Normal Distribution (Gaussian Distribution)

Probability and Statistics · FE Reference Handbook section

Probability and Statistics

28 formulas

10 exam-style examples

~60 min

All Probability and Statistics lectures

Learning objectives

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

This chapter section covers Normal Distribution (Gaussian Distribution) 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 normal distribution (gaussian distribution) describes physically and when it applies.
State every one of the 31 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. Normal Distribution (Gaussian Distribution) 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: normal distribution (gaussian distribution).
Capstone Studio instructional photograph

Probability and Statistics — Normal Distribution (Gaussian Distribution): 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 31 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

e 2 σ	Quantity produced by "e 2 σ" — read its definition and unit from the handbook line directly above the equation.
σ 2π	Quantity produced by "σ 2π" — read its definition and unit from the handbook line directly above the equation.
µ	Quantity produced by "µ = population mean" — read its definition and unit from the handbook line directly above the equation.
σ	Quantity produced by "σ = standard deviation of the population" — read its definition and unit from the handbook line directly above the equation.
When µ	Quantity produced by "When µ = 0 and σ2 = σ = 1, the distribution is called a standardized or unit normal distribution. Then" — read its definition and unit from the handbook line directly above the equation.
F(x)	Quantity produced by "F(x) = area under the curve from –∞ to x" — read its definition and unit from the handbook line directly above the equation.
R(x)	Quantity produced by "R(x) = area under the curve from x to ∞" — read its definition and unit from the handbook line directly above the equation.
W(x)	Quantity produced by "W(x) = area under the curve between –x and x" — read its definition and unit from the handbook line directly above the equation.
F(-x)	Quantity produced by "F(-x) = 1 - F(x)" — read its definition and unit from the handbook line directly above the equation.
z	Quantity produced by "z= v" — read its definition and unit from the handbook line directly above the equation.
Y	Quantity produced by "Y = X1 + X2 + ... Xn is approximately normal." — read its definition and unit from the handbook line directly above the equation.
ny	Quantity produced by "ny = 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.

of the mode). The averages of n observations tend to become normally distributed as n increases. The variate x is said to be
normally distributed if its density function f (x) is given by an expression of the form
x−µ n
where
-3 # x # 3
A unit normal distribution table is included at the end of this section. In the table, the following notations are utilized:
distribution can be used by utilizing the following transformation:
f(x) then becomes f(z), F(x) becomes F(z), etc.
The Central Limit Theorem
Let X1, X2, ..., Xn be a sequence of independent and identically distributed random variables each having mean µ and variance
and the standard deviation
Engineering Probability and Statistics
t-Distribution
Student's t-distribution has the probability density function given by:
Cc 2 m
v+1 v+1
2 − 2
vr C a 2 k
v v
where
xr - n
s/ n
χ2 - Distribution
If Z1, Z2, ..., Zn are independent unit normal random variables, then

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 — Normal Distribution (Gaussian Distribution)

Concrete strength is normal with μ = 4618 psi and σ = 300 psi. What fraction of cylinders fall below 4418 psi?

Given

μ = 4618 psi
σ = 300 psi
x = 4418 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
$z = (4418 - 4618) /300 = - 0.667$
Table lookup
$Φ (- 0.67) = 0.2525$
Result
$P (X < 4418) = 25.2$

Answer: ≈ 25.2% of cylinders

Why the other options are there

74.8% (upper tail reported)
0.67 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

Example 2

Normal probability below a limit — Normal Distribution (Gaussian Distribution) (2)

Concrete strength is normal with μ = 4740 psi and σ = 184 psi. What fraction of cylinders fall below 4336 psi?

Given

μ = 4740 psi
σ = 184 psi
x = 4336 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
$z = (4336 - 4740) /184 = - 2.196$
Table lookup
$Φ (- 2.20) = 0.0141$
Result
$P (X < 4336) = 1.4$

Answer: ≈ 1.4% of cylinders

Why the other options are there

98.6% (upper tail reported)
2.20 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

Example 3

Normal probability below a limit — Normal Distribution (Gaussian Distribution) (3)

Concrete strength is normal with μ = 3064 psi and σ = 429 psi. What fraction of cylinders fall below 2908 psi?

