Confidence Intervals, Sample Distributions and Sample Size

Probability and Statistics · FE Reference Handbook section

Probability and Statistics

4 formulas

10 exam-style examples

~53 min

All Probability and Statistics lectures

Learning objectives

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

This chapter section covers Confidence Intervals, Sample Distributions and Sample Size 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 confidence intervals, sample distributions and sample size 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. Confidence Intervals, Sample Distributions and Sample Size 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: confidence intervals, sample distributions and sample size.
Capstone Studio instructional photograph

Probability and Statistics — Confidence Intervals, Sample Distributions and Sample Size: 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Confidence Interval for the Mean µ of a Normal Distribution
(A) Standard deviation v is known
v v
X − Za/2 # n # X + Za/2
n n
(B) Standard deviation v is not known
s s
X − ta/2 # n # X + ta/2
n n
where ta/2 corresponds to n 1 degrees of freedom.
Engineering Probability and Statistics
Confidence Interval for the Difference Between Two Means µ1 and µ2
v12 v 22 v12 v 22
X1 − X2 − Za/2 n1 + n2 # n1 − n2 # X1 − X2 + Za/2 n1 + n2
c n1 + n1 m9_n1 − 1 is12 + _n2 − 1 is 22C c n1 + n1 m9_n1 − 1 is12 + _n2 − 1 is 22C
1 2 1 2
X1 − X2 − ta/2 n1 + n2 − 2 # n1 − n2 # X1 − X2 + ta/2 n1 + n2 − 2
where ta/2 corresponds to n1 + n2 − 2 degrees of freedom.
# v2 #
xa2/2, n - 1 x12 - a/2, n - 1

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

95% confidence interval on mean density

Nine field density tests give x̄ = 121.4 pcf with s = 3.6 pcf. Construct the 95% confidence interval on the mean (t₀.₀₂₅,₈ = 2.306).

Given

n = 9
x̄ = 121.4 pcf
s = 3.6 pcf
t = 2.306

Find

95% CI on μ

Start with the thinking

σ is unknown and n is small — use t, not z.
Degrees of freedom are n − 1 = 8.

Step-by-step solution

Standard error — SE = s/√n = 3.6/√9 = 1.20 pcf
Margin
$E = t \cdot S E = 2.306 (1.20) = 2.77 p c f$
Lower limit
$121.4 - 2.77 = 118.6 p c f$
Upper limit
$121.4 + 2.77 = 124.2 p c f$
Interval — 118.6 pcf ≤ μ ≤ 124.2 pcf

Answer: 118.6 to 124.2 pcf

Why the other options are there

119.1 to 123.7 (z = 1.96 used)
114.1 to 128.7 (√n omitted)

Reference: FE Reference Handbook — Probability and Statistics — Confidence intervals

Example 2

Normal probability below a limit — Confidence Intervals, Sample Distributions and Sample Size

Concrete strength is normal with μ = 4490 psi and σ = 276 psi. What fraction of cylinders fall below 4016 psi?

Given

μ = 4490 psi
σ = 276 psi
x = 4016 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
$z = (4016 - 4490) /276 = - 1.717$
Table lookup
$Φ (- 1.72) = 0.0430$
Result
$P (X < 4016) = 4.3$

Answer: ≈ 4.3% of cylinders

Why the other options are there

95.7% (upper tail reported)
1.72 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Confidence Intervals, Sample Distributions and Sample Size

Example 3

95% confidence interval on a mean — Confidence Intervals, Sample Distributions and Sample Size

A sample of 21 cores gives x̄ = 3070 psi and s = 250 psi. Construct the 95% confidence interval on the true mean (large-sample z).

Given

x̄ = 3070 psi
s = 250 psi
n = 21
z₀.₀₂₅ = 1.96

Find

95% CI for μ

Start with the thinking

The standard error, not s, sets the width.
Two-sided 95% splits α into two 2.5% tails.

Step-by-step solution

Standard error — SE = s/√n
Substituting
$S E = 250/ 21 = 54.55 p s i$
Margin — E = z·SE
Substituting
$E = 1.96 (54.55) = 106.9 p s i$
Interval — 3070 ± 106.9 → (2,963, 3,177) psi

Answer: (2,963, 3,177) psi

Why the other options are there

3070 ± 490.0 psi (√n omitted)
3070 ± 53.5 psi (one-tail margin halved)

Reference: FE Reference Handbook — Probability and Statistics → Confidence Intervals, Sample Distributions and Sample Size

Example 4

Sample size required for a target margin of error — Confidence Intervals, Sample Distributions and Sample Size

Concrete strength has a known standard deviation of 255.0 psi. How many cylinders must be tested so that the mean is estimated within ±54 psi at a z of 2.576?

Given

σ = 255.0 psi
E = 54 psi
z = 2.576

Find

Required sample size n

Start with the thinking

Sample size scales with the square of the ratio σ/E — halving the margin quadruples the testing.
Always round the computed sample size up to the next whole specimen.

Step-by-step solution

Formula
$n = (z σ / E)^{2}$
Substituting
$n = (2.576 \times 255.0/54)^{2}$
Inside the bracket — 12.1644
Evaluate
$n = 148.0 \to 148 s p ec im e n s$

Answer: n = 148 specimens

Why the other options are there

13 (forgot to square)
74 (halved the requirement)

Reference: FE Reference Handbook — Probability and Statistics → Confidence Intervals, Sample Distributions and Sample Size

Example 5

Normal probability below a limit — Confidence Intervals, Sample Distributions and Sample Size (2)

Concrete strength is normal with μ = 4363 psi and σ = 360 psi. What fraction of cylinders fall below 3864 psi?

