Expected Values
Probability and Statistics · FE Reference Handbook section
Learning objectives
What you must be able to do before leaving this section.
This chapter section covers Expected Values within Probability and Statistics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.
- Explain, in your own words, what expected values describes physically and when it applies.
- State every one of the 11 relations the handbook lists here and name each symbol with its unit.
- Select the correct relation from the wording of an exam stem within 20 seconds.
- Carry a complete solution from givens to a "most nearly" answer with the correct unit.
- Recognise the distractors generated by the unit trap: probabilities are dimensionless and must land in [0, 1].
Lecture
Why this section exists. Expected Values is the part of Probability and Statistics that lets you connect a sample of measurements from a construction or materials process to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.
How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.
How it is examined. Items from this page are written as one distribution or one counting rule, then a single probability or interval. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.
The habit that earns the points. Unit discipline. probabilities are dimensionless and must land in [0, 1]. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.
How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.
Photo 1. Where this shows up in practice: expected values.
Capstone Studio instructional photograph
Probability and Statistics — Expected Values: reference schematic for orienting the symbols used in this section.
Theory, developed
Read this before the equations — it is what makes them memorable.
The physical situation
Every item from this section describes a sample of measurements from a construction or materials process. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.
The governing principle
The 11 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.
Assumptions and limits of validity
Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.
Solution procedure you should automate
1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.
Photo 2. Probability and Statistics: the physical system the theory above idealises.
Capstone Studio instructional photograph
Notation used in this section
| f (xk), k | Quantity produced by "f (xk), k = 1, 2,..., n" — read its definition and unit from the handbook line directly above the equation. |
|---|---|
| k | Quantity produced by "k= 1" — read its definition and unit from the handbook line directly above the equation. |
Handbook notes for this section
Definitions and conditions exactly as the handbook states them.
- Let X be a discrete random variable having a probability
- mass function
- Engineering Probability and Statistics
- The expected value of X is defined as
- The variance of X is defined as
- n 2
- The expected value of Y is:
- The mean or expected value of the random variable X is now defined as
- while the variance is given by
- 2 2
- The standard deviation is given by
Core formulas for this FE topic
Definitions, applicability, units, assumptions and worked examples for each relation.
Worked exam-style examples
The four ways this section is written on the real exam — thoughts first, then equations, then substitution.
A culvert design costs $55,000 if no flood occurs and $256,000 if it does. The annual flood probability is 0.06. What is the expected annual cost?
Given
- C_normal = $55,000
- C_flood = $256,000
- p = 0.06
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.06($256,000) + 0.94($55,000)
Evaluate — E[C] = $15,360 + $51,700 = $67,060
Answer: E[C] ≈ $67,060
Why the other options are there
- $155,500 (simple average)
- $15,360 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
A culvert design costs $75,000 if no flood occurs and $371,000 if it does. The annual flood probability is 0.16. What is the expected annual cost?
Given
- C_normal = $75,000
- C_flood = $371,000
- p = 0.16
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.16($371,000) + 0.84($75,000)
Evaluate — E[C] = $59,360 + $63,000 = $122,360
Answer: E[C] ≈ $122,360
Why the other options are there
- $223,000 (simple average)
- $59,360 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
A culvert design costs $74,000 if no flood occurs and $319,000 if it does. The annual flood probability is 0.19. What is the expected annual cost?
Given
- C_normal = $74,000
- C_flood = $319,000
- p = 0.19
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.19($319,000) + 0.81($74,000)
Evaluate — E[C] = $60,610 + $59,940 = $120,550
Answer: E[C] ≈ $120,550
Why the other options are there
- $196,500 (simple average)
- $60,610 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
A culvert design costs $62,000 if no flood occurs and $321,000 if it does. The annual flood probability is 0.18. What is the expected annual cost?
Given
- C_normal = $62,000
- C_flood = $321,000
- p = 0.18
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.18($321,000) + 0.82($62,000)
Evaluate — E[C] = $57,780 + $50,840 = $108,620
Answer: E[C] ≈ $108,620
Why the other options are there
- $191,500 (simple average)
- $57,780 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
A culvert design costs $45,000 if no flood occurs and $184,000 if it does. The annual flood probability is 0.23. What is the expected annual cost?
