson, look for an average rate per interval. (6) In the actual board exam, allocate roughly 2 to 3 minutes per problem; if a distribution problem requires extensive calculation, attempt other sections first and return. (7) Memorize the 68-95-99.7 empirical rule so you can answer normal distribution questions WITHOUT a z-table when the problem asks about exactly ±1σ, ±2σ, or ±3σ intervals. (8) Check your answer for reasonableness — probabilities must be between 0 and 1, combinations must be positive integers, and standard deviations must be positive. (9) Practice converting between percentage and probability (e.g., 95% reliability = P = 0.95) — engineering problems often state reliabilities as percentages. (10) For the PRC CE board exam, this topic typically accounts for 5–10 items; securing all of them by mastering the fundamentals covered in these notes can make a decisive difference in your overall score.

Quick Review Questions

The compressive strength of 5 concrete cylinders (in MPa) is 24, 27, 30, 27, 32. What is the sample standard deviation?

Step 1: x̄ = (24+27+30+27+32)/5 = 140/5 = 28 MPa. Step 2: Σ(x−x̄)² = (−4)²+(−1)²+(2)²+(−1)²+(4)² = 16+1+4+1+16 = 38. Step 3: s² = 38/(5−1) = 38/4 = 9.5. Step 4: s = √9.5 ≈ 3.08 MPa. Note: Use n−1 because this is a sample.

A card is drawn at random from a standard 52-card deck. What is the probability of drawing a king or a heart?

P(King) = 4/52; P(Heart) = 13/52; P(King AND Heart) = 1/52 (king of hearts). By the addition rule: P(King ∪ Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13 ≈ 0.308. The events are NOT mutually exclusive — the king of hearts belongs to both sets.

In how many ways can 5 different engineering textbooks be arranged on a shelf?

This is a permutation of all 5 distinct books: P(5,5) = 5! = 5×4×3×2×1 = 120. Order matters because different arrangements are considered distinct.

A quality control inspector checks 10 products, each independently having a 20% chance of being defective. Find P(exactly 2 defective).

Binomial: n = 10, x = 2, p = 0.20. P(X=2) = C(10,2)(0.20)²(0.80)⁸ = 45 × 0.04 × 0.16777 = 45 × 0.006711 ≈ 0.3020.

Rebar yield strength is normally distributed with μ = 415 MPa and σ = 20 MPa. What percentage of rebars have yield strength between 395 MPa and 435 MPa?

z₁ = (395−415)/20 = −1.00; z₂ = (435−415)/20 = +1.00. The interval [395, 435] is exactly μ ± 1σ. By the Empirical Rule (68-95-99.7), approximately 68.27% of values fall within one standard deviation of the mean.

A committee of 3 engineers is to be selected from 8 candidates. How many different committees are possible?

Order does not matter (a committee, not a ranking). Use combinations: C(8,3) = 8!/(3!×5!) = (8×7×6)/(3×2×1) = 336/6 = 56.

An industrial pump has a reliability of 0.92. If 3 identical independent pumps are installed in parallel (system works if AT LEAST ONE pump works), what is the system reliability?

Use complement: P(system fails) = P(all 3 pumps fail) = (1−0.92)³ = (0.08)³ = 0.000512. P(system works) = 1 − 0.000512 = 0.999488 ≈ 99.95%. This demonstrates why parallel redundancy dramatically improves reliability.

On average, 3 trucks arrive at a weigh station per hour. Using the Poisson distribution, find P(exactly 5 trucks in one hour).

λ = 3, x = 5. P(X=5) = e^(−3) × 3⁵ / 5! = (0.04979 × 243) / 120 = 12.0997 / 120 ≈ 0.1008. Key check: this is a Poisson problem because we have a rate (average count per interval) with rare, independent events.

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