Given

μ = 4363 psi
σ = 360 psi
x = 3864 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
$z = (3864 - 4363) /360 = - 1.386$
Table lookup
$Φ (- 1.39) = 0.0829$
Result
$P (X < 3864) = 8.3$

Answer: ≈ 8.3% of cylinders

Why the other options are there

91.7% (upper tail reported)
1.39 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Confidence Intervals, Sample Distributions and Sample Size

Example 6

95% confidence interval on a mean — Confidence Intervals, Sample Distributions and Sample Size (2)

A sample of 28 cores gives x̄ = 3380 psi and s = 206 psi. Construct the 95% confidence interval on the true mean (large-sample z).

Given

x̄ = 3380 psi
s = 206 psi
n = 28
z₀.₀₂₅ = 1.96

Find

95% CI for μ

Start with the thinking

The standard error, not s, sets the width.
Two-sided 95% splits α into two 2.5% tails.

Step-by-step solution

Standard error — SE = s/√n
Substituting
$S E = 206/ 28 = 38.93 p s i$
Margin — E = z·SE
Substituting
$E = 1.96 (38.93) = 76.3 p s i$
Interval — 3380 ± 76.3 → (3,304, 3,456) psi

Answer: (3,304, 3,456) psi

Why the other options are there

3380 ± 403.8 psi (√n omitted)
3380 ± 38.2 psi (one-tail margin halved)

Reference: FE Reference Handbook — Probability and Statistics → Confidence Intervals, Sample Distributions and Sample Size

Example 7

Sample size required for a target margin of error — Confidence Intervals, Sample Distributions and Sample Size (2)

Concrete strength has a known standard deviation of 340.0 psi. How many cylinders must be tested so that the mean is estimated within ±39 psi at a z of 1.96?

Given

σ = 340.0 psi
E = 39 psi
z = 1.96

Find

Required sample size n

Start with the thinking

Sample size scales with the square of the ratio σ/E — halving the margin quadruples the testing.
Always round the computed sample size up to the next whole specimen.

Step-by-step solution

Formula
$n = (z σ / E)^{2}$
Substituting
$n = (1.96 \times 340.0/39)^{2}$
Inside the bracket — 17.0872
Evaluate
$n = 292.0 \to 292 s p ec im e n s$

Answer: n = 292 specimens

Why the other options are there

18 (forgot to square)
146 (halved the requirement)

Reference: FE Reference Handbook — Probability and Statistics → Confidence Intervals, Sample Distributions and Sample Size

Example 8

Normal probability below a limit — Confidence Intervals, Sample Distributions and Sample Size (3)

Concrete strength is normal with μ = 4033 psi and σ = 400 psi. What fraction of cylinders fall below 3545 psi?

Given

μ = 4033 psi
σ = 400 psi
x = 3545 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
$z = (3545 - 4033) /400 = - 1.220$
Table lookup
$Φ (- 1.22) = 0.1112$
Result
$P (X < 3545) = 11.1$

Answer: ≈ 11.1% of cylinders

Why the other options are there

88.9% (upper tail reported)
1.22 (z reported as a probability)

Reference: FE Reference Handbook — Probability and Statistics → Confidence Intervals, Sample Distributions and Sample Size

Example 9

95% confidence interval on a mean — Confidence Intervals, Sample Distributions and Sample Size (3)

A sample of 31 cores gives x̄ = 4403 psi and s = 228 psi. Construct the 95% confidence interval on the true mean (large-sample z).

Given

x̄ = 4403 psi
s = 228 psi
n = 31
z₀.₀₂₅ = 1.96

Find

95% CI for μ

Start with the thinking

The standard error, not s, sets the width.
Two-sided 95% splits α into two 2.5% tails.

Step-by-step solution

Standard error — SE = s/√n
Substituting
$S E = 228/ 31 = 40.95 p s i$
Margin — E = z·SE
Substituting
$E = 1.96 (40.95) = 80.3 p s i$
Interval — 4403 ± 80.3 → (4,323, 4,483) psi

Answer: (4,323, 4,483) psi

Why the other options are there

4403 ± 446.9 psi (√n omitted)
4403 ± 40.1 psi (one-tail margin halved)

Reference: FE Reference Handbook — Probability and Statistics → Confidence Intervals, Sample Distributions and Sample Size

Example 10

Sample size required for a target margin of error — Confidence Intervals, Sample Distributions and Sample Size (3)

Concrete strength has a known standard deviation of 195.0 psi. How many cylinders must be tested so that the mean is estimated within ±41 psi at a z of 1.645?

Given

σ = 195.0 psi
E = 41 psi
z = 1.645

Find

Required sample size n

Start with the thinking

Sample size scales with the square of the ratio σ/E — halving the margin quadruples the testing.
Always round the computed sample size up to the next whole specimen.

Step-by-step solution

Formula
$n = (z σ / E)^{2}$
Substituting
$n = (1.645 \times 195.0/41)^{2}$
Inside the bracket — 7.8238
Evaluate
$n = 61.21 \to 62 s p ec im e n s$

Answer: n = 62 specimens

Why the other options are there

8 (forgot to square)
31 (halved the requirement)

Reference: FE Reference Handbook — Probability and Statistics → Confidence Intervals, Sample Distributions and Sample Size

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

Confidence Intervals, Sample Distributions and Sample Size 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.