Given

μ = 3064 psi
σ = 429 psi
x = 2908 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
$z = (2908 - 3064) /429 = - 0.364$
Table lookup
$Φ (- 0.36) = 0.3581$
Result
$P (X < 2908) = 35.8$

Answer: ≈ 35.8% of cylinders

Why the other options are there

64.2% (upper tail reported)
0.36 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

Example 4

Normal probability below a limit — Normal Distribution (Gaussian Distribution) (4)

Concrete strength is normal with μ = 3397 psi and σ = 459 psi. What fraction of cylinders fall below 3188 psi?

Given

μ = 3397 psi
σ = 459 psi
x = 3188 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
$z = (3188 - 3397) /459 = - 0.455$
Table lookup
$Φ (- 0.46) = 0.3244$
Result
$P (X < 3188) = 32.4$

Answer: ≈ 32.4% of cylinders

Why the other options are there

67.6% (upper tail reported)
0.46 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

Example 5

Normal probability below a limit — Normal Distribution (Gaussian Distribution) (5)

Concrete strength is normal with μ = 4496 psi and σ = 185 psi. What fraction of cylinders fall below 3962 psi?

Given

μ = 4496 psi
σ = 185 psi
x = 3962 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
$z = (3962 - 4496) /185 = - 2.886$
Table lookup
$Φ (- 2.89) = 0.0019$
Result
$P (X < 3962) = 0.2$

Answer: ≈ 0.2% of cylinders

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

Example 6

Normal probability below a limit — Normal Distribution (Gaussian Distribution) (6)

Concrete strength is normal with μ = 4161 psi and σ = 176 psi. What fraction of cylinders fall below 3549 psi?

Given

μ = 4161 psi
σ = 176 psi
x = 3549 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
$z = (3549 - 4161) /176 = - 3.477$
Table lookup
$Φ (- 3.48) = 0.0003$
Result
$P (X < 3549) = 0.0$

Answer: ≈ 0.0% of cylinders

Why the other options are there

100.0% (upper tail reported)
3.48 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

Example 7

Normal probability below a limit — Normal Distribution (Gaussian Distribution) (7)

Concrete strength is normal with μ = 4374 psi and σ = 208 psi. What fraction of cylinders fall below 4210 psi?

Given

μ = 4374 psi
σ = 208 psi
x = 4210 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
$z = (4210 - 4374) /208 = - 0.788$
Table lookup
$Φ (- 0.79) = 0.2152$
Result
$P (X < 4210) = 21.5$

Answer: ≈ 21.5% of cylinders

Why the other options are there

78.5% (upper tail reported)
0.79 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

Example 8

Normal probability below a limit — Normal Distribution (Gaussian Distribution) (8)

Concrete strength is normal with μ = 4405 psi and σ = 394 psi. What fraction of cylinders fall below 4304 psi?

Given

μ = 4405 psi
σ = 394 psi
x = 4304 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
$z = (4304 - 4405) /394 = - 0.256$
Table lookup
$Φ (- 0.26) = 0.3988$
Result
$P (X < 4304) = 39.9$

Answer: ≈ 39.9% of cylinders

Why the other options are there

60.1% (upper tail reported)
0.26 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

Example 9

Normal probability below a limit — Normal Distribution (Gaussian Distribution) (9)

Concrete strength is normal with μ = 4678 psi and σ = 257 psi. What fraction of cylinders fall below 4212 psi?

Given

μ = 4678 psi
σ = 257 psi
x = 4212 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
$z = (4212 - 4678) /257 = - 1.813$
Table lookup
$Φ (- 1.81) = 0.0349$
Result
$P (X < 4212) = 3.5$

Answer: ≈ 3.5% of cylinders

Why the other options are there

96.5% (upper tail reported)
1.81 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

Example 10

Normal probability below a limit — Normal Distribution (Gaussian Distribution) (10)

Concrete strength is normal with μ = 4128 psi and σ = 209 psi. What fraction of cylinders fall below 3789 psi?

Given

μ = 4128 psi
σ = 209 psi
x = 3789 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
$z = (3789 - 4128) /209 = - 1.622$
Table lookup
$Φ (- 1.62) = 0.0524$
Result
$P (X < 3789) = 5.2$

Answer: ≈ 5.2% of cylinders

Why the other options are there

94.8% (upper tail reported)
1.62 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Normal Distribution (Gaussian Distribution)

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

Normal Distribution (Gaussian Distribution) contains 31 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.