Given
- C_normal = $45,000
- C_flood = $184,000
- p = 0.23
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.23($184,000) + 0.77($45,000)
Evaluate — E[C] = $42,320 + $34,650 = $76,970
Answer: E[C] ≈ $76,970
Why the other options are there
- $114,500 (simple average)
- $42,320 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
A culvert design costs $36,000 if no flood occurs and $248,000 if it does. The annual flood probability is 0.22. What is the expected annual cost?
Given
- C_normal = $36,000
- C_flood = $248,000
- p = 0.22
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.22($248,000) + 0.78($36,000)
Evaluate — E[C] = $54,560 + $28,080 = $82,640
Answer: E[C] ≈ $82,640
Why the other options are there
- $142,000 (simple average)
- $54,560 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
A culvert design costs $24,000 if no flood occurs and $188,000 if it does. The annual flood probability is 0.06. What is the expected annual cost?
Given
- C_normal = $24,000
- C_flood = $188,000
- p = 0.06
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.06($188,000) + 0.94($24,000)
Evaluate — E[C] = $11,280 + $22,560 = $33,840
Answer: E[C] ≈ $33,840
Why the other options are there
- $106,000 (simple average)
- $11,280 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
A culvert design costs $37,000 if no flood occurs and $341,000 if it does. The annual flood probability is 0.10. What is the expected annual cost?
Given
- C_normal = $37,000
- C_flood = $341,000
- p = 0.10
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.10($341,000) + 0.90($37,000)
Evaluate — E[C] = $34,100 + $33,300 = $67,400
Answer: E[C] ≈ $67,400
Why the other options are there
- $189,000 (simple average)
- $34,100 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
A culvert design costs $52,000 if no flood occurs and $146,000 if it does. The annual flood probability is 0.16. What is the expected annual cost?
Given
- C_normal = $52,000
- C_flood = $146,000
- p = 0.16
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.16($146,000) + 0.84($52,000)
Evaluate — E[C] = $23,360 + $43,680 = $67,040
Answer: E[C] ≈ $67,040
Why the other options are there
- $99,000 (simple average)
- $23,360 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
A culvert design costs $26,000 if no flood occurs and $284,000 if it does. The annual flood probability is 0.19. What is the expected annual cost?
Given
- C_normal = $26,000
- C_flood = $284,000
- p = 0.19
Find
E[C]
Start with the thinking
- Expected value weights each outcome by its probability.
- Probabilities must sum to one.
Step-by-step solution
Expectation — E[C] = p·C_flood + (1 − p)·C_normal
Substituting — E[C] = 0.19($284,000) + 0.81($26,000)
Evaluate — E[C] = $53,960 + $21,060 = $75,020
Answer: E[C] ≈ $75,020
Why the other options are there
- $155,000 (simple average)
- $53,960 (normal-case cost omitted)
Reference: FE Reference Handbook — Probability and Statistics → Expected Values
Self-check
Answer these without notes before moving on.
- Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
- Which assumption, if violated, makes the main relation of this section invalid?
- Given a sample of measurements from a construction or materials process, what is the first quantity you would compute, and why that one first?
- Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
- Rework Example 1 above from the givens alone, without reading the solution lines.
Chapter summary
- Expected Values contains 11 relations; you must be able to find this page in under 15 seconds.
- Exam style: one distribution or one counting rule, then a single probability or interval.
- Unit rule: probabilities are dimensionless and must land in [0, 1].
- Work the 10 examples until the solution path, not the answer, is automatic.
Common traps in this section
- probabilities are dimensionless and must land in [0, 1]
- Answering the intermediate quantity instead of the quantity requested.
- Rounding intermediate values before the final step.
- Using a relation from an adjacent handbook section that shares a symbol.
- Skipping the sketch — most lost points on this page start with a misread geometry.