Probability Functions, Distributions, and Expected Values

Probability and Statistics · FE Reference Handbook section

Probability and Statistics

3 formulas

10 exam-style examples

~51 min

All Probability and Statistics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Probability Functions, Distributions, and 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 probability functions, distributions, and expected values describes physically and when it applies.
State every one of the 3 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. Probability Functions, Distributions, and 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.

Three engineers in hard hats and safety vests reviewing drawings on a truck tailgate. — Photo 1. Where this shows up in practice: probability functions, distributions, and expected values.
Capstone Studio instructional photograph

Probability and Statistics — Probability Functions, Distributions, and 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 3 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.

Notation used in this section

X	Quantity produced by "X = x1, x2, x3, …, xn" — read its definition and unit from the handbook line directly above the equation.
f (xk)	Quantity produced by "f (xk) = P(X = xk), k = 1, 2, ..., n" — 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.

A random variable X has a probability associated with each of its possible values. The probability is termed a discrete
probability if X can assume only discrete values, or
random variable X is defined 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.

Example 1

Normal probability below a limit — Probability Functions, Distributions, and Expected Values

Concrete strength is normal with μ = 3135 psi and σ = 410 psi. What fraction of cylinders fall below 2592 psi?

Given

μ = 3135 psi
σ = 410 psi
x = 2592 psi

Find

P(X < x)

Start with the thinking

Standardise first; the table is always in z.
A negative z means the left tail — less than 0.5.

Step-by-step solution

Standardise
$z = (x - μ) / σ$
Substituting
$z = (2592 - 3135) /410 = - 1.324$
Table lookup
$Φ (- 1.32) = 0.0927$
Result
$P (X < 2592) = 9.3$

Answer: ≈ 9.3% of cylinders

Why the other options are there

90.7% (upper tail reported)
1.32 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 2

Expected cost of a risk event — Probability Functions, Distributions, and Expected Values

A culvert design costs $56,000 if no flood occurs and $337,000 if it does. The annual flood probability is 0.13. What is the expected annual cost?

Given

C_normal = $56,000
C_flood = $337,000
p = 0.13

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] = 0.13($337,000) + 0.87($56,000)
Evaluate — E[C] = $43,810 + $48,720 = $92,530

Answer: E[C] ≈ $92,530

Why the other options are there

$196,500 (simple average)
$43,810 (normal-case cost omitted)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 3

Normal probability below a limit — Probability Functions, Distributions, and Expected Values (2)

Concrete strength is normal with μ = 4785 psi and σ = 203 psi. What fraction of cylinders fall below 4137 psi?

Given

μ = 4785 psi
σ = 203 psi
x = 4137 psi

Find

P(X < x)

Start with the thinking

Standardise first; the table is always in z.
A negative z means the left tail — less than 0.5.

Step-by-step solution

Standardise
$z = (x - μ) / σ$
Substituting
$z = (4137 - 4785) /203 = - 3.192$
Table lookup
$Φ (- 3.19) = 0.0007$
Result
$P (X < 4137) = 0.1$

Answer: ≈ 0.1% of cylinders

Why the other options are there

99.9% (upper tail reported)
3.19 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 4

Expected cost of a risk event — Probability Functions, Distributions, and Expected Values (2)

A culvert design costs $48,000 if no flood occurs and $322,000 if it does. The annual flood probability is 0.23. What is the expected annual cost?

Given

C_normal = $48,000
C_flood = $322,000
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($322,000) + 0.77($48,000)
Evaluate — E[C] = $74,060 + $36,960 = $111,020

Answer: E[C] ≈ $111,020

Why the other options are there

$185,000 (simple average)
$74,060 (normal-case cost omitted)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 5

Normal probability below a limit — Probability Functions, Distributions, and Expected Values (3)

Concrete strength is normal with μ = 4410 psi and σ = 311 psi. What fraction of cylinders fall below 4078 psi?

