Cumulative Distribution Functions

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 Cumulative Distribution Functions 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 cumulative distribution functions describes physically and when it applies.
State every one of the 4 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. Cumulative Distribution Functions 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: cumulative distribution functions.
Capstone Studio instructional photograph

Probability and Statistics — Cumulative Distribution Functions: 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 4 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

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.

The cumulative distribution function, F, of a discrete random variable X that has a probability distribution described by P(xi) is
defined as
If X is continuous, the cumulative distribution function, F, 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 — Cumulative Distribution Functions

Concrete strength is normal with μ = 3979 psi and σ = 355 psi. What fraction of cylinders fall below 3807 psi?

Given

μ = 3979 psi
σ = 355 psi
x = 3807 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
$z = (3807 - 3979) /355 = - 0.485$
Table lookup
$Φ (- 0.48) = 0.3140$
Result
$P (X < 3807) = 31.4$

Answer: ≈ 31.4% of cylinders

Why the other options are there

68.6% (upper tail reported)
0.48 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 2

Working with a triangular probability density function — Cumulative Distribution Functions

A random variable has the probability density function f(x) = 2x/4² on 0 ≤ x ≤ 4 and zero elsewhere. Verify that it integrates to one, find P(2.0 ≤ X ≤ 3.0), the mean, and the median (0.50 fractile).

Given

f(x) = 2x/4², 0 ≤ x ≤ 4

Find

Total area, an interval probability, the mean and the median

Start with the thinking

A valid probability density function must integrate to exactly 1 over its support.
Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

Normalisation
$\int_{0}^{4} 2 x / 4^{2} d x = [x^{2} / 4^{2}]_{0}^{4} = 1 ✓$
Formula
$P (x_{1} \leq X \leq x_{2}) = F (x_{2}) - F (x_{1}) = (x_{2}^{2} - x_{1}^{2}) / b^{2}$
Substituting
$P = (3. 0^{2} - 2. 0^{2}) / 4^{2} = 0.3125$
Formula
$μ = \int x f (x) d x = 2 b /3$
Substituting
$μ = 2 (4) /3 = 2.667$
Median — F(m) = 0.5 → m²/4² = 0.5 → m = 4/√2 = 2.828

Answer: P = 0.313, μ = 2.667, median = 2.828

Why the other options are there

μ = 2.00 (assumed a uniform density)
P = 0.250 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 3

Normal probability below a limit — Cumulative Distribution Functions (2)

Concrete strength is normal with μ = 4494 psi and σ = 356 psi. What fraction of cylinders fall below 3943 psi?

Given

μ = 4494 psi
σ = 356 psi
x = 3943 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
$z = (3943 - 4494) /356 = - 1.548$
Table lookup
$Φ (- 1.55) = 0.0608$
Result
$P (X < 3943) = 6.1$

Answer: ≈ 6.1% of cylinders

Why the other options are there

93.9% (upper tail reported)
1.55 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 4

Working with a triangular probability density function — Cumulative Distribution Functions (2)

A random variable has the probability density function f(x) = 2x/8² on 0 ≤ x ≤ 8 and zero elsewhere. Verify that it integrates to one, find P(1.5 ≤ X ≤ 7.5), the mean, and the median (0.50 fractile).

Given

f(x) = 2x/8², 0 ≤ x ≤ 8

Find

Total area, an interval probability, the mean and the median

Start with the thinking

A valid probability density function must integrate to exactly 1 over its support.
Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

Normalisation
$\int_{0}^{8} 2 x / 8^{2} d x = [x^{2} / 8^{2}]_{0}^{8} = 1 ✓$
Formula
$P (x_{1} \leq X \leq x_{2}) = F (x_{2}) - F (x_{1}) = (x_{2}^{2} - x_{1}^{2}) / b^{2}$
Substituting
$P = (7. 5^{2} - 1. 5^{2}) / 8^{2} = 0.8438$
Formula
$μ = \int x f (x) d x = 2 b /3$
Substituting
$μ = 2 (8) /3 = 5.333$
Median — F(m) = 0.5 → m²/8² = 0.5 → m = 8/√2 = 5.657

Answer: P = 0.844, μ = 5.333, median = 5.657

Why the other options are there

μ = 4.00 (assumed a uniform density)
P = 0.750 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 5

Normal probability below a limit — Cumulative Distribution Functions (3)

Concrete strength is normal with μ = 4274 psi and σ = 196 psi. What fraction of cylinders fall below 3650 psi?

Given

μ = 4274 psi
σ = 196 psi
x = 3650 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
$z = (3650 - 4274) /196 = - 3.184$
Table lookup
$Φ (- 3.18) = 0.0007$
Result
$P (X < 3650) = 0.1$

Answer: ≈ 0.1% of cylinders

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 6

Working with a triangular probability density function — Cumulative Distribution Functions (3)

Given

f(x) = 2x/4², 0 ≤ x ≤ 4

Find

Total area, an interval probability, the mean and the median

Start with the thinking

A valid probability density function must integrate to exactly 1 over its support.
Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

Normalisation
$\int_{0}^{4} 2 x / 4^{2} d x = [x^{2} / 4^{2}]_{0}^{4} = 1 ✓$
Formula
$P (x_{1} \leq X \leq x_{2}) = F (x_{2}) - F (x_{1}) = (x_{2}^{2} - x_{1}^{2}) / b^{2}$
Substituting
$P = (3. 0^{2} - 2. 0^{2}) / 4^{2} = 0.3125$
Formula
$μ = \int x f (x) d x = 2 b /3$
Substituting
$μ = 2 (4) /3 = 2.667$
Median — F(m) = 0.5 → m²/4² = 0.5 → m = 4/√2 = 2.828

Answer: P = 0.313, μ = 2.667, median = 2.828

Why the other options are there

μ = 2.00 (assumed a uniform density)
P = 0.250 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 7

Normal probability below a limit — Cumulative Distribution Functions (4)

Concrete strength is normal with μ = 4163 psi and σ = 438 psi. What fraction of cylinders fall below 3891 psi?

Given

μ = 4163 psi
σ = 438 psi
x = 3891 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
$z = (3891 - 4163) /438 = - 0.621$
Table lookup
$Φ (- 0.62) = 0.2673$
Result
$P (X < 3891) = 26.7$

Answer: ≈ 26.7% of cylinders

Why the other options are there

73.3% (upper tail reported)
0.62 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 8

Working with a triangular probability density function — Cumulative Distribution Functions (4)

A random variable has the probability density function f(x) = 2x/3² on 0 ≤ x ≤ 3 and zero elsewhere. Verify that it integrates to one, find P(1.0 ≤ X ≤ 2.5), the mean, and the median (0.50 fractile).

Given

f(x) = 2x/3², 0 ≤ x ≤ 3

Find

Total area, an interval probability, the mean and the median

Start with the thinking

A valid probability density function must integrate to exactly 1 over its support.
Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

Normalisation
$\int_{0}^{3} 2 x / 3^{2} d x = [x^{2} / 3^{2}]_{0}^{3} = 1 ✓$
Formula
$P (x_{1} \leq X \leq x_{2}) = F (x_{2}) - F (x_{1}) = (x_{2}^{2} - x_{1}^{2}) / b^{2}$
Substituting
$P = (2. 5^{2} - 1. 0^{2}) / 3^{2} = 0.5833$
Formula
$μ = \int x f (x) d x = 2 b /3$
Substituting
$μ = 2 (3) /3 = 2.000$
Median — F(m) = 0.5 → m²/3² = 0.5 → m = 3/√2 = 2.121

Answer: P = 0.583, μ = 2.000, median = 2.121

Why the other options are there

μ = 1.50 (assumed a uniform density)
P = 0.500 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 9

Normal probability below a limit — Cumulative Distribution Functions (5)

Concrete strength is normal with μ = 4215 psi and σ = 320 psi. What fraction of cylinders fall below 3859 psi?

Given

μ = 4215 psi
σ = 320 psi
x = 3859 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
$z = (3859 - 4215) /320 = - 1.113$
Table lookup
$Φ (- 1.11) = 0.1330$
Result
$P (X < 3859) = 13.3$

Answer: ≈ 13.3% of cylinders

Why the other options are there

86.7% (upper tail reported)
1.11 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 10

Working with a triangular probability density function — Cumulative Distribution Functions (5)

A random variable has the probability density function f(x) = 2x/3² on 0 ≤ x ≤ 3 and zero elsewhere. Verify that it integrates to one, find P(1.5 ≤ X ≤ 2.0), the mean, and the median (0.50 fractile).

Given

f(x) = 2x/3², 0 ≤ x ≤ 3

Find

Total area, an interval probability, the mean and the median

Start with the thinking

A valid probability density function must integrate to exactly 1 over its support.
Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

Normalisation
$\int_{0}^{3} 2 x / 3^{2} d x = [x^{2} / 3^{2}]_{0}^{3} = 1 ✓$
Formula
$P (x_{1} \leq X \leq x_{2}) = F (x_{2}) - F (x_{1}) = (x_{2}^{2} - x_{1}^{2}) / b^{2}$
Substituting
$P = (2. 0^{2} - 1. 5^{2}) / 3^{2} = 0.1944$
Formula
$μ = \int x f (x) d x = 2 b /3$
Substituting
$μ = 2 (3) /3 = 2.000$
Median — F(m) = 0.5 → m²/3² = 0.5 → m = 3/√2 = 2.121

Answer: P = 0.194, μ = 2.000, median = 2.121

Why the other options are there

μ = 1.50 (assumed a uniform density)
P = 0.167 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

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

Cumulative Distribution Functions contains 4 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.