Given

μ = 4410 psi
σ = 311 psi
x = 4078 psi

Find

P(X < x)

Start with the thinking

Standardise first; the table is always in z.
A negative z means the left tail — less than 0.5.

Step-by-step solution

Standardise
$z = (x - μ) / σ$
Substituting
$z = (4078 - 4410) /311 = - 1.068$
Table lookup
$Φ (- 1.07) = 0.1429$
Result
$P (X < 4078) = 14.3$

Answer: ≈ 14.3% of cylinders

Why the other options are there

85.7% (upper tail reported)
1.07 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 6

Expected cost of a risk event — Probability Functions, Distributions, and Expected Values (3)

A culvert design costs $22,000 if no flood occurs and $310,000 if it does. The annual flood probability is 0.08. What is the expected annual cost?

Given

C_normal = $22,000
C_flood = $310,000
p = 0.08

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] = 0.08($310,000) + 0.92($22,000)
Evaluate — E[C] = $24,800 + $20,240 = $45,040

Answer: E[C] ≈ $45,040

Why the other options are there

$166,000 (simple average)
$24,800 (normal-case cost omitted)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 7

Normal probability below a limit — Probability Functions, Distributions, and Expected Values (4)

Concrete strength is normal with μ = 3133 psi and σ = 279 psi. What fraction of cylinders fall below 2731 psi?

Given

μ = 3133 psi
σ = 279 psi
x = 2731 psi

Find

P(X < x)

Start with the thinking

Standardise first; the table is always in z.
A negative z means the left tail — less than 0.5.

Step-by-step solution

Standardise
$z = (x - μ) / σ$
Substituting
$z = (2731 - 3133) /279 = - 1.441$
Table lookup
$Φ (- 1.44) = 0.0748$
Result
$P (X < 2731) = 7.5$

Answer: ≈ 7.5% of cylinders

Why the other options are there

92.5% (upper tail reported)
1.44 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 8

Expected cost of a risk event — Probability Functions, Distributions, and Expected Values (4)

A culvert design costs $62,000 if no flood occurs and $390,000 if it does. The annual flood probability is 0.20. What is the expected annual cost?

Given

C_normal = $62,000
C_flood = $390,000
p = 0.20

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] = 0.20($390,000) + 0.80($62,000)
Evaluate — E[C] = $78,000 + $49,600 = $127,600

Answer: E[C] ≈ $127,600

Why the other options are there

$226,000 (simple average)
$78,000 (normal-case cost omitted)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 9

Normal probability below a limit — Probability Functions, Distributions, and Expected Values (5)

Concrete strength is normal with μ = 4654 psi and σ = 152 psi. What fraction of cylinders fall below 4205 psi?

Given

μ = 4654 psi
σ = 152 psi
x = 4205 psi

Find

P(X < x)

Start with the thinking

Standardise first; the table is always in z.
A negative z means the left tail — less than 0.5.

Step-by-step solution

Standardise
$z = (x - μ) / σ$
Substituting
$z = (4205 - 4654) /152 = - 2.954$
Table lookup
$Φ (- 2.95) = 0.0016$
Result
$P (X < 4205) = 0.2$

Answer: ≈ 0.2% of cylinders

Why the other options are there

99.8% (upper tail reported)
2.95 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 10

Expected cost of a risk event — Probability Functions, Distributions, and Expected Values (5)

A culvert design costs $58,000 if no flood occurs and $346,000 if it does. The annual flood probability is 0.09. What is the expected annual cost?

Given

C_normal = $58,000
C_flood = $346,000
p = 0.09

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] = 0.09($346,000) + 0.91($58,000)
Evaluate — E[C] = $31,140 + $52,780 = $83,920

Answer: E[C] ≈ $83,920

Why the other options are there

$202,000 (simple average)
$31,140 (normal-case cost omitted)

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and 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

Probability Functions, Distributions, and Expected Values contains 3 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.